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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:
$\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big),-1<\text{x}<1$
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Answer

Let $\text{y}=\sin^{-1}\Big(\frac{\text{x}+\sqrt{1-\text{x}^3}}{\sqrt{2}}\Big)$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\sin\theta\Big(\frac{1}{\sqrt{2}}\Big)+\cos\theta\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\sin^{-1}\Big\{\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow\Big(-\frac{\pi}{2}+\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}+\theta\Big)<\frac{3\pi}{4}$
So, from equation (i),
$\text{y}=\theta+\frac{\pi}{4}\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ as }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\sin^{-1}\text{x}+\frac{\pi}{4} \big[\text{Since},\sin\theta=\text{x}\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+0$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$

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