Calculus 8th Edition

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