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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\sin^{-1}\big(1-2\text{x}^2\big),0<\text{x}<1$
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Answer

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

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