মনে করি, \(f(x)=x^{n}\) \(\therefore f(x+h)=(x+h)^{n}\) আমরা জানি, \[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{x^{n}\}=\lim_{h \rightarrow 0}\frac{(x+h)^{n}-x^{n}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{x^{n}\left(1+\frac{h}{x}\right)^{n}-x^{n}}{h}\]
\[=x^{n}\lim_{h \rightarrow 0}\frac{\left(1+\frac{h}{x}\right)^{n}-1}{h}\]
যেহেতু, \(h \rightarrow 0\) সুতরাং \(|h|\) কে \(|x|\) অপেক্ষা ক্ষুদ্রতর মনে করা যায়, তখন \(\frac{h}{x}\) এর সাংখ্যিক মাণ \(1\) অপেক্ষা ক্ষুদ্রতর । সুতরাং \(\left(1+\frac{h}{x}\right)^n\) কে দ্বিপদী উপপাদ্যের সাহায্যে বিস্তার করা যায়।
\[\therefore \frac{d}{dx}\{x^{n}\}=x^{n}\lim_{h \rightarrow 0}\frac{1+n.\frac{h}{x}+\frac{n(n-1)}{2!}.\left(\frac{h}{x}\right)^{2}+....\infty -1}{h}\] ➜ Note \[\because (1+x)^n=1+nx+\frac{n(n-1)}{2!}x^{2}+....\infty \]
\[=x^{n}\lim_{h \rightarrow 0}\frac{\frac{nh}{x}+\frac{n(n-1)}{2!}.\frac{h^2}{x^2}+....\infty}{h}\]
\[=x^{n}\lim_{h \rightarrow 0}\frac{h\left(\frac{n}{x}+\frac{n(n-1)}{2!}.\frac{h}{x^2}+....\infty \right)}{h}\]
\[=x^{n}\lim_{h \rightarrow 0}\left(\frac{n}{x}+\frac{n(n-1)}{2!}.\frac{h}{x^2}+....\infty \right)\]
\[=x^{n}.\frac{n}{x}\] ➜ Note লিমিট ব্যবহার করে।
\[=nx^{n-1}\]

মনে করি,
\(f(x)=ax^{2}+bx+c\)
\(\therefore f(x+h)=a(x+h)^{2}+b(x+h)+c\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(ax^{2}+bx+c)=\lim_{h \rightarrow 0}\frac{a(x+h)^{2}+b(x+h)+c-ax^{2}-bx-c}{h}\]
\[=\lim_{h \rightarrow 0}\frac{ax^2+2axh+ah^{2}+bx+bh+c-ax^{2}-bx-c}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2axh+ah^{2}+bh}{h}\]
\[=\lim_{h \rightarrow 0}\frac{h(2ax+ah+b)}{h}\]
\[=\lim_{h \rightarrow 0}(2ax+ah+b)\]
\[=2ax+0+b\] ➜ Note লিমিট ব্যবহার করে।
\[=2ax+b\]

মনে করি,
\(f(x)=\sin x\)
\(\therefore f(x+h)=\sin (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sin x)=\lim_{h \rightarrow 0}\frac{\sin (x+h)-\sin x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{x+h+x}{2}\sin \frac{x+h-x}{2}}{h}\] ➜ Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2x+h}{2}\sin \frac{h}{2}}{h}\] \[=2\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\] \[=\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\] \[=\cos \frac{2x}{2}\times 1\] ➜ Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=\cos x\]

মনে করি,
\(f(x)=\cos x\)
\(\therefore f(x+h)=\cos (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cos x)=\lim_{h \rightarrow 0}\frac{\cos (x+h)-\cos x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{x+h+x}{2}\sin \frac{x-x-h}{2}}{h}\] ➜ Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2x+h}{2}\sin \frac{-h}{2}}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\]
\[=-\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\]
\[=-\sin \frac{2x}{2}\times 1\] ➜ Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-\sin x\]

মনে করি,
\(f(x)=\tan x\)
\(\therefore f(x+h)=\tan (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\tan x)=\lim_{h \rightarrow 0}\frac{\tan (x+h)-\tan x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}}{h}\] ➜ Note \(\because \tan x=\frac{\sin x}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h-x)}{\cos (x+h)\cos x}}{h}\] ➜ Note \(\because \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin h}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\times \frac{\sin h}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\]
\[=\frac{1}{\cos x.\cos x}\times 1\] ➜ Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=\frac{1}{\cos^2 x}\]
\[=\sec^2 x\]

মনে করি,
\(f(x)=\csc x\)
\(\therefore f(x+h)=\csc (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে

\[\Rightarrow \frac{d}{dx}(\csc x)=\lim_{h \rightarrow 0}\frac{\csc (x+h)-\csc x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\sin (x+h)}-\frac{1}{\sin x}}{h}\] ➜ Note \(\because \csc x=\frac{1}{\sin x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin x-\sin (x+h)}{\sin (x+h)\sin x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\sin x-\sin (x+h)}{h\sin (x+h)\sin x}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{x+x+h}{2}\sin \frac{x-x-h}{2}}{h\sin (x+h)\sin x}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\cos \frac{2x+h}{2}\sin \frac{-h}{2}}{h\sin (x+h)\sin x}\]
\[=-2\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\times \frac{1}{\sin (x+h)\sin x}\]
\[=-\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{\sin (x+h)\sin x}\]
\[=-\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\sin (x+h)\sin x}\]
\[=-\cos \frac{2x}{2}\times 1\times \frac{1}{\sin x.\sin x}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-\cos x\times \frac{1}{\sin^2 x}\]
\[=-\frac{1}{\sin x}\times \frac{\cos x}{\sin x}\]
\[=-\csc x\times \cot x\] ➜Note \(\because \frac{1}{\sin x}=\csc x, \frac{\cos x}{\sin x}=\cot x\)
\[=-\csc x\cot x\]

মনে করি,
\(f(x)=\sec x\)
\(\therefore f(x+h)=\sec (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sec x)=\lim_{h \rightarrow 0}\frac{\sec (x+h)-\sec x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{h}\] ➜Note \(\because \sec x=\frac{1}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos x-\cos (x+h)}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\cos x-\cos (x+h)}{h\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{x+x+h}{2}\sin \frac{x+h-x}{2}}{h\sin (x+h)\sin x}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\sin \frac{2x+h}{2}\sin \frac{h}{2}}{h\cos (x+h)\cos x}\]
\[=2\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\times \frac{1}{\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\]
\[=\sin \frac{2x}{2}\times 1\times \frac{1}{\cos x.\cos x}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=\sin x\times \frac{1}{\cos^2 x}\]
\[=\frac{1}{\cos x}\times \frac{\sin x}{\cos x}\]
\[=\sec x\times \tan x\] ➜Note \(\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\)
\[=\sec x\tan x\]

মনে করি,
\(f(x)=\cot x\)
\(\therefore f(x+h)=\cot (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cot x)=\lim_{h \rightarrow 0}\frac{\cot (x+h)-\cot x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}}{h}\] ➜Note \(\because \cot x=\frac{\cos x}{\sin x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin x\cos (x+h)-\cos x\sin (x+h)}{\sin (x+h)\sin x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x-x-h)}{\sin (x+h)\sin x}}{h}\] ➜Note \(\because \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (-h)}{\sin (x+h)\sin x}}{h}\]
\[=-\lim_{h \rightarrow 0}\frac{1}{\sin (x+h)\sin x}\times \frac{\sin h}{h}\]
\[=-\lim_{h \rightarrow 0}\frac{1}{\sin (x+h)\sin x}\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\]
\[=-\frac{1}{\sin x.\sin x}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-\frac{1}{\sin^2 x}\]
\[=-\csc^2 x\] ➜Note \(\because \frac{1}{\sin^2 x}=\csc^2 x\)

মনে করি,
\(f(x)=\sin ax\)
\(\therefore f(x+h)=\sin a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sin ax)=\lim_{h \rightarrow 0}\frac{\sin a(x+h)-\sin ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{a(x+h)+ax}{2}\sin \frac{a(x+h)-ax}{2}}{h}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{ax+ah+ax}{2}\sin \frac{ax+ah-ax}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2ax+ah}{2}\sin \frac{ah}{2}}{h}\]
\[=2\lim_{h \rightarrow 0}\cos \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\]
\[=a\lim_{h \rightarrow 0}\cos \frac{2ax+ah}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\]
\[=a\cos \frac{2ax}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=a\cos ax\]

মনে করি,
\(f(x)=\cos ax\)
\(\therefore f(x+h)=\cos a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cos ax)=\lim_{h \rightarrow 0}\frac{\cos a(x+h)-\cos ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{a(x+h)+ax}{2}\sin \frac{ax-a(x+h)}{2}}{h}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{ax+ah+ax}{2}\sin \frac{ax-ax-ah}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2ax+ah}{2}\sin \frac{-ah}{2}}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\]
\[=-a\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\]
\[=-a\sin \frac{2ax}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-a\sin ax\]

মনে করি,
\(f(x)=\tan ax\)
\(\therefore f(x+h)=\tan a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\tan ax)=\lim_{h \rightarrow 0}\frac{\tan a(x+h)-\tan ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin a(x+h)}{\cos a(x+h)}-\frac{\sin ax}{\cos ax}}{h}\] ➜Note \(\because \tan x=\frac{\sin x}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin a(x+h)\cos ax-\cos a(x+h)\sin ax}{\cos a(x+h)\cos ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (ax+ah)\cos ax-\cos (ax+ah)\sin ax}{\cos (ax+ah)\cos ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (ax+ah-ax)}{\cos (ax+ah)\cos ax}}{h}\] ➜Note \(\because \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin ah}{\cos (ax+ah)\cos ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (ax+ah)\cos ax}\times \frac{\sin ah}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (ax+ah)\cos ax}\times \lim_{h \rightarrow 0}\frac{\sin ah}{ah}\times a\]
\[=a\frac{1}{\cos ax.\cos ax}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=a\frac{1}{\cos^2 ax}\]
\[=a\sec^2 ax\]

মনে করি,
\(f(x)=e^{x}\)
\(\therefore f(x+h)=e^{x+h}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{e^{x}\}=\lim_{h \rightarrow 0}\frac{e^{x+h}-e^{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{x}.e^{h}-e^{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{x}\left(e^{h}-1\right)}{h}\]
\[=e^{x}\lim_{h \rightarrow 0}\frac{e^{h}-1}{h}\]
\[=e^{x}\lim_{h \rightarrow 0}\frac{1+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{x}\lim_{h \rightarrow 0}\frac{\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+.......\infty}{h}\]
\[=e^{x}\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+.......\infty \right)}{h}\]
\[=e^{x}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+.......\infty \right)\]
\[=e^{x}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=e^{x}\times (1+0)\]
\[=e^{x}\times 1\]
\[=e^{x}\]

মনে করি,
\(f(x)=e^{mx}\)
\(\Rightarrow f(x+h)=e^{m(x+h)}\)
\(\therefore f(x+h)=e^{mx+mh}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{e^{mx}\}=\lim_{h \rightarrow 0}\frac{e^{mx+mh}-e^{mx}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{mx}.e^{mh}-e^{mx}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{mx}\left(e^{mh}-1\right)}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{e^{mh}-1}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{1+\frac{mh}{1!}+\frac{m^2h^2}{2!}+\frac{m^3h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{\frac{mh}{1!}+\frac{m^2h^2}{2!}+\frac{m^3h^3}{3!}+.......\infty}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{mh\left(\frac{1}{1!}+\frac{mh}{2!}+\frac{m^2h^2}{3!}+.......\infty \right)}{h}\]
\[=me^{mx}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{mh}{2!}+\frac{m^2h^2}{3!}+.......\infty \right)\]
\[=me^{mx}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=me^{mx}\times (1+0)\]
\[=me^{mx}\times 1\]
\[=me^{mx}\]

মনে করি,
\(f(x)=a^{x}\)
\(\therefore f(x+h)=a^{x+h}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{a^{x}\}=\lim_{h \rightarrow 0}\frac{a^{x+h}-a^{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{a^{x}.a^{h}-a^{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{a^{x}\left(a^{h}-1\right)}{h}\]
\[=a^{x}\lim_{h \rightarrow 0}\frac{a^{h}-1}{h}\]
\[=a^{x}\lim_{h \rightarrow 0}\frac{1+\frac{h}{1!}\ln a+\frac{h^2}{2!}(\ln a)^2+\frac{h^3}{3!}(\ln a)^3+.......\infty -1}{h}\] ➜Note \[\because a^{x}=1+\frac{x}{1!}\ln a+\frac{x^2}{2!}(\ln a)^2+\frac{x^3}{3!}(\ln a)^3+.......\infty \]
\[=a^{x}\lim_{h \rightarrow 0}\frac{\frac{h}{1!}\ln a+\frac{h^2}{2!}(\ln a)^2+\frac{h^3}{3!}(\ln a)^3+.......\infty}{h}\]
\[=a^{x}\lim_{h \rightarrow 0}\frac{h\ln a\left(\frac{1}{1!}+\frac{h}{2!}\ln a+\frac{h^2}{3!}(\ln a)^2+.......\infty \right)}{h}\]
\[=a^{x}\ln a\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{h}{2!}\ln a+\frac{h^2}{3!}(\ln a)^2+.......\infty \right)\]
\[=a^{x}\ln a\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=a^{x}\ln a\times (1+0)\]
\[=a^{x}\ln a\times 1\]
\[=a^{x}\ln a\]

মনে করি,
\(f(x)=\ln x\)
\(\therefore f(x+h)=\ln (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{\ln x\}=\lim_{h \rightarrow 0}\frac{\ln (x+h)-\ln x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \frac{(x+h)}{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{h}{x}-\frac{1}{2}.\frac{h^2}{x^2}+\frac{1}{3}.\frac{h^3}{x^3}-......\infty}{h}\] ➜Note \[\because \ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+.......\infty \]
\[=\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)}{h}\]
\[=\lim_{h \rightarrow 0}\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)\]
\[=\frac{1}{x}-0+0-......\] ➜Note লিমিট ব্যবহার করে।
\[=\frac{1}{x}\]

মনে করি,
\(f(x)=\log_ax\)
\(\Rightarrow f(x)=\ln x \times \log_ae\)
\(\therefore f(x+h)=\ln (x+h)\times \log_ae\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{\log_ax\}=\lim_{h \rightarrow 0}\frac{\ln (x+h)\times \log_ae-\ln x\times \log_ae}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\log_ae\{\ln (x+h)-\ln x\}}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\ln \frac{(x+h)}{x}}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\frac{h}{x}-\frac{1}{2}.\frac{h^2}{x^2}+\frac{1}{3}.\frac{h^3}{x^3}-......\infty}{h}\] ➜Note \[\because \ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+.......\infty \]
\[=\log_ae\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)\]
\[=\log_ae\left(\frac{1}{x}-0+0-......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=\log_ae\times \frac{1}{x}\]
\[=\frac{1}{x}\log_ae\]
\[=\frac{1}{x\ln a}\]

\((a)\) \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)

দেওয়া আছে,
\(u=f(x)\) এবং \(v=\phi(x)\) উভয়ে \(x\)-এর ফাংশন।
\(\therefore y=u+v ......(1)\), \(x\)-এর ফাংশন হবে।
আবার,
\(y+\delta y=u+\delta u+v+\delta v ......(2)\)
\((2)-(1)\)-এর সাহায্যে,
\(\delta y=\delta u+\delta v\)
\(\Rightarrow \frac{\delta y}{\delta x}=\frac{\delta u}{\delta x}+\frac{\delta v}{\delta x}\) ➜Note উভয়পার্শে \[\delta x\] ভাগ করে।
\[\Rightarrow \lim_{\delta x \rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x \rightarrow 0}\frac{\delta u}{\delta x}+\lim_{\delta x \rightarrow 0}\frac{\delta v}{\delta x}\] ➜Note উভয়পার্শে \[\lim_{\delta x \rightarrow 0}\] লিমিট সংযোজন করে।
\[\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\] ➜Note \[\because \lim_{\delta x \rightarrow 0}\frac{\delta y}{\delta x}=\frac{d}{dx}(y) \]
\[\therefore \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\] ➜Note \[\because y=u+v \]
(proved)

\((b)\) \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)

দেওয়া আছে,
\(u=f(x)\) এবং \(v=\phi(x)\) উভয়ে \(x\)-এর ফাংশন।
\(\therefore y=u-v ......(1)\), \(x\)-এর ফাংশন হবে।
আবার,
\(y+\delta y=u+\delta u-v-\delta v ......(2)\)
\((2)-(1)\)-এর সাহায্যে,
\(\delta y=\delta u-\delta v\)
\(\Rightarrow \frac{\delta y}{\delta x}=\frac{\delta u}{\delta x}-\frac{\delta v}{\delta x}\) ➜Note উভয়পার্শে \[\delta x\] ভাগ করে।
\[\Rightarrow \lim_{\delta x \rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x \rightarrow 0}\frac{\delta u}{\delta x}-\lim_{\delta x \rightarrow 0}\frac{\delta v}{\delta x}\] ➜Note উভয়পার্শে \[\lim_{\delta x \rightarrow 0}\] লিমিট সংযোজন করে।
\[\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\] ➜Note \[\because \lim_{\delta x \rightarrow 0}\frac{\delta y}{\delta x}=\frac{d}{dx}(y) \]
\[\therefore \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\] ➜Note \[\because y=u-v \]
(proved)

\((c)\) \(\frac{d}{dx}(c)=0\) যখন, \(c\) স্থির রাশি।

ধরি,
\(x\)-এর সকল মানের জন্য \(f(x)=c\)
\(\therefore f(x+h)=c\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(c)=\lim_{h \rightarrow 0}\frac{c-c}{h}\] ➜Note \[\because f(x)=c, f(x+h)=c\]
\[=\lim_{h \rightarrow 0}\frac{0}{h}\]
\[=\lim_{h \rightarrow 0}0\]
\[=0\]

\((d)\) \(\frac{d}{dx}(cu)=c\frac{d}{dx}(u)\) যখন, \(u=f(x)\) এবং \(c\) স্থির রাশি।

দেওয়া আছে,
\(u=f(x)\) এবং \(c\) স্থির রাশি।
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{cf(x)\}=\lim_{h \rightarrow 0}\frac{cf(x+h)-cf(x)}{h}\] ➜Note উভয় পার্শে \[c\] গুণ করে।
\[=\lim_{h \rightarrow 0}\frac{c\{f(x+h)-f(x)\}}{h}\]
\[=\lim_{h \rightarrow 0}c\times \frac{f(x+h)-f(x)}{h}\]
\[=\lim_{h \rightarrow 0}c\times \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[=c\times \frac{d}{dx}\{f(x)\}\] ➜Note \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[=c\frac{d}{dx}(u)\] ➜Note \[\because u=f(x)\]

ধরি,
\(y=x^7\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^7)\)
\(=7x^{7-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=7x^{6}\)

ধরি,
\(x=\sqrt[3]{y}\)
\(\Rightarrow \frac{d}{dy}(x)=\frac{d}{dy}(\sqrt[3]{y})\)
\(=\frac{d}{dy}(y^{\frac{1}{3}})\)
\(=\frac{1}{3}.y^{\frac{1}{3}-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=\frac{1}{3}.y^{\frac{1-3}{3}}\)
\(=\frac{1}{3}.y^{\frac{-2}{3}}\)
\(=\frac{1}{3}.y^{-\frac{2}{3}}\)

ধরি,
\(y=\frac{1}{t^3}\)
\(\Rightarrow \frac{d}{dt}(y)=\frac{d}{dt}(\frac{1}{t^3})\)
\(=\frac{d}{dt}(t^{-3})\)
\(=-3t^{-3-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=-3t^{-4}\)
\(=-\frac{3}{t^{4}}\)

ধরি,
\(y=x^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-\frac{1}{2}})\)
\(=-\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(=-\frac{1}{2}x^{\frac{-3}{2}}\)
\(=-\frac{1}{2}x^{-\frac{3}{2}}\)

ধরি,
\(y=\sin x+3\cos x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin x+3\cos x)\)
\(=\frac{d}{dx}\sin x+\frac{d}{dx}(3\cos x)\)
\(=\frac{d}{dx}\sin x+3\frac{d}{dx}\cos x\)
\(=\cos x-3\sin x\) ➜Note \[\because \frac{d}{dx}\sin x=\cos x, \frac{d}{dx}\cos x=-\sin x\]

ধরি,
\(y=\sec \theta+\tan \theta\)
\(\Rightarrow \frac{d}{d\theta}(y)=\frac{d}{d\theta}(\sec \theta+\tan \theta)\)
\(=\frac{d}{d\theta}\sec \theta+\frac{d}{d\theta}\tan \theta\)
\(=\sec \theta\tan \theta+\sec^2 \theta\) ➜Note \[\because \frac{d}{d\theta}\sec \theta=\sec \theta\tan \theta, \frac{d}{d\theta}\tan \theta=\sec^2 \theta\]
\(= \sec \theta(\tan \theta+\sec \theta)\)

ধরি,
\(y=5\ln x-\cot x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5\ln x-\cot x)\)
\(=\frac{d}{dx}(5\ln x)-\frac{d}{dx}\cot x\)
\(=5\frac{d}{dx}\ln x-\frac{d}{dx}\cot x\)
\(=5\frac{1}{x}+\csc^2 x\) ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\cot x)=-\csc^2 x\]
\(=\frac{5}{x}+\csc^2 x\)

ধরি,
\(y=x^{a+1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{a+1})\)
\(=(a+1)x^{a+1-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=(a+1)x^{a}\)

ধরি,
\(y=(a+1)^{x-1}\)
\(\Rightarrow y=(a+1)^x.(a+1)^{-1}\)
\(\Rightarrow y=\frac{(a+1)^x}{(a+1)}\)
\(\Rightarrow y=(a+1)^{-1}(a+1)^x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a+1)^{-1}(a+1)^x\}\)
\(=(a+1)^{-1}\frac{d}{dx}\{(a+1)^x\}\)
\(=(a+1)^{-1}.(a+1)^x\ln (a+1)\) ➜Note \[\because \frac{d}{dx}(a^x)=a^x\ln a\]
\(=(a+1)^{x-1}\ln (a+1)\)

ধরি,
\(y=\ln x^{a}\)
\(\Rightarrow y=a\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a\ln x)\)
\(=a\frac{d}{dx}\ln x\)
\(=a\frac{1}{x}\) ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}\]

ধরি,
\(y=\log x^{a}\)
\(\Rightarrow y=a\log x\)
\(\Rightarrow y=a\ln x\times \ln_{10}e\)
\(\Rightarrow y=a\ln x\times \frac{1}{\ln 10}\)
\(\Rightarrow y=\frac{a}{\ln 10}\times\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{a}{\ln 10}\times\ln x)\)
\(=\frac{a}{\ln 10}\frac{d}{dx}\ln x\)
\(=\frac{a}{\ln 10}\times \frac{1}{x}\) ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}\]
\(=\frac{a}{x\ln 10}\)

ধরি,
\(y=\log_ax^{a}\)
\(\Rightarrow y=a\log_ax\)
\(\Rightarrow y=a\ln x\times \ln_{a}e\)
\(\Rightarrow y=a\ln x\times \frac{1}{\ln a}\)
\(\Rightarrow y=\frac{a}{\ln a}\times\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{a}{\ln a}\times\ln x)\)
\(=\frac{a}{\ln a}\frac{d}{dx}\ln x\)
\(=\frac{a}{\ln a}\times \frac{1}{x}\) ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}\]
\(=\frac{a}{x\ln a}\)

ধরি,
\(y=e^{-a\ln x}\)
\(\Rightarrow y=e^{\ln x^{-a}}\)
\(\Rightarrow y=x^{-a}\) ➜Note \[\because e^{\ln x}=x\]
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-a})\)
\(=-ax^{-a-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]

ধরি,
\(y=(ax)^{a}\)
\(\Rightarrow y=a^{a}.x^{a}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a^{a}.x^{a})\)
\(=a^{a}\frac{d}{dx}x^{a}\)
\(=a^{a}.a.x^{a-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=a^{a+1}x^{a-1}\)

ধরি,
\(y=5x^{3}-3x^2+7x-9\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5x^{3}-3x^2+7x-9)\)
\(=\frac{d}{dx}(5x^{3})-\frac{d}{dx}(3x^2)+\frac{d}{dx}(7x)-\frac{d}{dx}(9)\)
\(=5\frac{d}{dx}(x^{3})-3\frac{d}{dx}(x^2)+7\frac{d}{dx}(x)-\frac{d}{dx}(9)\)
\(=5.3x^{2}-3.2x+7.1-0\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0 \]
\(=15x^{2}-6x+7\)

ধরি,
\(y=\frac{x^7+4x^3}{x^5}\)
\(\Rightarrow y=\frac{x^7}{x^5}+\frac{4x^3}{x^5}\)
\(\Rightarrow y=x^2+\frac{4}{x^2}\)
\(\Rightarrow y=x^2+4x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+4x^{-2})\)
\(=\frac{d}{dx}(x^{2})+\frac{d}{dx}(4x^{-2})\)
\(=\frac{d}{dx}(x^{2})+4\frac{d}{dx}(x^{-2})\)
\(=2x+4.(-2).x^{-2-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=2x-\frac{8}{x^3}\)
\(=2\left(x-\frac{4}{x^3}\right)\)

ধরি,
\(y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\Rightarrow y=x^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=x^{\frac{1}{2}}+x^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{1}{2}}+x^{-\frac{1}{2}})\)
\(=\frac{d}{dx}(x^{\frac{1}{2}})+\frac{d}{dx}(x^{-\frac{1}{2}})\)
\(=\frac{1}{2}x^{\frac{1}{2}-1}-\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\(=\frac{1}{2}x^{\frac{1-2}{2}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(=\frac{1}{2}x^{\frac{-1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}\)
\(=\frac{1}{2x^{\frac{1}{2}}}-\frac{1}{2x^{\frac{3}{2}}}\)
\(=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x^{3}}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x^{3}}}\right)\)

ধরি,
\(y=\frac{x^4-9}{x^2-3}\)
\(\Rightarrow y=\frac{(x^2)^2-3^2}{x^2-3}\)
\(\Rightarrow y=\frac{(x^2+3)(x^2-3)}{x^2-3}\)
\(\Rightarrow y=x^2+3\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+3)\)
\(=\frac{d}{dx}(x^2)+\frac{d}{dx}(3)\)
\(=2x+0\) ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\]
\(=2x\)

ধরি,
\(y=(1-\sqrt{x})^2\)
\(\Rightarrow y=1-2\sqrt{x}+(\sqrt{x})^2\)
\(\Rightarrow y=1-2\sqrt{x}+x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(1-2\sqrt{x}+x)\)
\(=\frac{d}{dx}(1)-\frac{d}{dx}(2\sqrt{x})+\frac{d}{dx}(x)\)
\(=0-2\frac{d}{dx}(\sqrt{x})+1\) ➜Note \[\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\]
\(=-2\times \frac{1}{2\sqrt{x}}+1\) ➜Note \[\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\]
\(=-\frac{1}{\sqrt{x}}+1\)

ধরি,
\(y=\ln x-5\sec x+2\cot x-b^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln x-5\sec x+2\cot x-b^{x})\)
\(=\frac{d}{dx}(\ln x)-\frac{d}{dx}(5\sec x)+\frac{d}{dx}(2\cot x)-\frac{d}{dx}(b^{x})\)
\(=\frac{d}{dx}(\ln x)-5\frac{d}{dx}(\sec x)+2\frac{d}{dx}(\cot x)-\frac{d}{dx}(b^{x})\)
\(=\frac{1}{x}-5\sec x\tan x-2\csc^2 x-b^{x}\ln b\) ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\sec x)=\sec x\tan x\], \[\frac{d}{dx}(\cot x)=-\csc^2 x, \frac{d}{dx}(a^{x})=a^x\ln a\]

মনে করি,
\(f(x)=e^{ax}\)
\(\therefore f(x+h)=e^{a(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(e^{ax})=\lim_{h \rightarrow 0}\frac{e^{a(x+h)}-e^{ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{ax+ah}-e^{ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{ax}.e^{ah}-e^{ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{ax}\left(e^{ah}-1\right)}{h}\]
\[=e^{ax}\lim_{h \rightarrow 0}\frac{e^{ah}-1}{h}\]
\[=e^{ax}\lim_{h \rightarrow 0}\frac{1+\frac{ah}{1!}+\frac{a^2h^2}{2!}+\frac{a^3h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{ax}\lim_{h \rightarrow 0}\frac{\frac{ah}{1!}+\frac{a^2h^2}{2!}+\frac{a^3h^3}{3!}+.......\infty}{h}\]
\[=e^{ax}\lim_{h \rightarrow 0}\frac{ah\left(\frac{1}{1!}+\frac{ah}{2!}+\frac{a^2h^2}{3!}+.......\infty \right)}{h}\]
\[=ae^{ax}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{ah}{2!}+\frac{a^2h^2}{3!}+.......\infty \right)\]
\[=ae^{ax}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=ae^{ax}\times (1+0)\]
\[=ae^{ax}\times 1\]
\[=ae^{ax}\]

মনে করি,
\(f(x)=e^{5x}\)
\(\therefore f(x+h)=e^{5(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(e^{5x})=\lim_{h \rightarrow 0}\frac{e^{5(x+h)}-e^{5x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{5x+5h}-e^{5x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{5x}.e^{5h}-e^{5x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{5x}\left(a^{5h}-1\right)}{h}\]
\[=e^{5x}\lim_{h \rightarrow 0}\frac{e^{5h}-1}{h}\]
\[=e^{5x}\lim_{h \rightarrow 0}\frac{1+\frac{5h}{1!}+\frac{5^2h^2}{2!}+\frac{5^3h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{5x}\lim_{h \rightarrow 0}\frac{\frac{5h}{1!}+\frac{5^2h^2}{2!}+\frac{5^3h^3}{3!}+.......\infty}{h}\]
\[=e^{5x}\lim_{h \rightarrow 0}\frac{5h\left(\frac{1}{1!}+\frac{5h}{2!}+\frac{5^2h^2}{3!}+.......\infty \right)}{h}\]
\[=5e^{5x}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{5h}{2!}+\frac{5^2h^2}{3!}+.......\infty \right)\]
\[=5e^{5x}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=5e^{5x}\times (1+0)\]
\[=5e^{5x}\times 1\]
\[=5e^{5x}\]

মনে করি,
\(f(x)=\ln ax\)
\(\therefore f(x+h)=\ln a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{\ln ax\}=\lim_{h \rightarrow 0}\frac{\ln a(x+h)-\ln ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \frac{a(x+h)}{ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \frac{ax+ah}{ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{h}{x}-\frac{1}{2}.\frac{h^2}{x^2}+\frac{1}{3}.\frac{h^3}{x^3}-......\infty}{h}\] ➜Note \[\because \ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+.......\infty \]
\[=\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)}{h}\]
\[=\lim_{h \rightarrow 0}\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)\]
\[=\frac{1}{x}-0+0-......\] ➜Note লিমিট ব্যবহার করে।
\[=\frac{1}{x}\]

মনে করি,
\(f(x)=\cos 2x\)
\(\therefore f(x+h)=\cos 2(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cos 2x)=\lim_{h \rightarrow 0}\frac{\cos 2(x+h)-\cos 2x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2(x+h)+2x}{2}\sin \frac{2x-2(x+h)}{2}}{h}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2x+2h+2x}{2}\sin \frac{2x-2x-2h}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{4x+2h}{2}\sin \frac{-2h}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2(2x+h)}{2}\sin (-h)}{h}\]
\[=-2\lim_{h \rightarrow 0}\frac{\sin (2x+h)\sin h}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin (2x+h)\times \frac{\sin h}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin (2x+h)\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\]
\[=-2\sin 2x\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-2\sin 2x\]

মনে করি,
\(f(x)=\cos 3x\)
\(\therefore f(x+h)=\cos 3(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cos 3x)=\lim_{h \rightarrow 0}\frac{\cos 3(x+h)-\cos 3x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{3(x+h)+3x}{2}\sin \frac{3x-3(x+h)}{2}}{h}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{3x+3h+3x}{2}\sin \frac{3x-3x-3h}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{6x+3h}{2}\sin \frac{-3h}{2}}{h}\]
\[=-2\lim_{h \rightarrow 0}\frac{\sin \frac{6x+3h}{2}\sin \frac{3h}{2}}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin \frac{6x+3h}{2}\times \frac{\sin \frac{3h}{2}}{\frac{3h}{2}}\times \frac{3}{2}\]
\[=-3\lim_{h \rightarrow 0}\sin \frac{6x+3h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{3h}{2}}{\frac{3h}{2}}\]
\[=-3\sin \frac{6x}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-3\sin 3x\]

মনে করি,
\(f(x)=\sin 2x\)
\(\therefore f(x+h)=\sin 2(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sin 2x)=\lim_{h \rightarrow 0}\frac{\sin 2(x+h)-\sin 2x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2(x+h)+2x}{2}\sin \frac{2(x+h)-2x}{2}}{h}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2x+2h+2x}{2}\sin \frac{2x+2h-2x}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{4x+2h}{2}\sin \frac{2h}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2(2x+h)}{2}\sin h}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos (2x+h)\sin h}{h}\]
\[=2\lim_{h \rightarrow 0}\cos (2x+h)\times \frac{\sin h}{h}\]
\[=2\lim_{h \rightarrow 0}\cos (2x+h)\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\]
\[=2\cos 2x\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=2\cos 2x\]

মনে করি,
\(f(x)=\sin 3x\)
\(\therefore f(x+h)=\sin 3(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sin 3x)=\lim_{h \rightarrow 0}\frac{\sin 3(x+h)-\sin 3x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{3(x+h)+3x}{2}\sin \frac{3(x+h)-3x}{2}}{h}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{3x+3h+3x}{2}\sin \frac{3x+3h-3x}{2}}{h}\]
\[=2\lim_{h \rightarrow 0}\frac{\cos \frac{6x+3h}{2}\sin \frac{3h}{2}}{h}\]
\[=2\lim_{h \rightarrow 0}\cos \frac{6x+3h}{2}\times \frac{\sin \frac{3h}{2}}{\frac{3h}{2}}\times \frac{3}{2}\]
\[=3\lim_{h \rightarrow 0}\cos \frac{6x+3h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{3h}{2}}{\frac{3h}{2}}\]
\[=3\cos \frac{6x}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=3\cos 3x\]

মনে করি,
\(f(x)=\sec ax\)
\(\therefore f(x+h)=\sec a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sec ax)=\lim_{h \rightarrow 0}\frac{\sec a(x+h)-\sec ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\cos a(x+h)}-\frac{1}{\cos ax}}{h}\] ➜Note \(\because \sec x=\frac{1}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos ax-\cos a(x+h)}{\cos a(x+h)\cos ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\cos ax-\cos a(x+h)}{h\cos a(x+h)\cos ax}\]
\[=\lim_{h \rightarrow 0}\frac{\cos ax-\cos (ax+ah)}{h\cos (ax+ah)\cos ax}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{ax+ax+ah}{2}\sin \frac{ax+ah-ax}{2}}{h\sin (ax+ah)\sin ax}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\sin \frac{2ax+ah}{2}\sin \frac{ah}{2}}{h\cos (ax+ah)\cos ax}\]
\[=2\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\times \frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\sin \frac{2ax}{2}\times 1\times \frac{1}{\cos ax.\cos ax}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=a\sin ax\times \frac{1}{\cos^2 ax}\]
\[=a\frac{1}{\cos ax}\times \frac{\sin ax}{\cos ax}\]
\[=a\sec ax\times \tan ax\] ➜Note \(\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\)
\[=a\sec ax\tan ax\]

মনে করি,
\(f(x)=\tan 2x\)
\(\therefore f(x+h)=\tan 2(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\tan 2x)=\lim_{h \rightarrow 0}\frac{\tan 2(x+h)-\tan 2x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin 2(x+h)}{\cos 2(x+h)}-\frac{\sin 2x}{\cos 2x}}{h}\] ➜Note \(\because \tan x=\frac{\sin x}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin 2(x+h)\cos 2x-\cos 2(x+h)\sin 2x}{\cos 2(x+h)\cos 2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (2x+2h)\cos 2x-\cos (2x+2h)\sin 2x}{\cos (2x+2h)\cos 2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (2x+2h-2x)}{\cos (2x+2h)\cos 2x}}{h}\] ➜Note \(\because \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin 2h}{\cos (2x+2h)\cos 2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (2x+2h)\cos 2x}\times \frac{\sin 2h}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (2x+2h)\cos 2x}\times \lim_{h \rightarrow 0}\frac{\sin 2h}{2h}\times 2\]
\[=2\frac{1}{\cos 2x.\cos 2x}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=2\frac{1}{\cos^2 2x}\]
\[=2\sec^2 2x\]

মনে করি,
\(f(x)=\sqrt{x}\)
\(\therefore f(x+h)=\sqrt{x+h}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sqrt{x})=\lim_{h \rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\] ➜Note লব ও হরের সহিত \((\sqrt{x+h}+\sqrt{x})\) গুণ করে।
\[=\lim_{h \rightarrow 0}\frac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}\]
\[=\lim_{h \rightarrow 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]
\[=\lim_{h \rightarrow 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\]
\[=\frac{1}{\sqrt{x}+\sqrt{x}}\] ➜Note লিমিট ব্যবহার করে।
\[=\frac{1}{2\sqrt{x}}\]

মনে করি,
\(f(x)=\cos ax\)
\(\therefore f(x+h)=\cos a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\cos ax)=\lim_{h \rightarrow 0}\frac{\cos a(x+h)-\cos ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{a(x+h)+ax}{2}\sin \frac{ax-a(x+h)}{2}}{h}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{ax+ah+ax}{2}\sin \frac{ax-ax-ah}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2ax+ah}{2}\sin \frac{-ah}{2}}{h}\]
\[=-2\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\]
\[=-a\lim_{h \rightarrow 0}\sin \frac{2ax+ah}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\]
\[=-a\sin \frac{2ax}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-a\sin ax\]

মনে করি,
\(f(x)=e^{2x}\)
\(\therefore f(x+h)=e^{2(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{e^{2x}\}=\lim_{h \rightarrow 0}\frac{e^{2x+2h}-e^{2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{2x}.e^{2h}-e^{2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{2x}\left(e^{2h}-1\right)}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{e^{2h}-1}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{1+\frac{2h}{1!}+\frac{2^2h^2}{2!}+\frac{2^3h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{2x}\lim_{h \rightarrow 0}\frac{\frac{2h}{1!}+\frac{2^2h^2}{2!}+\frac{2^3h^3}{3!}+.......\infty}{h}\]
\[=e^{2x}\lim_{h \rightarrow 0}\frac{2h\left(\frac{1}{1!}+\frac{2h}{2!}+\frac{2^2h^2}{3!}+.......\infty \right)}{h}\]
\[=2e^{2x}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{2h}{2!}+\frac{2^2h^2}{3!}+.......\infty \right)\]
\[=2e^{2x}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=2e^{2x}\times (1+0)\]
\[=2e^{2x}\times 1\]
\[=2e^{2x}\]

মনে করি,
\(f(x)=e^{\sin x}\)
\(\therefore f(x+h)=e^{\sin (x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(e^{\sin x})=\lim_{h \rightarrow 0}\frac{e^{\sin (x+h)}-e^{\sin x}}{h}\]
ধরি,
\[\sin x=z.....(1)\]
এবং
\[\sin (x+h)=z+k.....(2)\]
\[(2)-(1)\]-এর সাহায্যে
\[\sin (x+h)-\sin x=k .....(3)\]
এখন,
\[h\rightarrow 0\Rightarrow k\rightarrow 0\]
\[\Rightarrow \frac{d}{dx}(e^{\sin x})=\lim_{k \rightarrow 0}\frac{e^{z+k}-e^{z}}{h}\] ➜Note \[\because \sin (x+h)=z+k, \sin x=z\]
\[=\lim_{k \rightarrow 0}\frac{e^{z+k}-e^{z}}{k}\times \frac{k}{h}\]
\[=\lim_{k \rightarrow 0}\frac{e^{z+k}-e^{z}}{k}\times \lim_{h \rightarrow 0}\frac{\sin (x+h)-\sin x}{h}\] ➜Note \[\because \sin (x+h)-\sin x=k\]
\[=\lim_{k \rightarrow 0}\frac{e^{z+k}-e^{z}}{k}\times \lim_{h \rightarrow 0}\frac{\sin (x+h)-\sin x}{h}\]
\[=\lim_{k \rightarrow 0}\frac{e^{z}e^{k}-e^{z}}{k}\]\[\times \lim_{h \rightarrow 0}\frac{2\cos \frac{(x+h)+x}{2}\sin \frac{(x+h)-x}{2}}{h}\] ➜Note \[\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\]
\[=\lim_{k \rightarrow 0}\frac{e^{z}(e^{k}-1)}{k}\]\[\times 2\lim_{h \rightarrow 0}\frac{\cos \frac{x+h+x}{2}\sin \frac{x+h-x}{2}}{h}\]
\[=e^{z}\lim_{k \rightarrow 0}\frac{(e^{k}-1)}{k}\]\[\times 2\lim_{h \rightarrow 0}\frac{\cos \frac{2x+h}{2}\sin \frac{h}{2}}{h}\]
\[=e^{z}\lim_{k \rightarrow 0}\frac{\left(1+\frac{k}{1!}+\frac{k^2}{2!}+\frac{k^3}{3!}+.......\infty-1\right)}{k}\]\[\times 2\lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{z}\lim_{k \rightarrow 0}\frac{\left(\frac{k}{1!}+\frac{k^2}{2!}+\frac{k^3}{3!}+.......\infty \right)}{k}\]\[\times \lim_{h \rightarrow 0}\cos \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\]
\[=e^{z}\lim_{k \rightarrow 0}\frac{k\left(\frac{1}{1!}+\frac{k}{2!}+\frac{k^2}{3!}+.......\infty \right)}{k}\]\[\times \cos \frac{2x}{2}\times 1\] ➜Note \[\because \frac{\sin x}{x}=1 \]
\[=e^{z}\lim_{k \rightarrow 0}\left(\frac{1}{1!}+\frac{k}{2!}+\frac{k^2}{3!}+.......\infty \right)\times \cos x\]
\[=e^{z}\left(\frac{1}{1}+0+0+.......0\right)\times \cos x\] ➜Note লিমিট ব্যবহার করে।
\[=\cos x e^{z}(1+0)\]
\[=\cos x e^{z}\]
\[=\cos x e^{\sin x}\] ➜Note \[\because z=\sin x\]

মনে করি,
\(f(x)=\csc ax\)
\(\therefore f(x+h)=\csc a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\csc ax)=\lim_{h \rightarrow 0}\frac{\csc a(x+h)-\csc ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\sin a(x+h)}-\frac{1}{\sin ax}}{h}\] ➜Note \(\because \csc x=\frac{1}{\sin x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin ax-\sin a(x+h)}{\sin a(x+h)\sin ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\sin ax-\sin a(x+h)}{h\sin a(x+h)\sin ax}\]
\[=\lim_{h \rightarrow 0}\frac{\sin ax-\sin (ax+ah)}{h\sin (ax+ah)\sin ax}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{ax+ax+ah}{2}\sin \frac{ax-ax-ah}{2}}{h\sin a(x+h)\sin ax}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\cos \frac{2ax+ah}{2}\sin \frac{-ah}{2}}{h\sin a(x+h)\sin ax}\]
\[=-2\lim_{h \rightarrow 0}\cos \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\times \frac{1}{\sin a(x+h)\sin ax}\]
\[=-a\lim_{h \rightarrow 0}\cos \frac{2ax+ah}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{1}{\sin a(x+h)\sin ax}\]
\[=-a\lim_{h \rightarrow 0}\cos \frac{2ax+ah}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\sin a(x+h)\sin ax}\]
\[=-a\cos \frac{2ax}{2}\times 1\times \frac{1}{\sin ax.\sin ax}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=-a\cos ax\times \frac{1}{\sin^2 ax}\]
\[=-a\frac{1}{\sin ax}\times \frac{\cos ax}{\sin ax}\]
\[=-a\csc ax\times \cot ax\] ➜Note \(\because \frac{1}{\sin x}=\csc x, \frac{\cos x}{\sin x}=\cot x\)
\[=-a\csc ax\cot ax\]

মনে করি,
\(f(x)=\log_ax\)
\(\Rightarrow f(x)=\ln x \times \log_ae\)
\(\therefore f(x+h)=\ln (x+h)\times \log_ae\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{\log_ax\}=\lim_{h \rightarrow 0}\frac{\ln (x+h)\times \log_ae-\ln x\times \log_ae}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\log_ae\{\ln (x+h)-\ln x\}}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\ln \frac{(x+h)}{x}}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\frac{\frac{h}{x}-\frac{1}{2}.\frac{h^2}{x^2}+\frac{1}{3}.\frac{h^3}{x^3}-......\infty}{h}\] ➜Note \[\because \ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+.......\infty \]
\[=\log_ae\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)}{h}\]
\[=\log_ae\lim_{h \rightarrow 0}\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)\]
\[=\log_ae\left(\frac{1}{x}-0+0-......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=\log_ae\times \frac{1}{x}\]
\[=\frac{1}{x}\log_ae\]
\[=\frac{1}{x\ln a}\]

মনে করি,
\(f(x)=\ln x\)
\(\therefore f(x+h)=\ln (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{\ln x\}=\lim_{h \rightarrow 0}\frac{\ln (x+h)-\ln x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \frac{(x+h)}{x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{h}{x}-\frac{1}{2}.\frac{h^2}{x^2}+\frac{1}{3}.\frac{h^3}{x^3}-......\infty}{h}\] ➜Note \[\because \ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+.......\infty \]
\[=\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)}{h}\]
\[=\lim_{h \rightarrow 0}\left(\frac{1}{x}-\frac{1}{2}.\frac{h}{x^2}+\frac{1}{3}.\frac{h^2}{x^3}-......\infty\right)\]
\[=\frac{1}{x}-0+0-......\] ➜Note লিমিট ব্যবহার করে।
\[=\frac{1}{x}\]

মনে করি,
\(f(x)=\sec 2x\)
\(\therefore f(x+h)=\sec 2(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sec 2x)=\lim_{h \rightarrow 0}\frac{\sec 2(x+h)-\sec 2x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\cos 2(x+h)}-\frac{1}{\cos 2x}}{h}\] ➜Note \(\because \sec x=\frac{1}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos 2x-\cos 2(x+h)}{\cos 2(x+h)\cos 2x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\cos 2x-\cos (2x+2h)}{h\cos (2x+2h)\cos 2x}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{2x+2x+2h}{2}\sin \frac{2x+2h-2x}{2}}{h\sin (2x+2h)\sin 2x}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\sin \frac{4x+2h}{2}\sin \frac{2h}{2}}{h\cos (2x+2h)\cos 2x}\]
\[=2\lim_{h \rightarrow 0}\sin \frac{2(2x+h)}{2}\times \frac{\sin h}{h}\times \frac{1}{\cos (2x+2h)\cos 2x}\]
\[=2\lim_{h \rightarrow 0}\sin (2x+h)\times \frac{\sin h}{h}\times \frac{1}{\cos (2x+2h)\cos 2x}\]
\[=2\lim_{h \rightarrow 0}\sin (2x+h)\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\times \lim_{h \rightarrow 0}\frac{1}{\cos (2x+2h)\cos 2x}\]
\[=2\sin 2x\times 1\times \frac{1}{\cos 2x.\cos 2x}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=2\sin 2x\times \frac{1}{\cos^2 2x}\]
\[=2\frac{1}{\cos 2x}\times \frac{\sin 2x}{\cos 2x}\]
\[=2\sec 2x\times \tan 2x\] ➜Note \(\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\)
\[=2\sec 2x\tan 2x\]

মনে করি,
\(f(x)=\sec ax\)
\(\therefore f(x+h)=\sec a(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sec ax)=\lim_{h \rightarrow 0}\frac{\sec a(x+h)-\sec ax}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\cos a(x+h)}-\frac{1}{\cos ax}}{h}\] ➜Note \(\because \sec x=\frac{1}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos ax-\cos a(x+h)}{\cos a(x+h)\cos ax}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\cos ax-\cos (ax+ah)}{h\cos (ax+ah)\cos ax}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{ax+ax+ah}{2}\sin \frac{ax+ah-ax}{2}}{h\sin (ax+ah)\sin ax}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\sin \frac{2ax+2h}{2}\sin \frac{ah}{2}}{h\cos (ax+ah)\cos ax}\]
\[=2\lim_{h \rightarrow 0}\sin \frac{2(ax+h)}{2}\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{a}{2}\times \frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\lim_{h \rightarrow 0}\sin (2x+h)\times \frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\lim_{h \rightarrow 0}\sin (ax+h)\times \lim_{h \rightarrow 0}\frac{\sin \frac{ah}{2}}{\frac{ah}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\cos (ax+ah)\cos ax}\]
\[=a\sin ax\times 1\times \frac{1}{\cos ax.\cos ax}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=a\sin ax\times \frac{1}{\cos^2 ax}\]
\[=a\frac{1}{\cos ax}\times \frac{\sin ax}{\cos ax}\]
\[=a\sec ax\times \tan ax\] ➜Note \(\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\)
\[=a\sec ax\tan ax\]

মনে করি,
\(f(x)=5x^2-2x+9\)
\(\therefore f(x+h)=5(x+h)^2-2(x+h)+9\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(5x^2-2x+9)=\lim_{h \rightarrow 0}\frac{5(x+h)^2-2(x+h)+9-(5x^2-2x+9)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{5(x^2+2xh+h^2)-2x-2h+9-5x^2+2x-9}{h}\]
\[=\lim_{h \rightarrow 0}\frac{5x^2+10xh+5h^2-2x-2h+9-5x^2+2x-9}{h}\]
\[=\lim_{h \rightarrow 0}\frac{10xh+5h^2-2h}{h}\]
\[=\lim_{h \rightarrow 0}\frac{h(10x+5h-2)}{h}\]
\[=\lim_{h \rightarrow 0}(10x+5h-2)\]
\[=10x+0-2\] ➜Note লিমিট ব্যবহার করে।
\[=10x-2\]

মনে করি,
\(f(x)=e^{x}\cos x\)
\(\therefore f(x+h)=e^{x+h}\cos (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(e^{x}\cos x)=\lim_{h \rightarrow 0}\frac{e^{x+h}\cos (x+h)-e^{x}\cos x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^x.e^h\cos (x+h)-e^{x}\cos x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^x\{e^h\cos (x+h)-\cos x\}}{h}\]
\[=e^x\lim_{h \rightarrow 0}\frac{e^h\cos (x+h)-\cos x}{h}\]
\[=e^x\lim_{h \rightarrow 0}\frac{\left(1+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+ .....\infty \right)\cos (x+h)-\cos x}{h}\] ➜Note \(\because e^{x}=1+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+ .....\infty\)
\[=e^x\lim_{h \rightarrow 0}\frac{\cos (x+h)+\left(\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+ .....\infty \right)\cos (x+h)-\cos x}{h}\]
\[=e^x\lim_{h \rightarrow 0}\frac{\cos (x+h)-\cos x+\left(\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+ .....\infty \right)\cos (x+h)}{h}\]
\[=e^x\lim_{h \rightarrow 0}\frac{2\sin \frac{(x+h)+x}{2}\sin \frac{x-(x+h)}{2}+h\left(\frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+ .....\infty \right)\cos (x+h)}{h}\]
\[=e^x\left[\lim_{h \rightarrow 0}\frac{2\sin \frac{x+h+x}{2}\sin \frac{x-x-h}{2}}{h}+\lim_{h \rightarrow 0}\frac{h\left(\frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+ .....\infty \right)\cos (x+h)}{h}\right]\]
\[=e^x\left[\lim_{h \rightarrow 0}2\sin \frac{2x+h}{2}\times \frac{\sin \frac{-h}{2}}{h}+\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+ .....\infty \right)\cos (x+h)\right]\]
\[=e^x\left[\lim_{h \rightarrow 0}\{-2\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\}+\lim_{h \rightarrow 0}\left(\frac{1}{1}+\frac{h}{2!}+\frac{h^2}{3!}+ .....\infty \right)\cos (x+h)\right]\]
\[=e^x\left[\lim_{h \rightarrow 0}\{-\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\}+\lim_{h \rightarrow 0}\left(1+\frac{h}{2!}+\frac{h^2}{3!}+ .....\infty \right)\cos (x+h)\right]\]
\[=e^x\left[-\sin \frac{2x}{2}\times 1+(1+0+0+ .....)\cos x\right]\] ➜Note লিমিট ব্যবহার করে।
\[=e^x\left[-\sin x+(1+0)\cos x\right]\]
\[=e^x(-\sin x+\cos x)\]
\[=e^x(\cos x-\sin x)\]

মনে করি,
\(f(x)=\sin bx\)
\(\therefore f(x+h)=\sin b(x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sin bx)=\lim_{h \rightarrow 0}\frac{\sin b(x+h)-\sin bx}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{b(x+h)+ax}{2}\sin \frac{b(x+h)-ax}{2}}{h}\] ➜Note \(\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\)
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{bx+bh+bx}{2}\sin \frac{bx+bh-bx}{2}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{2\cos \frac{2bx+bh}{2}\sin \frac{bh}{2}}{h}\]
\[=2\lim_{h \rightarrow 0}\cos \frac{2bx+bh}{2}\times \frac{\sin \frac{bh}{2}}{\frac{bh}{2}}\times \frac{b}{2}\]
\[=b\lim_{h \rightarrow 0}\cos \frac{2bx+bh}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{bh}{2}}{\frac{bh}{2}}\]
\[=a\cos \frac{2bx}{2}\times 1\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=b\cos bx\]

মনে করি,
\(y=x^{-7}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{-7})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=-7x^{-7-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=-7x^{-8}\]

মনে করি,
\(y=\frac{1}{5}x^{5}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\frac{1}{5}x^{5})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{1}{5}\frac{d}{dx}(x^{5})\]
\[=\frac{1}{5}\times 5x^{5-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=x^{4}\]

মনে করি,
\(y=\frac{2}{3}x^{9}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\frac{2}{3}x^{9})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{2}{3}\frac{d}{dx}(x^{9})\]
\[=\frac{2}{3}\times 9x^{9-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=2\times 3x^{8}\]
\[=6x^{8}\]

মনে করি,
\(y=\frac{6}{x^{4}}\)
\(\Rightarrow y=6x^{-4}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(6x^{-4})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=6\frac{d}{dx}(x^{-4})\]
\[=6(-4)x^{-4-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=-24x^{-5}\]
\[=-\frac{24}{x^{5}}\]

মনে করি,
\(y=x^{\frac{1}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{1}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{1}{3}x^{\frac{1}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{1}{3}x^{\frac{1-3}{3}}\]
\[=\frac{1}{3}x^{\frac{-2}{3}}\]
\[=\frac{1}{3}x^{-\frac{2}{3}}\]
\[=\frac{1}{3x^{\frac{2}{3}}}\]
\[=\frac{1}{3\sqrt[3]{x^{2}}}\]

মনে করি,
\(y=3x^{-\frac{2}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(3x^{-\frac{2}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=3\frac{d}{dx}(x^{-\frac{2}{3}})\]
\[=3\times -\frac{2}{3}x^{-\frac{2}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=-2x^{\frac{-2-3}{3}}\]
\[=-2x^{\frac{-5}{3}}\]
\[=-2x^{-\frac{5}{3}}\]
\[=-\frac{2}{x^{\frac{5}{3}}}\]
\[=-\frac{2}{\sqrt[3]{x^{5}}}\]

মনে করি,
\(y=\sqrt[3]{x^{2}}\)
\(\Rightarrow y=x^{\frac{2}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{2}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{2}{3}x^{\frac{2}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{2}{3}x^{\frac{2-3}{3}}\]
\[=\frac{2}{3}x^{\frac{-1}{3}}\]
\[=\frac{2}{3}x^{-\frac{1}{3}}\]
\[=\frac{2}{3x^{\frac{1}{3}}}\]
\[=\frac{2}{3\sqrt[3]{x}}\]

মনে করি,
\(y=\sqrt[3]{4x}\)
\(\Rightarrow y=\sqrt[3]{4}\sqrt[3]{x}\)
\(\Rightarrow y=\sqrt[3]{4}x^{\frac{1}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt[3]{4}x^{\frac{1}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\sqrt[3]{4}\frac{d}{dx}(x^{\frac{1}{3}})\]
\[=\sqrt[3]{4}\times \frac{1}{3}x^{\frac{1}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{\sqrt[3]{4}}{3}x^{\frac{1-3}{3}}\]
\[=\frac{\sqrt[3]{4}}{3}x^{\frac{-2}{3}}\]
\[=\frac{\sqrt[3]{4}}{3}x^{-\frac{2}{3}}\]
\[=\frac{\sqrt[3]{4}}{3x^{\frac{2}{3}}}\]
\[=\frac{\sqrt[3]{4}}{3\sqrt[3]{x^{2}}}\]

মনে করি,
\(y=ax^2-2bx+c\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(ax^2-2bx+c)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(ax^2)-\frac{d}{dx}(2bx)+\frac{d}{dx}(c)\]
\[=a\frac{d}{dx}(x^2)-2b\frac{d}{dx}(x)+\frac{d}{dx}(c)\]
\[=a.2x-2b+0\] ➜Note \[\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\]
\[=2ax-2b\]
\[=2(ax-b)\]

মনে করি,
\(y=\frac{x^9+x^3}{x^6}\)
\(\Rightarrow y=\frac{x^9}{x^6}+\frac{x^3}{x^6}\)
\(\Rightarrow y=x^3+\frac{1}{x^3}\)
\(\Rightarrow y=x^3+x^{-3}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^3+x^{-3})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^3)+\frac{d}{dx}(x^{-3})\]
\[=3x^{3-1}-3x^{-3-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=3x^{2}-3x^{-4}\]
\[=3x^{2}-\frac{3}{x^{4}}\]
\[= 3\left(x^2-\frac{1}{x^4}\right)\]

মনে করি,
\(y=(2x)^n-b^n\)
\(\Rightarrow y=2^n.x^n-b^n\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(2^n.x^n-b^n)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(2^n.x^n)-\frac{d}{dx}(b^n)\]
\[=2^n\frac{d}{dx}(x^n)-\frac{d}{dx}(b^n)\]
\[=2^n.nx^{n-1}-0\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\]
\[=2^{n}nx^{n-1}\]

মনে করি,
\(y=x^{3}(x^{2}-x-2)\)
\(\Rightarrow y=x^{5}-x^{4}-2x^{3}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{5}-x^{4}-2x^{3})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{5})-\frac{d}{dx}(x^{4})-\frac{d}{dx}(2x^{3})\]
\[=\frac{d}{dx}(x^{5})-\frac{d}{dx}(x^{4})-2\frac{d}{dx}(x^{3})\]
\[=5x^{5-1}-4x^{4-1}-2.3x^{3-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=5x^{4}-4x^{3}-6x^{2}\]
\[= x^2(5x^{2}-4x-6)\]

মনে করি,
\(y=2x^{3}-4x^{\frac{5}{2}}+\frac{7}{2}x^{-\frac{2}{3}}+7\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(2x^{3}-4x^{\frac{5}{2}}+\frac{7}{2}x^{-\frac{2}{3}}+7)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(2x^{3})-\frac{d}{dx}(4x^{\frac{5}{2}})-\frac{d}{dx}(\frac{7}{2}x^{-\frac{2}{3}})+\frac{d}{dx}(7)\]
\[=2\frac{d}{dx}(x^{3})-4\frac{d}{dx}(x^{\frac{5}{2}})-\frac{7}{2}\frac{d}{dx}(x^{-\frac{2}{3}})+\frac{d}{dx}(7)\]
\[=2.3x^{3-1}-4\times \frac{5}{2}x^{\frac{5}{2}-1}-\frac{7}{2}\times -\frac{2}{3}x^{-\frac{2}{3}-1}+0\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\]
\[=6x^{2}-10x^{\frac{5-2}{2}}+\frac{7}{3}x^{\frac{-2-3}{3}}\]
\[=6x^{2}-10x^{\frac{3}{2}}+\frac{7}{3}x^{\frac{-5}{3}}\]
\[=6x^{2}-10x^{\frac{3}{2}}+\frac{7}{3}x^{-\frac{5}{3}}\]
\[=6x^{2}-10x^{\frac{3}{2}}+\frac{7}{3x^{\frac{5}{3}}}\]
\[=6x^{2}-10x^{\frac{3}{2}}+\frac{7}{3\sqrt[3]{x^{5}}}\]

মনে করি,
\(y=(x^{2}-5)^{2}\)
\(\Rightarrow y=(x^{2})^{2}-2.x^2.5+5^{2}\) ➜Note \[\because (a-b)^2=a^2-2ab+b^2\]
\(\Rightarrow y=x^{4}-10x^2+25\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{4}-10x^2+25)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{4})-\frac{d}{dx}(10x^2)+\frac{d}{dx}(25)\]
\[=\frac{d}{dx}(x^{4})-10\frac{d}{dx}(x^2)+\frac{d}{dx}(25)\]
\[=4x^{4-1}-10.2x+0\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\]
\[=4x^{4-1}-10.2x+0\]
\[=4x^{3}-20x\]
\[=4x(x^{2}-5)\]

মনে করি,
\(y=\left(x+\frac{1}{x}\right)^{2}\)
\(\Rightarrow y=x^{2}+2.x.\frac{1}{x}+\left(\frac{1}{x}\right)^{2}\) ➜Note \[\because (a+b)^2=a^2+2ab+b^2\]
\(\Rightarrow y=x^{2}+2+\frac{1}{x^2}\)
\(\Rightarrow y=x^{2}+2+x^{-2}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{2}+2+x^{-2})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{2})+\frac{d}{dx}(2)+\frac{d}{dx}(x^{-2})\]
\[=2x+0-2x^{-2-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\]
\[=2x-2x^{-3}\]
\[=2x-\frac{2}{x^{3}}\]
\[=2\left(x-\frac{1}{x^{3}}\right)\]

মনে করি,
\(y=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\)
\(\Rightarrow y=x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\)
\(\Rightarrow y=x^{\frac{1}{3}}+x^{-\frac{1}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{1}{3}}+x^{-\frac{1}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{\frac{1}{3}})+\frac{d}{dx}(x^{-\frac{1}{3}})\]
\[=\frac{1}{3}x^{\frac{1}{3}-1}-\frac{1}{3}x^{-\frac{1}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{1}{3}x^{\frac{1-3}{3}}-\frac{1}{3}x^{\frac{-1-3}{3}}\]
\[=\frac{1}{3}x^{\frac{-2}{3}}-\frac{1}{3}x^{\frac{-4}{3}}\]
\[=\frac{1}{3}x^{-\frac{2}{3}}-\frac{1}{3}x^{-\frac{4}{3}}\]
\[=\frac{1}{3x^{\frac{2}{3}}}-\frac{1}{3x^{\frac{4}{3}}}\]
\[=\frac{1}{3\sqrt[3]{x^{2}}}-\frac{1}{3\sqrt[3]{x^{4}}}\]
\[=\frac{1}{3}\left(\frac{1}{\sqrt[3]{x^{2}}}-\frac{1}{\sqrt[3]{x^{4}}}\right)\]

মনে করি,
\(y=\frac{5}{x}+\ln x\)
\(\Rightarrow y=5x^{-1}+\ln x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(5x^{-1}+\ln x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(5x^{-1})+\frac{d}{dx}(\ln x)\]
\[=5\frac{d}{dx}(x^{-1})+\frac{d}{dx}(\ln x)\]
\[=5\times -1x^{-1-1}+\frac{1}{x}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\ln x)=\frac{1}{x}\]
\[=-5x^{-2}+\frac{1}{x}\]
\[=-\frac{5}{x^{2}}+\frac{1}{x}\]
\[=\frac{1}{x}\left(1-\frac{5}{x}\right)\]

মনে করি,
\(y=\sqrt{x^5}\)
\(\Rightarrow y=x^{\frac{5}{2}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{5}{2}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{5}{2}x^{\frac{5}{2}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{5}{2}x^{\frac{5-2}{2}}\]
\[=\frac{5}{2}x^{\frac{3}{2}}\]
\[=\frac{5}{2}\sqrt{x^{3}}\]

মনে করি,
\(y=\frac{1}{\sqrt{x^{-3}}}\)
\(\Rightarrow y=\frac{1}{x^{-\frac{3}{2}}}\)
\(\Rightarrow y=x^{\frac{3}{2}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{3}{2}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{3}{2}x^{\frac{3}{2}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{3}{2}x^{\frac{3-2}{2}}\]
\[=\frac{3}{2}x^{\frac{1}{2}}\]
\[=\frac{3}{2}\sqrt{x}\]

মনে করি,
\(y=\frac{x^3-x^7}{\sqrt{x}}\)
\(\Rightarrow y=\frac{x^3-x^7}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=\frac{x^3}{x^{\frac{1}{2}}}-\frac{x^7}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=x^{3-\frac{1}{2}}-x^{7-\frac{1}{2}}\)
\(\Rightarrow y=x^{\frac{6-1}{2}}-x^{\frac{14-1}{2}}\)
\(\Rightarrow y=x^{\frac{5}{2}}-x^{\frac{13}{2}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{5}{2}}-x^{\frac{13}{2}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{\frac{5}{2}})-\frac{d}{dx}(x^{\frac{13}{2}})\]
\[=\frac{5}{2}x^{\frac{5}{2}-1}-\frac{13}{2}x^{\frac{13}{2}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{5}{2}x^{\frac{5-2}{2}}-\frac{13}{2}x^{\frac{13-2}{2}}\]
\[=\frac{5}{2}x^{\frac{3}{2}}-\frac{13}{2}x^{\frac{11}{2}}\]
\[=\frac{5}{2}\sqrt{x^{3}}-\frac{13}{2}\sqrt{x^{11}}\]
\[=\frac{1}{2}(5\sqrt{x^{3}}-13\sqrt{x^{11}})\]

মনে করি,
\(y=\frac{(x-1)^2}{\sqrt[3]{x}}\)
\(\Rightarrow y=\frac{x^2-2x+1}{x^{\frac{1}{3}}}\)
\(\Rightarrow y=\frac{x^2}{x^{\frac{1}{3}}}-2\frac{x}{x^{\frac{1}{3}}}+\frac{1}{x^{\frac{1}{3}}}\)
\(\Rightarrow y=x^{2-\frac{1}{3}}-2x^{1-\frac{1}{3}}+x^{-\frac{1}{3}}\)
\(\Rightarrow y=x^{\frac{6-1}{3}}-2x^{\frac{3-1}{3}}+x^{-\frac{1}{3}}\)
\(\Rightarrow y=x^{\frac{5}{3}}-2x^{\frac{2}{3}}+x^{-\frac{1}{3}}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^{\frac{5}{3}}-2x^{\frac{2}{3}}+x^{-\frac{1}{3}})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^{\frac{5}{3}})-\frac{d}{dx}(2x^{\frac{2}{3}})+\frac{d}{dx}(x^{-\frac{1}{3}})\]
\[=\frac{d}{dx}(x^{\frac{5}{3}})-2\frac{d}{dx}(x^{\frac{2}{3}})+\frac{d}{dx}(x^{-\frac{1}{3}})\]
\[=\frac{5}{3}x^{\frac{5}{3}-1}-2\frac{2}{3}x^{\frac{2}{3}-1}-\frac{1}{3}x^{-\frac{1}{3}-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=\frac{5}{3}x^{\frac{5-3}{3}}-\frac{4}{3}x^{\frac{2-3}{3}}-\frac{1}{3}x^{\frac{-1-3}{3}}\]
\[=\frac{5}{3}x^{\frac{2}{3}}-\frac{4}{3}x^{\frac{-1}{3}}-\frac{1}{3}x^{\frac{-4}{3}}\]
\[=\frac{5}{3}x^{\frac{2}{3}}-\frac{4}{3}x^{-\frac{1}{3}}-\frac{1}{3}x^{-\frac{4}{3}}\]
\[=\frac{5}{3}x^{\frac{2}{3}}-\frac{4}{3x^{\frac{1}{3}}}-\frac{1}{3x^{\frac{4}{3}}}\]
\[=\frac{5}{3}\sqrt[3]{x^{2}}-\frac{4}{3\sqrt[3]{x}}-\frac{1}{3\sqrt[3]{x^{4}}}\]
\[=\frac{1}{3}\left(5\sqrt[3]{x^{2}}-\frac{4}{\sqrt[3]{x}}-\frac{1}{\sqrt[3]{x^{4}}}\right)\]

মনে করি,
\(y=\frac{x^2-25}{x+5}\)
\(\Rightarrow y=\frac{x^2-5^2}{x+5}\)
\(\Rightarrow y=\frac{(x+5)(x-5)}{x+5}\)
\(\Rightarrow y=x-5\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x-5)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x)-\frac{d}{dx}(5)\]
\[=1-0\] ➜Note \[\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\]
\[=1\]

মনে করি,
\(y=4\sin x-\cos x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(4\sin x-\cos x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(4\sin x)-\frac{d}{dx}(\cos x)\]
\[=4\frac{d}{dx}(\sin x)-\frac{d}{dx}(\cos x)\]
\[=4\cos x+\sin x)\] ➜Note \[\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\cos x)=-\sin x\]

মনে করি,
\(y=5\cos x+\ln x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(5\cos x+\ln x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(5\cos x)+\frac{d}{dx}(\ln x)\]
\[=5\frac{d}{dx}(\cos x)+\frac{d}{dx}(\ln x)\]
\[=-5\sin x+\frac{1}{x}\] ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\cos x)=-\sin x\]

মনে করি,
\(y=3e^{x}-5\ln x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(3e^{x}-5\ln x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(3e^{x})-\frac{d}{dx}(5\ln x)\]
\[=3\frac{d}{dx}(e^{x})-5\frac{d}{dx}(\ln x)\]
\[=3e^{x}-5\frac{1}{x}\] ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(e^x)=e^x\]
\[=3e^{x}-\frac{5}{x}\]

মনে করি,
\(y=x-3\log_ax+5\cos x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x-3\log_ax+5\cos x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x)-\frac{d}{dx}(3\log_ax)+\frac{d}{dx}(5\cos x)\]
\[=\frac{d}{dx}(x)-3\frac{d}{dx}(\log_ax)+5\frac{d}{dx}(\cos x)\]
\[=1-3\frac{1}{x\ln a}-5\sin x\] ➜Note \[\because \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(x)=1, \frac{d}{dx}(\cos x)=-\sin x\]
\[=1-\frac{3}{x\ln a}-5\sin x\]

মনে করি,
\(y=2x^{a}-5e^{x}+b\sin x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(2x^{a}-5e^{x}+b\sin x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(2x^{a})-\frac{d}{dx}(5e^{x})+\frac{d}{dx}(b\sin x)\]
\[=2\frac{d}{dx}(x^{a})-5\frac{d}{dx}(e^{x})+b\frac{d}{dx}(\sin x)\]
\[=2ax^{a-1}-5e^{x}+b\cos x\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\sin x)=\cos x\]

মনে করি,
\(y=ax^{4}-4\log_ax\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(ax^{4}-4\log_ax)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(ax^{4})-\frac{d}{dx}(4\log_ax)\]
\[=a\frac{d}{dx}(x^{4})-4\frac{d}{dx}(\log_ax)\]
\[=a.4x^{4-1}-4\frac{1}{x\ln x}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln x}\]
\[=4(ax^{3}-\frac{1}{x\ln x})\]

মনে করি,
\(y=7\log_ax-5\ln x+4\cos x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(7\log_ax-5\ln x+4\cos x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(7\log_ax)-\frac{d}{dx}(5\ln x)+\frac{d}{dx}(4\cos x)\]
\[=7\frac{d}{dx}(\log_ax)-5\frac{d}{dx}(\ln x)+4\frac{d}{dx}(\cos x)\]
\[=7\frac{1}{x\ln x}-5\frac{1}{x}-4\sin x\] ➜Note \[\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln x}, \frac{d}{dx}(\cos x)=-\sin x\]
\[=\frac{7}{x\ln x}-\frac{5}{x}-4\sin x\]

মনে করি,
\(y=t^{3}+10t^2-5t\)
\[\therefore \frac{d}{dt}(y)=\frac{d}{dt}(t^{3}+10t^2-5t)\] ➜Note \[t\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dt}(t^{3})+\frac{d}{dt}(10t^2)-\frac{d}{dt}(5t)\]
\[=\frac{d}{dt}(t^{3})+10\frac{d}{dt}(t^2)-5\frac{d}{dt}(t)\]
\[=3t^{3-1}+10.2t^{2-1}-5.1\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1 \]
\[=3t^{2}+20t^{1}-5.1\]
\[=3t^{2}+20t-5\]

মনে করি,
\(x=y^{3}+\frac{1}{y^{3}}\)
\(\Rightarrow x=y^{3}+y^{-3}\)
\[\therefore \frac{d}{dy}(x)=\frac{d}{dy}(y^{3}+y^{-3})\] ➜Note \[y\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dy}(y^{3})+\frac{d}{dy}(y^{-3})\]
\[=3y^{3-1}-3y^{-3-1}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}\]
\[=3y^{2}-3y^{-4}\]
\[=3y^{2}-\frac{3}{y^{4}}\]
\[=3\left(y^{2}-\frac{1}{y^{4}}\right)\]

মনে করি,
\(y=t^{5}+5t^3-9\)
\[\therefore \frac{d}{dt}(y)=\frac{d}{dt}(t^{5}+5t^3-9)\] ➜Note \[t\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dt}(t^{5})+\frac{d}{dt}(5t^3)-\frac{d}{dt}(9)\]
\[=\frac{d}{dt}(t^{5})+5\frac{d}{dt}(t^3)-\frac{d}{dt}(9)\]
\[=5t^{5-1}+5.3t^{3-1}-0\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0 \]
\[=5t^{4}+15t^{2}\]
\[=5t^2(t^{2}+3)\]

মনে করি,
\(y=(x^2+2)(x-1)\)
\(\Rightarrow y=x^3+2x-x^2-2\)
\(\Rightarrow y=x^3-x^2+2x-2\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^3-x^2+2x-2)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(x^3)-\frac{d}{dx}(x^2)+\frac{d}{dx}(2x)-\frac{d}{dx}(2)\]
\[=\frac{d}{dx}(x^3)-\frac{d}{dx}(x^2)+2\frac{d}{dx}(x)-\frac{d}{dx}(2)\]
\[=3x^{3-1}-2x^{2-1}+2-0\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0 \]
\[=3x^{2}-2x^{1}+2\]
\[=3x^{2}-2x+2\]

মনে করি,
\(y=z(z+1)^3\)
\(\Rightarrow y=z(z^3+3z^2+3z+1)\) ➜Note \[\because (a+b)^3=a^3+3a^2b+3ab^2+b^3\]
\(\Rightarrow y=z^4+3z^3+3z^2+z\)
\[\therefore \frac{d}{dz}(y)=\frac{d}{dz}(z^4+3z^3+3z^2+z)\] ➜Note \[z\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dz}(z^4)+\frac{d}{dz}(3z^3)+\frac{d}{dz}(3z^2)+\frac{d}{dz}(z)\]
\[=\frac{d}{dz}(z^4)+3\frac{d}{dz}(z^3)+3\frac{d}{dz}(z^2)+\frac{d}{dz}(z)\]
\[=4z^{4-1}+3.3z^{3-1}+3.2z+1\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1 \]
\[=4z^{3}+9z^{2}+6z+1\]

মনে করি,
\(y=ax^4-4\log_ax\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(ax^4-4\log_ax)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(ax^4)-\frac{d}{dx}(4\log_ax)\]
\[=a\frac{d}{dx}(x^4)-4\frac{d}{dx}(\log_ax)\]
\[=a.4x^{4-1}-4\frac{1}{x\ln a}\] ➜Note \[\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}\]
\[=4ax^{3}-\frac{4}{x\ln a}\]
\[=4\left(ax^{3}-\frac{1}{x\ln a}\right)\]

মনে করি,
\(y=5e^{x}-6a^{x}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(5e^{x}-6a^{x})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(5e^{x})-\frac{d}{dx}(6a^{x})\]
\[=5\frac{d}{dx}(e^{x})-6\frac{d}{dx}(a^{x})\]
\[=5e^{x}-6a^{x}\ln a\] ➜Note \[\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(a^x)=a^x\ln a\]

মনে করি,
\(y=5\cos x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(5\cos x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=5\frac{d}{dx}(\cos x)\]
\[=-5\sin x\] ➜Note \[\because \frac{d}{dx}(\cos x)=-\sin x\]

মনে করি,
\(y=9\sin x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(9\sin x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=9\frac{d}{dx}(\sin x)\]
\[=9\cos x\] ➜Note \[\because \frac{d}{dx}(\sin x)=\cos x\]

মনে করি,
\(y=2\cos \theta+9\sec \theta\)
\[\therefore \frac{d}{d\theta}(y)=\frac{d}{d\theta}(2\cos \theta+9\sec \theta)\] ➜Note \[\theta\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{d\theta}(2\cos \theta)+\frac{d}{d\theta}(9\sec \theta)\]
\[=2\frac{d}{d\theta}(\cos \theta)+9\frac{d}{d\theta}(\sec \theta)\]
\[=-2\sin \theta+9\sec \theta\tan \theta\] ➜Note \[\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\sec x)=\sec x\tan x\]

মনে করি,
\(y=\sec x+\cot x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\sec x+\cot x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(\sec x)+\frac{d}{dx}(\cot x)\]
\[=\sec x\tan x-\csc^2 x\] ➜Note \[\because \frac{d}{dx}(\cot x)=-\csc^2 x, \frac{d}{dx}(\sec x)=\sec x\tan x \]

মনে করি,
\(y=8\cos t+\ln t+5t\)
\[\therefore \frac{d}{dt}(y)=\frac{d}{dt}(8\cos t+\ln t+5t)\] ➜Note \[t\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dt}(8\cos t)+\frac{d}{dt}(\ln t)+\frac{d}{dt}(5t)\]
\[=8\frac{d}{dt}(\cos t)+\frac{d}{dt}(\ln t)+5\frac{d}{dt}(t)\]
\[=-8\sin t+\frac{1}{t}+5.1\] ➜Note \[\because \frac{d}{dx}(\cot x)=-\csc^2 x, \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(x)=1\]
\[=-8\sin t+\frac{1}{t}+5\]

মনে করি,
\(y=\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}\)
\(\Rightarrow y=\sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}}\) ➜Note \[\because 1-\cos 2A=2\sin^2 A, 1+\cos 2A=2\cos^2 A\]
\(\Rightarrow y=\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}}\)
\(\Rightarrow y=\frac{\sin \theta}{\cos \theta}\)
\(\Rightarrow y=\tan \theta\)
\[\therefore \frac{d}{d\theta}(y)=\frac{d}{d\theta}(\tan \theta)\] ➜Note \[\theta\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\sec^2 \theta\] ➜Note \[\because \frac{d}{dx}(\tan x)=\sec^2 x\]

মনে করি,
\(y=\frac{2\tan \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}\)
\(\Rightarrow y=\tan 2\frac{\theta}{2}\) ➜Note \[\because \tan 2A=\frac{2\tan A}{1+\tan^2 A}\]
\(\Rightarrow y=\tan \theta\)
\[\therefore \frac{d}{d\theta}(y)=\frac{d}{d\theta}(\tan \theta)\] ➜Note \[\theta\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\sec^2 \theta\] ➜Note \[\because \frac{d}{dx}(\tan x)=\sec^2 x\]

মনে করি,
\(y=\frac{\sin x+\cos x}{\sqrt{1+sin 2x}}\)
\(\Rightarrow y=\frac{\sin x+\cos x}{\sqrt{\sin^2 x+\cos^2 x+sin 2x}}\) ➜Note \[\because \sin^2 A+\cos^2 A=1\]
\(\Rightarrow y=\frac{\sin x+\cos x}{\sqrt{\sin^2 x+\cos^2 x+2sin x\cos x}}\) ➜Note \[\because \sin 2A=2sin x\cos x\]
\(\Rightarrow y=\frac{\sin x+\cos x}{\sqrt{\sin^2 x+2sin x\cos x+\cos^2 x}}\)
\(\Rightarrow y=\frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^2}}\) ➜Note \[\because (a+b)^2=a^2+2ab+b^2\]
\(\Rightarrow y=\frac{\sin x+\cos x}{\sin x+\cos x}\)
\(\Rightarrow y=1\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(1)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=0\] ➜Note \[\because \frac{d}{dx}(c)=0\]

মনে করি,
\(y=\tan^{-1}(\sec x+\tan x)\)
\(\Rightarrow y=\tan^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)\) ➜Note \[\because \sec x=\frac{1}{\cos x}, \tan x=\frac{\sin x}{\cos x}\]
\(\Rightarrow y=\tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)\)
\(\Rightarrow y=\tan^{-1}\left(\frac{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}\right)\) ➜Note \[\because \sin^2 A+\cos^2 A=1, \cos 2A=\cos^2 A-\sin^2 A, \sin 2A=2\sin A\cos A\]
\(\Rightarrow y=\tan^{-1}\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\) ➜Note \[\because a^2-b^2=(a+b)(a-b), (a+b)^2=a^2+b^2+2ab\]
\(\Rightarrow y=\tan^{-1}\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\)
\(\Rightarrow y=\tan^{-1}\frac{\tan \frac{x}{2}+1}{1-\tan \frac{x}{2}}\) ➜Note লব ও হর কে \[\cos \frac{x}{2}\] দ্বারা ভাগ করে।
\(\Rightarrow y=\tan^{-1}\frac{\tan \frac{x}{2}+\tan \frac{\pi}{4}}{1-\tan \frac{x}{2}\tan \frac{\pi}{4}}\) ➜Note\[\because \tan \frac{\pi}{4}=1\]
\(\Rightarrow y=\tan^{-1}\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\) ➜Note\[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]
\(\Rightarrow y=\frac{x}{2}+\frac{\pi}{4}\)
\(\Rightarrow y=\frac{1}{2}x+\frac{\pi}{4}\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\frac{1}{2}x+\frac{\pi}{4})\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(\frac{1}{2}x)+\frac{d}{dx}(\frac{\pi}{4})\]
\[=\frac{1}{2}\frac{d}{dx}(x)+\frac{d}{dx}(\frac{\pi}{4})\]
\[=\frac{1}{2}+0\] ➜Note \[\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\]
\[=\frac{1}{2}\]

মনে করি,
\(y=\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)\)
\(\Rightarrow y=\tan^{-1}\left(\frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}\right)\) ➜Note \[\because \sin^2 A+\cos^2 A=1, \cos 2A=\cos^2 A-\sin^2 A, \sin 2A=2\sin A\cos A\]
\(\Rightarrow y=\tan^{-1}\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}\) ➜Note \[\because a^2-b^2=(a+b)(a-b), (a+b)^2=a^2+b^2+2ab\]
\(\Rightarrow y=\tan^{-1}\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}}\)
\(\Rightarrow y=\tan^{-1}\frac{1-\tan \frac{x}{2}}{\tan \frac{x}{2}+1}\) ➜Note লব ও হর কে \[\cos \frac{x}{2}\] দ্বারা ভাগ করে।
\(\Rightarrow y=\tan^{-1}\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4}\tan \frac{x}{2}}\) ➜Note\[\because \tan \frac{\pi}{4}=1\]
\(\Rightarrow y=\tan^{-1}\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\) ➜Note\[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]
\(\Rightarrow y=\frac{\pi}{4}-\frac{x}{2}\)
\(\Rightarrow y=\frac{\pi}{4}-\frac{1}{2}x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\frac{\pi}{4}-\frac{1}{2}x\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(\frac{\pi}{4})-\frac{d}{dx}(\frac{1}{2}x)\]
\[=\frac{d}{dx}(\frac{\pi}{4})-\frac{1}{2}\frac{d}{dx}(x)\]
\[=0-\frac{1}{2}\] ➜Note \[\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\]
\[=-\frac{1}{2}\]

মনে করি,
\(y=\frac{\cos x-\cos 2x}{1-\cos x}\)
\(\Rightarrow y=\frac{2\sin \frac{x+2x}{2}\sin \frac{2x-x}{2}}{2\sin^2 \frac{x}{2}}\) ➜Note \[\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}, 1-\cos A=2\sin^2 \frac{A}{2}\]
\(\Rightarrow y=\frac{2\sin \frac{3x}{2}\sin \frac{x}{2}}{2\sin^2 \frac{x}{2}}\)
\(\Rightarrow y=\frac{\sin \frac{3x}{2}}{\sin \frac{x}{2}}\)
\(\Rightarrow y=\frac{\sin 3\frac{x}{2}}{\sin \frac{x}{2}}\)
\(\Rightarrow y=\frac{3\sin \frac{x}{2}-4\sin^3 \frac{x}{2}}{\sin \frac{x}{2}}\) ➜Note \[\because \sin 3A=3\sin A-4\sin^3 A\]
\(\Rightarrow y=\frac{\sin \frac{x}{2}\left(3-4\sin^2 \frac{x}{2}\right)}{\sin \frac{x}{2}}\)
\(\Rightarrow y=3-4\sin^2 \frac{x}{2}\)
\(\Rightarrow y=3-2.2\sin^2 \frac{x}{2}\)
\(\Rightarrow y=3-2\left(1-\cos 2\frac{x}{2}\right)\) ➜Note \[\because 2\sin^2 A=1-\cos 2A\]
\(\Rightarrow y=3-2(1-\cos x)\)
\(\Rightarrow y=3-2+2\cos x\)
\(\Rightarrow y=1+2\cos x\)
\[\therefore \frac{d}{dx}(y)=\frac{d}{dx}(1+2\cos x)\] ➜Note \[x\]-এর সাপেক্ষে অন্তরীকরণ করে।
\[=\frac{d}{dx}(1)+\frac{d}{dx}(2\cos x)\]
\[=\frac{d}{dx}(1)+2\frac{d}{dx}(\cos x)\]
\[=0-2\sin x\] ➜Note \[\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos x)=-\sin x\]
\[=-2\sin x\]

দেওয়া আছে,
\[f(x)=\frac{1}{\sin x}, g(x)=\frac{1}{\tan x}\]
মনে করি,
\(F(x)=\frac{f(x)}{g(x)}\)
\(\Rightarrow F(x)=\frac{\frac{1}{\sin x}}{\frac{1}{\tan x}}\)
\(\Rightarrow F(x)=\frac{\frac{1}{\sin x}}{\cot x}\) ➜Note \[\because \frac{1}{\tan A}=\cot A\]
\(\Rightarrow F(x)=\frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x}}\) ➜Note \[\because \cot A=\frac{\cos A}{\sin A}\]
\(\Rightarrow F(x)=\frac{1}{\sin x}\times \frac{\sin x}{\cos x}\)
\(\Rightarrow F(x)=\frac{1}{\cos x}\)
\(\Rightarrow F(x)=\sec x\)
\(\therefore F(x+h)=\sec (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{F(x)\}=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\sec x)=\lim_{h \rightarrow 0}\frac{\sec (x+h)-\sec x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{h}\] ➜Note \(\because \sec x=\frac{1}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\cos x-\cos (x+h)}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\cos x-\cos (x+h)}{h\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin \frac{x+x+h}{2}\sin \frac{x+h-x}{2}}{h\sin (x+h)\sin x}\] ➜Note \(\because \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}\)
\[=2\lim_{h \rightarrow 0}\frac{\sin \frac{2x+h}{2}\sin \frac{h}{2}}{h\cos (x+h)\cos x}\]
\[=2\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{2}\times \frac{1}{\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \frac{1}{\cos (x+h)\cos x}\]
\[=\lim_{h \rightarrow 0}\sin \frac{2x+h}{2}\times \lim_{h \rightarrow 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\times \lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\]
\[=\sin \frac{2x}{2}\times 1\times \frac{1}{\cos x.\cos x}\] ➜Note \(\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\)
\[=\sin x\times \frac{1}{\cos^2 x}\]
\[=\frac{1}{\cos x}\times \frac{\sin x}{\cos x}\]
\[=\sec x\times \tan x\] ➜Note \(\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\)
\[=\sec x\tan x\]

মনে করি,
\(f(p)=e^{-2p}\)
\(\therefore f(p+h)=e^{-2(p+h)}\)
আমরা জানি,
\[\frac{d}{dp}\{f(p)\}=\lim_{h \rightarrow 0}\frac{f(p+h)-f(p)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dp}(e^{-2p})=\lim_{h \rightarrow 0}\frac{e^{-2(p+h)}-e^{-2p}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{-2p-2h}-e^{-2p}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{-2p}e^{-2h}-e^{-2p}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{-2p}(e^{-2h}-1)}{h}\]
\[=e^{-2p}\lim_{h \rightarrow 0}\frac{e^{-2h}-1}{h}\]
\[=e^{-2p}\lim_{h \rightarrow 0}\frac{1-\frac{2h}{1!}+\frac{2^2h^2}{2!}-\frac{2^3h^3}{3!}+......\infty -1}{h}\] ➜Note \(\because e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+ .......\infty\)
\[=e^{-2p}\lim_{h \rightarrow 0}\frac{-\frac{2h}{1!}+\frac{2^2h^2}{2!}-\frac{2^3h^3}{3!}+......\infty}{h}\]
\[=e^{-2p}\lim_{h \rightarrow 0}\frac{-2h\left(\frac{1}{1!}-\frac{2h}{2!}+\frac{2^2h^2}{3!}-......\infty\right)}{h}\]
\[=-2e^{-2p}\lim_{h \rightarrow 0}\left(\frac{1}{1!}-\frac{2h}{2!}+\frac{2^2h^2}{3!}-......\infty\right)\]
\[=-2e^{-2p}\left(\frac{1}{1}-0+0-0......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=-2e^{-2p}(1)\]
\[=-2e^{-2p}\]

মনে করি,
\(f(x)=x^{5}\)
\(\therefore f(x+h)=(x+h)^{5}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(x^{5})=\lim_{h \rightarrow 0}\frac{(x+h)^{5}-x^{5}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{x^5+^5C_1x^4h+^5C_2x^3h^2+^5C_3x^2h^3+...... -x^{5}}{h}\] ➜Note \(\because (a+x)^n=a^n+^nC_1a^{n-1}x+^nC_2a^{n-2}x^2+^nC_3a^{n-3}x^3+...... \)
\[=\lim_{h \rightarrow 0}\frac{^5C_1x^4h+^5C_2x^3h^2+^5C_3x^2h^3+...... }{h}\]
\[=\lim_{h \rightarrow 0}\frac{h(^5C_1x^4+^5C_2x^3h+^5C_3x^2h^2+...... )}{h}\]
\[=\lim_{h \rightarrow 0}(^5C_1x^4+^5C_2x^3h+^5C_3x^2h^2+...... )\]
\[=^5C_1x^4+0+0+0..... \]
\[=^5C_1x^4+0\]
\[=\frac{5!}{(5-1)!1!}x^4\] ➜Note \(\because ^nC_r=\frac{n!}{(n-r)!r!}\)
\[=\frac{5\times 4!}{4!1!}x^4\]
\[=\frac{5}{1}x^4\]
\[=5x^4\]
এখন,
\[x=2\] বিন্দুতে অন্তরজ,
\[=5.2^4\]
\[=5.16\]
\[=80\]

মনে করি,
\(f(x)=e^{mx}\)
\(\Rightarrow f(x+h)=e^{m(x+h)}\)
\(\therefore f(x+h)=e^{mx+mh}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}\{e^{mx}\}=\lim_{h \rightarrow 0}\frac{e^{mx+mh}-e^{mx}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{mx}.e^{mh}-e^{mx}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{e^{mx}\left(e^{mh}-1\right)}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{e^{mh}-1}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{1+\frac{mh}{1!}+\frac{m^2h^2}{2!}+\frac{m^3h^3}{3!}+.......\infty -1}{h}\] ➜Note \[\because e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\infty \]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{\frac{mh}{1!}+\frac{m^2h^2}{2!}+\frac{m^3h^3}{3!}+.......\infty}{h}\]
\[=e^{mx}\lim_{h \rightarrow 0}\frac{mh\left(\frac{1}{1!}+\frac{mh}{2!}+\frac{m^2h^2}{3!}+.......\infty \right)}{h}\]
\[=me^{mx}\lim_{h \rightarrow 0}\left(\frac{1}{1!}+\frac{mh}{2!}+\frac{m^2h^2}{3!}+.......\infty \right)\]
\[=me^{mx}\times \left(\frac{1}{1}+0+0+.......\right)\] ➜Note লিমিট ব্যবহার করে।
\[=me^{mx}\times (1+0)\]
\[=me^{mx}\times 1\]
\[=me^{mx}\]
এখন,
\[x=a\] বিন্দুতে অন্তরজ,
\[=me^{ma}\]

মনে করি,
\(f(x)=\tan x\)
\(\therefore f(x+h)=\tan (x+h)\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে
\[\Rightarrow \frac{d}{dx}(\tan x)=\lim_{h \rightarrow 0}\frac{\tan (x+h)-\tan x}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}}{h}\] ➜Note \(\because \tan x=\frac{\sin x}{\cos x}\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin (x+h-x)}{\cos (x+h)\cos x}}{h}\] ➜Note \(\because \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\[=\lim_{h \rightarrow 0}\frac{\frac{\sin h}{\cos (x+h)\cos x}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\times \frac{\sin h}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{\cos (x+h)\cos x}\times \lim_{h \rightarrow 0}\frac{\sin h}{h}\]
\[=\frac{1}{\cos x.\cos x}\times 1\] ➜Note \[\because \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\]
\[=\frac{1}{\cos^2 x}\]
\[=\sec^2 x\]
এখন,
\[x=\frac{\pi}{4}\] বিন্দুতে অন্তরজ,
\[=\sec^2 \frac{\pi}{4}\]
\[=(\sqrt{2})^2\] ➜Note \(\because \sec \frac{\pi}{4}=\sqrt{2}\)
\[=2\]