Answer
$f''(x)=\sec^3{x}+\sec{x}\tan^2{x}$
Work Step by Step
First Derivative
By Theorem $2.9$: $f'(x)=\sec{x}\tan{x}$
Second Derivative:
Product Rule:
$f’'(x)=\frac{d}{dx}f'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x)$
$u(x)=\sec{x} ;u’(x)=\sec{x}\tan{x} $
$v(x)=tan{x} ;v’(x)=\sec^2{x} $
$f''(x)=\sec^3{x}+\sec{x}\tan^2{x}$
