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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions from first principles:
$\log\text{cosec x}$
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Answer

Let $\text{f(x)}=\log\text{cosec x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\text{cosec x}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\text{cosec}(\text{x}+\text{h})-\log\text{cosec x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{\frac{\text{cosec}(\text{x}+\text{h})}{\text{cosec x}}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}}{\sin(\text{x}+\text{h})}-1\Big)\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\begin{Bmatrix}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big)\Big\}}{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}} \end{Bmatrix}\frac{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{\text{x}+\text{x}+\text{h}}{2}\Big)\sin\Big(\frac{\text{x}-\text{x}-\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})\text{h}}$
$\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\text{ and }\sin\text{A}-\sin\text{B} \\ =2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})(-2)}\bigg\{\frac{\sin\big(-\frac{\text{h}}{2}\big)}{-\frac{\text{h}}{2}}\bigg\}$
$=-\cot\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}(\log\text{cosec x})=-\cot\text{x}$

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