Question

find the derivation of​

Answer 1

Step-by-step explanation:

Let $\rm\:y=tan^{-1}\bigg(\frac{3x-x^{3}}{1-3x^{2}}\bigg)$ and

$\rm\:z=sin^{-1}\bigg(\frac{2x}{1+x^{2}}\bigg)$

Put $\rm\:x=tan(\alpha)$

Now,

$\rm \: \rm\:y=tan^{-1}\bigg(\frac{3 \tan( \alpha ) - tan^{3}( \alpha )}{1-3tan^{2}( \alpha )}\bigg)$

$\rm \: \implies \rm\:y=tan^{-1}(\tan( 3\alpha ) )$

$\rm \: \implies \rm\:y= 3\alpha$

Differenatiating both sides w.r.t $\alpha$

$\rm \: \implies \rm\: \frac{dy}{d \alpha } = 3 \: \: ....(i)$

And,

$\rm \: \implies z=sin^{-1}( \sin( 2\alpha ))$

$\rm \: \implies z= 2\alpha$

$\rm \: \implies \frac{dz}{d \alpha } = 2 \: \: ....(ii)$

So,

$\rm \: \implies \frac{dy}{d z} = \frac{3}{2} \: \:$