#### Answer

$f'(x) = 2x sin (x) + x^{2}cos (x)$

#### Work Step by Step

$f(x) = x^{2}sin (x)$
Lets use the formula $(fg)' = f'g+ fg'$
$f'(x) = (x^{2}sin (x))' = (x^{2})'sin (x) + x^{2}(sin (x))'$
We know that $(sin (x))' = cos (x)$
So:
$f'(x) = (x^{2}sin (x))' = 2x sin (x) + x^{2}cos (x)$