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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
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Answer

We have, $\text{y}=\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$

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