How do you differentiate ln(f(x))? Tricks like these occur commonly in STEP questions (including one I was looking at earlier today).

Assumed knowledge: The chain rule, implicit differentiation and being able to differentiate exp(x).Chain rule: d/dx = d/dy * dy/dxImplicit differentiation: f(y) = g(x) => f'(y)*dy/dx = g'(x) (where f'(y) represents df/dy and g'(x) represents dg/dx)d(exp(t))/dt = exp(t)exp(x) is the inverse of ln(x): that is exp(ln(x))=ln(exp(x))=x Note: exp(x) is sometimes written e^x

So we let y = ln(f(x)). (We note that we need f(x)>0 as ln(t) is defined only for t>0)We want this in a form that we are able to differentiate.So we take exponents.=> exp(y) = exp(ln(f(x))) = f(x) (keep note that exp(y) = f(x) ) Now it is in a form where we can differentiate implicitly=> d(exp(y))/dy * dy/dx = f'(x) (using the chain rule)=> exp(y) * dy/dx = f'(x) (using that we know the derivative of exp(t) with respect to some variable t is just exp(t) )So now we can manipulate this to get an expression for dy/dx=> dy/dx = f'(x)/exp(y)But we already have exp(y) in terms of x=> dy/dx = f'(x)/f(x)Hence we can write that the derivative of ln(f(x)) is f'(x)/f(x)Or more formallyd(ln(f(x))/dx = f'(x)/f(x) Which is what we were trying to find!
Quick note: being able to differentiate functions using tricks like this is very useful. An example being question 4 from STEP 3 2017 which asked us to integrate f'(x)/(f(x)*ln(x)) with respect to x. Knowing the above result was key!