Sure! :) So you’re gonna have to use the product rule for this one, which is (f * g)’ = (f ‘ * g) + (f * g’ ). In this case, your f is x, and your g is ln^3(x). So if we take the first part of the product rule, we want to multiply the derivative of f, which is 1, times g, so you have (1)(ln^3(x)) = ln^3(x). For the second part, you want to multiply f by the derivative of g, and to get the derivative of g, you need to use the chain rule. The chain rule is basically you just differentiate the outermost action first, keeping what’s inside as it is, then differentiate the second part, and so forth. So in this case, the first thing we have to do with ln^3(x) is take the derivative of the power, so you’ll get 3ln^2(x). Now you have to take the derivative of what’s inside, which is ln(x), and the derivative of that is 1/x. So putting it all together, g’ is 3ln^2(x)*(1/x). Now you just plug all of this in to the product rule equation, so you have (f ‘ * g) + (f * g’ ) = (1)(ln^3(x))+(x)((3ln^2(x))(1/x)), and the x’s cancel out in the second half, so you have ln^3(x) + 3ln^2(x), and that’s your answer. :)