Question

Find $\frac{dy}{dx}$in the following questions:

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

Answer 1

Substitute $x=tan\theta =ta{n}^{-1}$

$y=co{s}^{-1}\left(\frac{2tant\theta }{1+ta{n}^2\theta }\right)$

$\Rightarrow y=co{s}^{-1}\left(sin2\theta \right)(\because sin2\theta =\frac{2tant\theta }{1+ta{n}^2\theta })$

$\Rightarrow y=co{s}^{-1}\left[cos(\frac{\pi }{x}-2\theta )\right](\because sin2\theta =cos(\frac{\pi }{2}-2\theta ))$

$\Rightarrow y=\frac{\pi }{2}-2\theta \Rightarrow y=\frac{\pi }{2}-2ta{n}^{-1}x\because \theta =tan{t}^{-1}x$

Differentiating both sides w.r.t x, we get

$\frac{dy}{dx}=0-\frac{2}{1+x^2}\Rightarrow \frac{dy}{dx}=\frac{-2}{1+x^2}[\because (ta{n}^{-1}x=\frac{1}{1+x^2}\left)\right]$