Chain Rule: The General Logarithm Rule - Problem 1
We’re talking about how to differentiate a special kind of composite function. This kind; natural log of g(x). Now because the derivative of natural log is 1 over x, the derivative quantity of the chain, would be 1 over g(x) times g'(x). So let’s differentiate one of these kinds of functions. H(x) equals natural log of 1 plus e to the x.
I have it written up here; h' is going to be 1 over the inside stuff, 1 over 1 plus e to the x times the derivative, with respect to x of 1 plus e to the x. The derivative of the inside. Of course that’s just going to be zero plus e to the x. So 1 plus e to the x here. And that’s just e to the x over 1 plus e to the x. that’s the derivative of natural log of 1 plus e to the x.