## Calculus 10th Edition

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