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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:
$\log\cos\text{x}$
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Answer

Let $\text{f(x)} = \log \cos \text{x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\cos(\text{x}+\text{h})$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\cos(\text{x}+\text{h})-\log\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log^{\log(\text{x}+\text{h})}_{\cos\text{x}}}{\text{h}}\ \Big[\because\ \log\text{A}-\log\text{B}=\log\Big(\frac{\text{A}}{\text{B}}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big[1+\left\{\frac{\cos(\text{z}+\text{h})}{\cos\text{z}}-1\right\}\Big]}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\left\{1+\frac{\cos(\text{x}+\text{h})-\cos\text{z}}{\cos\text{z}}\right\}}{\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}}\times\lim\limits_{\text{h}\rightarrow0}\left\{\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}}\right\}$
$=1\times\lim\limits_{\text{h}\rightarrow0}\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\cos\text{x}\times\text{h}}\ \Big[\because\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-2\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)\sin\Big(\frac{\text{x}+\text{h}+\text{x}}{2}\Big)}{\cos\text{x}\times\text{h}}$
$=-2\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\sin\big(\frac{\text{h}}{2}\big)}{2\cos\text{x}\times\big(\frac{\text{x}}{2}\big)}$
$=\frac{-2\sin\text{x}}{2\cos\text{x}}\Big[\because \lim\limits\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=-\tan\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\cos\text{x})=-\tan\text{x}$

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