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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
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Answer

Here, $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big)$
Put $\text{x}=\cos2\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2}(\cos\theta-\sin\theta)}{\sqrt{2}(\cos\theta+\sin\theta)}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta-\sin\theta}{\cos\theta}}{\frac{\cos\theta+\sin\theta}{\cos\theta}}\bigg)$
[Dividing numerator and denomainator by $\cos\theta$]
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}\bigg)$
$=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\times\tan\theta}\bigg)$
$=\tan^{-1}\Big[\tan\big(\frac{\pi}{4}-\theta\big)\Big]$
$=\frac{\pi}{4}-\theta$
$=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\ (\text{Using x}=\cos2\theta)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$

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