Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 16

The Derivative is: $y'=-x^2\sin x$

$y=x^2\cos x- 2x\sin x -2\cos x$ Applying Derivative rules: $y'=f'(x)+g'(x)$ and $h'(x)=v'(x)\cdot u(x)+v(x)\cdot u'(x)$ $y'=((2)x^{2-1}(\cos x)+x^2(-\sin x))-2(x^{1-1}(\sin x)+x(\cos x))-2(-\sin x)$ $y'=x\cos x-x^2\sin x-2\sin x-x\cos x+2\sin x$ $y'=-x^2\sin x$