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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate $\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$ with respect to $\sqrt{1+\text{a}^2\text{x}^2}$
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Answer

Let, $\text{u}=\tan^{-1}\Big(\frac{1+\text{ax}}{1-\text{ax}}\Big)$
Put $\text{ax}= \tan\theta$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1+\tan\theta}{1-\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}\bigg)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\theta\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}+\theta$
$\Rightarrow\text{u}=\frac{\pi}{4}+\tan^{-1}(\text{ax}) [\text{since,}\tan\theta=\text{ax}]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0+\frac{1}{1+(\text{ax}^2)}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\ .....(\text{i})$
Now,
Let, $\text{v}=\sqrt{1+\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}\frac{\text{d}}{\text{dx}}(1+\text{a}^2\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+\text{a}^2\text{x}^2}}(2\text{a}^2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{a}^2\text{x}}{\sqrt{1+\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Dividing equation (i) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\text{a}}{1+\text{a}^2\text{x}^2}\times\frac{\sqrt{1+\text{a}^2\text{x}^2}}{\text{a}^2\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{ax}\sqrt{1+\text{a}^2\text{x}^2}}$

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