#### Answer

$f'(u) = \pi sec^2(\pi x)$

#### Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$.
Original expression: $y = tan(\pi x)$
$u = g(x) = \pi x$
$y = f(u) = tan(u)$
Apply the chain rule to find the derivative:
$f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$ => $\frac{dy}{du}tan(u) \times\frac{du}{dx}(\pi x) $
Derive (Note: $\pi $ is a constant and is moved to the outside of the derivative of $u$)
$sec^2(u)\times \pi$
$\pi sec^2(\pi x)$