Question

Let $y={\sin }^{-1}(3x-4x^3)$ then find $\frac{dy}{dx}$.

Answer 1

Given,

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

or, $y=3{\sin }^{-1}x$.

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{3}{\sqrt{1-x^2}}$.