If y= ^ -1 (2x)+2 ^ -1 1-4x^2 , -1 2 <x<0, find dy dx . - Vidyadip

Question

If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, -\frac{1}{2}<\text{x}<0,$ find $\frac{\text{dy}}{\text{dx}}.$

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Answer

Here, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$=\cos^2(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\theta\Big)\Big)\ .....(\text{i})$
Now, $-\frac{1}{2}<\text{x}<0$
$\Rightarrow -1<2\text{x}<0$
$\Rightarrow -1<\cos\theta<0$
$\Rightarrow \frac{\pi}{2}<\theta<\pi$
And
$\Rightarrow -\frac{\pi}{2}>-\theta>-\pi$
$\Rightarrow \Big(\frac{\pi}{2}-\frac{\pi}{2}\Big)>\Big(\frac{\pi}{2}-\theta\Big)>\Big(\frac{\pi}{2}-\pi\Big)$
$\Rightarrow 0>\Big(\frac{\pi}{2}-\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+2\Big[-\Big(\frac{\pi}{2}-\theta\Big)\Big]$
$\begin{bmatrix} \text{Since}, \cos^{-1}\cos(\theta)=\theta, \text{if }\theta\in[0,\pi] \\ \cos^{-1}\cos(\theta)=-\theta, \text{if }\theta\in[-\pi,0] \end{bmatrix}$
$\text{y}=\theta-2\times\frac{\pi}{2}+2\theta$
$\text{y}=-\pi+3\theta$
$\text{y}=-\pi+3\cos^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0+3\Big(\frac{1}{\sqrt{1-(2\text{x})^2}}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{-3}{\sqrt{1-4\text{x}^2}}(2)$
$\frac{\text{dy}}{\text{dx}}=-\frac{6}{\sqrt{1-4\text{x}^2}}$

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