[Differentiate y = ln(x)] This is an example of many situations in maths where you need to solve something that is similar to what you can solve, but not in its current form. A good idea, then, is to see what you can do to get into a form where you can use what you already know. Consider: y = ln(x) e^y = x This is something that you can differentiate: dx/dy = e^y Then, get this back into the form that you want: dx/dy = x dy/dx = 1/x