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