Answer
The Derivative is:
$y'=3x^2\sin x\cos x+x^3\cos^2 x-x^3\sin^2 x$
Work Step by Step
$y=x^3\sin x\cos x$
Applying Derivative Rules:
$y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
$y'=(3x^{3-1})(\sin x\cos x)+(x^3)((\cos x)(\cos x)+(\sin x)(-\sin x))$
$y'=3x^2\sin x\cos x+x^3\cos^2 x-x^3\sin^2 x$
