Find d y d x if y = ^ –1 (2 x 1+x^ 2 ),-1

Question

Find $\frac{d y}{d x}$ if $y = \cos^{–1}\left(\frac{2 x}{1+x^{2}}\right),-1$

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Answer

It is given that $y = \cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$\Rightarrow \cos y=\frac{2 x}{1+x^{2}}$
Differentiating both sides w.r.t. x, we get,
$-\sin y \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow -\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \times 2-2 x \cdot 2 x}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \sqrt{1-\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{-2\left(1-\mathrm{x}^{2}\right)}{~~~~~~\left(1+\mathrm{x}^{2}\right)^{2}}\right]$
$\Rightarrow \sqrt{\frac{\left(1+x^{2}\right)^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \sqrt{\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \frac{1-x^{2}}{1+x^{2}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}$

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