## Calculus 8th Edition

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