## Calculus (3rd Edition)

$f'(g(x)) = 2cos(2x+1)$
$f(u) = sin(u), g(x) = 2x+1$ $f(g(x)) = sin(2x+1)$ $f'(u) = cos(u)$ and $g'(x) = 2$ $f'(g(x)) = g'(x)f'(u)$ $f'(g(x)) = 2(cos(u))$ We know that $u = g(x) = 2x+1$ so: $f'(g(x)) = 2cos(2x+1)$