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Model Paper 6 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 65 Marks
Question
Differentiate log sin $x$ from first principles.
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Answer

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

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