Answer
$f'(x) = \dfrac{x+1}{\sqrt{x^2+2x+9}}(\cos {\sqrt{x^2+2x+9}})$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$u = \sqrt{x^2+2x+9}$
$f(u) = \sin u$
Derivate the function:
$f'(u) = \cos u$
Now let's find u'
$u' = \dfrac{x+1}{\sqrt{x^2+2x+1}}$
Then undo the substitution, simplify and get the answer:
$f'(x) = \dfrac{x+1}{\sqrt{x^2+2x+9}}(\cos {\sqrt{x^2+2x+9}})$
