Question

Differentiate w.r.t. $x$ the function: $sin^{-1}(x\sqrt{x}),0≤x≤1$

Answer 1

The correct answer is $=\frac{3}{2}\sqrt{\frac{x}{1-x^3}}$
Let $y=sin^{-1}(x\sqrt{x})$
Using chain rule,we obtain
$\frac{dy}{dx}=\frac{d}{dx}(sin^{-1}(x\sqrt{x}))$
$=\frac{1}{\sqrt{1-(x\sqrt{x})^2}}\times\frac{d}{dx}(x\sqrt{x})$
$=\frac{1}{\sqrt{1-x^3}}.\frac{d}{dx}(x^{\frac{3}{2}})$
$=\frac{1}{\sqrt{1-x^3}}\times\frac{3}{2}.x^{\frac{1}{2}}$
$=\frac{3\sqrt{x}}{2\sqrt{1-x^3}}$
$=\frac{3}{2}\sqrt{\frac{x}{1-x^3}}$