Answer
$f''(x)=2\cos{x^2}-4x^2\sin{x^2}.$
Work Step by Step
Using the Chain Rule:
$u=x^2$; $\dfrac{du}{dx}=2x$
$\dfrac{d}{du}f(u)=\cos{u}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=2x\cos{x^2}$
Product Rule $(f’'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=2x ;u’(x)=2 $
$v(x)=\cos{x^2} ;v’(x)=-2x\sin{x^2} $
$f''(x)=(2)(\cos{x^2})+(-2x\sin{x^2})(2x)$
$=2\cos{x^2}-4x^2\sin{x^2}.$
