Question

Let $y={\sin }^{-1}{x}^2+\sin x^3$ then find $\frac{dy}{dx}$.

Answer 1

Given,

$y={\sin }^{-1}{x}^2+\sin x^3$.

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

$\frac{dy}{dx}=\frac{2x}{\sqrt{1-x^4}}+\cos x^3.\left(3x^2\right)$

or, $\frac{dy}{dx}=\frac{2x}{\sqrt{1-x^4}}+3x^2.\cos x^3$