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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
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Answer

Let $\text{y}=\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\}$
Put $\text{x}=\cos2\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{2}\Big\}$
$=\sin^{-1}\Big\{\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{2}\Big\}$
$=\sin^{-1}\Big\{\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)+\Big(\frac{1}{\sqrt{2}}\Big)\sin\theta\Big\}$
$=\sin^{-1}\Big\{\cos\theta\sin\Big(\frac{\pi}{4}\Big)+\cos\frac{\pi}{4}\sin\theta\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow 0<\cos2\theta<1$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0 < \theta < \frac{\pi}{4}$
$\Rightarrow \frac{\pi}{4}<\Big(\theta+\frac{\pi}{4}\Big)<\frac{\pi}{2}$
So from eqaution (i),
$\text{y}=\theta+\frac{\pi}{4}\ \Big[\text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}+\frac{\pi}{4}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)+0$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{1-\text{x}^2}}$

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