Answer
f'($\theta$)=sec($\theta$)($tan^{2}$($\theta$)+$sec^{2}$($\theta$))
Work Step by Step
The product rule states that if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x).
If f($\theta$)=tan($\theta$)sec($\theta$), then we can set h(x)=tan($\theta$) and g(x)=sec($\theta$), and compute the derivative of the function f($\theta$).
f'($\theta$)=tan($\theta$)$\frac{d}{d\theta}$[sec($\theta$)]+sec($\theta$)$\frac{d}{d\theta}$[tan($\theta$)].
$\frac{d}{d\theta}$[sec($\theta$)]=sec($\theta$)tan($\theta$)
$\frac{d}{d\theta}$[tan($\theta$)]=$sec^{2}$($\theta$)
The above derivatives could be derived using the quotient rule, but it'd be a good idea to have them memorized.
We'll then find that f'($\theta$)=tan($\theta$)sec($\theta$)tan($\theta$)+sec($\theta$)$sec^{2}$($\theta$)
We'll then factor out sec($\theta$) and get
f'($\theta$)=sec($\theta$)($tan^{2}$($\theta$)+$sec^{2}$($\theta$))
