## Calculus, 10th Edition (Anton)

$f'(x) = \dfrac{-5\sin(5x)}{2\sqrt{\cos(5x)}}$
First, lets make an «u» substitution in order to make it easier. $f(u) = \sqrt{u}$ $u = \cos(5x)$ Then lets derivate using the chain rule $f'(u) = \dfrac{1}{2\sqrt{u}} \times u'$ Now let's find u' $u' = -5\sin(5x)$ Now let's undo the substitution and simplify $f'(x) = \dfrac{1}{2\sqrt{\cos(5x)}} \times -5\sin(5x)$ And you got the answer: $f'(x) = \dfrac{-5\sin(5x)}{2\sqrt{\cos(5x)}}$