Differentiate the following functions with respect to x: ^ -1 (2x… - Vidyadip

Question

Differentiate the following functions with respect to x:
$\sin^{-1}\big(2\text{x}^2-1\big),0<\text{x}<1$

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Answer

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

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