এ অধ্যায়ের পাঠ্যসূচী
- দুইটি ফাংশনের গুনফলের অন্তরীকরণ (Differentiation of multiplication and division of two functions)
- \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
- দুইটি ফাংশনের ভাগফলের অন্তরীকরণ (Differentiation of multiplication and division of two functions)
- \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
- অনুসিদ্ধান্ত (Illustration)
- অধ্যায় \(ix.C\)-এর উদাহরণসমুহ
- অধ্যায় \(ix.C\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(ix.C\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(ix.C\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(ix.C\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
ফাংশনের গুণফলের অন্তরীকরণ।
Differentiation of multiplication and division of two functions
যখন, \(u=u(x)\) এবং \(v=v(x)\) অর্থাৎ উভয়ে \(x\)-এর ফাংশন।
\((a)\) \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\) ফাংশনের ভাগফলের অন্তরীকরণ।
Differentiation of multiplication and division of two functions
যখন, \(u=u(x)\) এবং \(v=v(x)\) অর্থাৎ উভয়ে \(x\)-এর ফাংশন।
\((b)\) \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\) অনুসিদ্ধান্ত
Illustration
\(\frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
×
যখন, \(u=u(x)\) এবং \(v=v(x)\) অর্থাৎ উভয়ে \(x\)-এর ফাংশন।
\((a)\) \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
proof
দেওয়া আছে,
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=uv=u(x)v(x)\)
\(\therefore f(x+h)=u(x+h)v(x+h)\)
আমরা জানি,
\(\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\) ➜ মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(uv)=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\left[\frac{u(x+h)v(x+h)-u(x+h)v(x)}{h}+\frac{u(x+h)v(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\left[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}+v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}\]
\(=u(x)\frac{d}{dx}\{v(x)\}+v(x)\frac{d}{dx}\{u(x)\}\) ➜ \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\(=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\therefore \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
(proved)
\(\therefore \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=uv=u(x)v(x)\)
\(\therefore f(x+h)=u(x+h)v(x+h)\)
আমরা জানি,
\(\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\) ➜ মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(uv)=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\left[\frac{u(x+h)v(x+h)-u(x+h)v(x)}{h}+\frac{u(x+h)v(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\left[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}+v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}\]
\(=u(x)\frac{d}{dx}\{v(x)\}+v(x)\frac{d}{dx}\{u(x)\}\) ➜ \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\(=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\therefore \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
(proved)
×
যখন, \(u=u(x)\) এবং \(v=v(x)\) অর্থাৎ উভয়ে \(x\)-এর ফাংশন।
\((b)\) \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
proof
দেওয়া আছে,
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=\frac{u}{v}=\frac{u(x)}{v(x)}\)
\(\therefore f(x+h)=\frac{u(x+h)}{v(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(\frac{u}{v})=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)v(x)-v(x+h)u(x)}{v(x+h)v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-v(x+h)u(x)}{hv(x+h)v(x)}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-v(x+h)u(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-u(x)v(x)-v(x+h)u(x)+u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{\{u(x+h)v(x)-u(x)v(x)\}-\{v(x+h)u(x)-u(x)v(x)\}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\frac{u(x+h)v(x)-u(x)v(x)}{h}-\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-u(x)v(x)}{h}-\lim_{h \rightarrow 0}\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\]
\[=\frac{1}{v(x).v(x)}\times \left[v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}-u(x)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}\right]\]
\[=\frac{1}{v.v}\times \left[v(x)\frac{d}{dx}\{u(x)\}-u(x)\frac{d}{dx}\{v(x)\}\right]\] ➜ \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[=\frac{1}{v^2}\times \left[v\frac{d}{dx}(u)-u\frac{d}{dx}(v)\right]\]
\[=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
\[\therefore \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
(proved)
\(\therefore \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=\frac{u}{v}=\frac{u(x)}{v(x)}\)
\(\therefore f(x+h)=\frac{u(x+h)}{v(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(\frac{u}{v})=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)v(x)-v(x+h)u(x)}{v(x+h)v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-v(x+h)u(x)}{hv(x+h)v(x)}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-v(x+h)u(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-u(x)v(x)-v(x+h)u(x)+u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{\{u(x+h)v(x)-u(x)v(x)\}-\{v(x+h)u(x)-u(x)v(x)\}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\frac{u(x+h)v(x)-u(x)v(x)}{h}-\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-u(x)v(x)}{h}-\lim_{h \rightarrow 0}\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\]
\[=\frac{1}{v(x).v(x)}\times \left[v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}-u(x)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}\right]\]
\[=\frac{1}{v.v}\times \left[v(x)\frac{d}{dx}\{u(x)\}-u(x)\frac{d}{dx}\{v(x)\}\right]\] ➜ \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[=\frac{1}{v^2}\times \left[v\frac{d}{dx}(u)-u\frac{d}{dx}(v)\right]\]
\[=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
\[\therefore \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
(proved)
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