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

Question

Differentiate the following functions with respect to x:
$\cos^{-1}\big\{2\text{x}\sqrt{1-\text{x}^2}\big\},\frac{1}{\sqrt{2}}<\text{x}<1$

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Answer

Let $\text{y}=\cos^{-1}\Big\{2\text{x}\sqrt{1-\text{x}^2}\Big\}$
Put $\text{x}=\cos\theta$
$\text{y}=\cos^{-1}\Big\{2\cos\sqrt{1-\cos^2\theta}\Big\}$
$=\cos^{-1}\big\{2\cos\theta\sin\theta\big\}$
$\text{y}=\cos^{-1}\big\{\sin2\theta\big\}$
$\big[\text{Since}, \sin2\theta=2\sin\theta\cos\theta,\sin^2\theta+\cos^2\theta=1\big]$
$\text{y}=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)\Big]\ .....(\text{i})$
Now,
$\frac{1}{\sqrt{2}}<\text{x}<1$
$\Rightarrow\frac{1}{\sqrt{2}}<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0>-2\theta>-\frac{\pi}{2}$
$\Rightarrow\frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>0$
Hence, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\big[\text{Sicne}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=0-2\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$

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