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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following w.r.t. x:
$\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg),-1<\text{x}<1,\text{ x}\neq0$
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Answer

Let $\text{y}=\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg)$
Substituting $\text{x}^2=\cos2\theta,$ we get
$\text{y}=\tan^{-1}\bigg(\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}\bigg)$
$=\tan^{-1}\bigg(\frac{\sqrt{1+2\cos^2\theta-1}+\sqrt{1-1+2\sin^2\theta}}{\sqrt{1+2\cos^2\theta-1}-\sqrt{1-1+2\sin^2\theta}}\bigg)$
$=\tan^{-1}\bigg(\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}\bigg)$
$=\tan^{-1}\bigg[\frac{\sqrt{2}(\cos\theta+\sin\theta)}{\sqrt{2}(\cos\theta-\sin\theta)}\bigg]$
$=\tan^{-1}\Big(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\Big)$
$=\tan^{-1}\Bigg(\frac{\frac{\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}\Bigg)$
$=\tan^{-1}\Big(\frac{1+\tan\theta}{1-\tan\theta}\Big)$
$=\tan^{-1}\tan\Big(\frac{\pi}{4}+\theta\Big)$ $\Big[\because\ \tan(\text{a+b})=\frac{\tan\text{a}+\tan\text{b}}{1-\tan\text{a}\cdot\tan\text{b}}\Big]$
$\therefore\ \text{y}=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}^2$ $\Big[\because\ 2\theta=\cos^{-1}\text{x}^2\Rightarrow\theta=\frac{1}{2}\cos^{-1}\text{x}^2\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=0+\frac{1}{2}\cdot\frac{-1}{\sqrt{1-\text{x}^4}}\cdot\frac{\text{d}}{\text{dx}}\text{x}^2$
$=\frac{1}{2}\cdot\frac{-2\text{x}}{\sqrt{1-\text{x}^4}}=\frac{-\text{x}}{\sqrt{1-\text{x}^4}}$
Find $\frac{\text{dy}}{\text{dx}}$ of each of the functions expressed in parametric form.

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