Answer
$f''(0)=0.$
Work Step by Step
$u=x^2$; $\dfrac{du}{dx}=2x$
$\dfrac{d}{du}f(u)=-\sin{u}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=-2x\sin{x^2}.$
Product Rule $(f’'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=-2x ;u’(x)=-2$
$v(x)=\sin{x^2} ;v’(x)=2x\cos{x^2} $
$f''(x)=(-2)(\sin{x^2})(-2x)(2x\cos{x^2})$
$=-2(\sin{x^2}+2x^2\cos{x^2}).$
$f''(0)=-2(\sin{0^2}+2(0)^2\cos{0^2})=0$.
