f(x)=(e^(squar root of x))*ln( (squar root of x ) +5 )
So here we are going to need a combination of the product rule and the chain rule. i.e the derivative of the first times the 2nd plut the 1st times the derivative of the 2nd.
$\displaystyle f'(x)=\left( \frac{1}{2\sqrt{x}}\right) e^{\sqrt{x}}\ln(\sqrt{x}+5)+e^{\sqrt{x}}\left( \frac{1}{\sqrt{x}+5}\right) \left(\frac{1}{2\sqrt{x}}\right) $
It just needs to be simplified from here