Question

Find $\frac{dy}{dx}:y=co{s}^{-1}\left(\frac{2x}{1+x^2}\right),-1<x<1$

Answer 1

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

we know is continuous and difft,

i.e., we can write as

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

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

differentiating w.r to $x$

$\frac{d}{dx}\left(\cos y\right)=2\frac{d}{dx}\left[\frac{x}{1+x^2}\right]$ Using $d(x,y)=xdy+ydx\left)\right)$

$-\sin y\frac{dy}{dx}=2[\frac{1}{1+x^2}-\frac{x}{{(1+x^2)}^2}.2x]$

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

we know that

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

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

$\Rightarrow \frac{dy}{dx}=-2\frac{{(1-x^2)}^{1/2}}{{(1-x^2)}^{3/2}}$