Differentiate log sin x from first principles. - Vidyadip

Question

Differentiate log sin x from first principles.

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Answer

Let f (x) = log sin x. Then, f (x + h) = log sin (x + h)
$\therefore \frac{d}{d x}(f(x))=\underset{{h \rightarrow 0}}{\lim} \frac{f(x+h)-f(x)}{h}$
$\begin{array}{l}\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \sin (x+h)-\log \sin x}{h} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{\frac{\sin (x+h)}{\sin x}\right\}}{h} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)}{\sin x}-1\right\}}{h} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\lim _h \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{\sin x}\right\}}{h}\end{array}$
$\begin{array}{l}\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{\sin x}\right\}}{h\left\{\frac{\sin (x+h)-\sin x}{\sin x}\right\}} \times\left\{\frac{\sin (x+h)-\sin x}{\sin x}\right\} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \frac{\sin (x+h)-\sin x}{h} \times \frac{1}{\sin x} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \underset{{h \rightarrow 0}}{\lim} \frac{2 \sin \frac{h}{2} \cos \left(x+\frac{h}{2}\right)}{h} \times \frac{1}{\sin x} \\ \Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(\frac{h}{2}\right) \cos \left(x+\frac{h}{2}\right)}{\frac{h}{2}} \times \frac{1}{\sin x} \\ \Rightarrow \frac{d}{d x}( f ( x ))=1 \times \cos x \times \frac{1}{\sin x}=\cot x .\end{array}$

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