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Question:

Find $\frac{d y}{d x}$ :

$y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$

Solution:

The given relationship is $y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$\Rightarrow \cos y=\frac{2 x}{1+x^{2}}$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}(\cos y)=\frac{d}{d x} \cdot\left(\frac{2 x}{1+x^{2}}\right)$’

$\Rightarrow-\sin y \cdot \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow-\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \times 2-2 x \cdot 2 x}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow\left[\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}\right] \frac{d y}{d x}=-\left[\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$

$\Rightarrow \sqrt{\frac{\left(1+x^{2}\right)^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow \sqrt{\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow \frac{1-x^{2}}{1+x^{2}} \cdot \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}$

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