Question

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

$y=ta{n}^{-1}\frac{3x-x^3}{1-3x^2},-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}.$

Answer 1

Substitute $ta{n}^{-1}x=\theta i.e.,x=tan\theta$

$\therefore y=ta{n}^{-1}\frac{3x-x^3}{1-3x^2}=ta{n}^{-1}\frac{3tan\theta -ta{n}^3\theta }{1-3ta{n}^2\theta }(\because tan3\theta =\frac{3tan\theta -ta{n}^3\theta }{1-3ta{n}^2\theta })$

$y=ta{n}^{-1}\left(tan3\theta \right)=3\theta =3ta{n}^{-1}x$

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

$\Rightarrow \frac{dy}{dx}=3\frac{d}{dx}\left(ta{n}^{-1}x\right)=\frac{3}{1+x^2}(\because \frac{d}{dx}(ta{n}^{-1}x)=\frac{1}{1+x^2})$