# What is the derivative of g(x)= x^2*sqrt(1-x^2)?

The answer is: $y ' = \frac{2 x - 3 {x}^{3}}{\sqrt{1 - {x}^{2}}}$.
$y ' = 2 x \cdot \sqrt{1 - {x}^{2}} + {x}^{2} \cdot \frac{1}{2 \sqrt{1 - {x}^{2}}} \cdot \left(- 2 x\right) =$
$= 2 x \cdot \sqrt{1 - {x}^{2}} - {x}^{3} / \sqrt{1 - {x}^{2}} = \frac{2 x \left(1 - {x}^{2}\right) - {x}^{3}}{\sqrt{1 - {x}^{2}}} =$
$= \frac{2 x - 3 {x}^{3}}{\sqrt{1 - {x}^{2}}}$.