Chapter 3 - Applications of Differentiation - 3.3 How Derivatives Affect the Shape of a Graph - 3.3 Exercises - Page 230: 63

See proof

$y$ = $x\sin x$ $y'$ = $x\cos x+\sin x$ $y''$ = $-x\sin x+2\cos x$ $y''$ = $0$ $2\cos x$ = $x\sin x$ $(2\cos x)^{2}$ = $(x \sin x)^{2}$ $4\cos^{2}x$ = $x^{2}\sin^{2}x$ $4\cos^{2}x$ = $x^{2}(1-\cos^{2}x)$ $4\cos^{2}x(x^{2}+4)$ = $4x^{2}$ $y^{2}(x^{2}+4)$ = $4x^{2}$ since $y$ = $2\cos x$ when $y''$ = $0$