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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:
$\text{e}^{\sqrt{\cot\text{x}}}$
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Answer

Let $\text{f(x)} =\text{e}^{\sqrt{\cot\text{x}}}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{\cot\text{x}}}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})=\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}}-\text{e}^{\sqrt{\cot\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\cot\text{x}}}\Big(\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1\Big)}{\text{h}}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{e}^{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}-1}{{\sqrt{\cot(\text{x}+\text{h})}}-\sqrt{\cot\text{x}}}\bigg)\times\bigg(\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\cot(\text{x}+\text{h})}-\sqrt{\cot\text{x}}}{\text{h}}\times\frac{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}{\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}}$
$\Big[\text{Since, } \lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\text{ and rationalizing numerator}\Big]$
$\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})-\cot\text{x}}{\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(\text{x}+\text{h}-\text{x})}}{\text{x}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$\Big[\text{Since,} \cot(\text{A}-\text{B})=\frac{\cot\text{A}\cot\text{B}+1}{\cot\text{A}-\cot\text{B}}\Big]$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\cot(-\text{h})\times\text{h}\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\text{e}^{\sqrt{\cot\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\cot(\text{x}+\text{h})\cot\text{x}+1}{\Big(\frac{\text{h}}{\cot\text{h}}\Big)\big(\sqrt{\cot(\text{x}+\text{h})}+\sqrt{\cot\text{x}}\big)}$
$=\frac{\text{e}^\sqrt{\cot\text{x}}\times(\cot^2\text{x}+1)}{2\sqrt{\cot\text{x}}}\ \Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}\ \big[\because(1+\cot^2\text{x})=\text{cosec}^2\text{x}\big]$
$\therefore\ \frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot}\text{x}}$

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