Question

Find $\frac{dy}{dx}$, where $y={\sin }^{-1}(3x-4x^3)$

Answer 1

Given

$y={\sin }^{-1}(3x-4x^3)$

Let $x=\sin \theta$

$dx=\cos \theta d\theta \Rightarrow \frac{d\theta }{dx}=\frac{1}{\sqrt{1-x^2}}$

$y={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )$

$={\sin }^{-1}\left(\sin 3\theta \right)$

$=n\pi +{(-1)}^{n}3\theta$

$\frac{dy}{dx}={(-1)}^{n}3\frac{d\theta }{dx}$

$\therefore \frac{dy}{dx}=\frac{{(-1)}^{n}3}{\sqrt{1-x^2}},nϵz$