Can someone help me differentiate this?

f(x)= 2cos(5 ln(x))
So what is f'(x)?

You'll need the chain rule, which says

dy/dx = dy/du . du/dx
Set 5ln(x) as u, then you proceed with
f'(x) = 2cos(u).du/dx
noting that d(ln(x))/dx = 1/x
Can you take it from here?