In Session 17, we are working on d/dx e^(x ln a). We already know that d/dx e^x = e^x, and Professor Jerison claims that d/dx e^(x ln a) = (ln a)e^(x ln a), but here is where I am confused:

Via the Chain Rule, if we consider that f(x) = e^x and g(x) = x ln a, and that we are really looking for d/dx f(g(x)), shouldn't the solution be f'(g(x))(g'(x)), which would be [e^(x ln a)][1(ln a) + x (1/a)(da/dx)]?

Question

In Session 17, we are working on d/dx e^(x ln a). We already know that d/dx e^x = e^x, and Professor Jerison claims that d/dx e^(x ln a) = (ln a)e^(x ln a), but here is where I am confused:

Via the Chain Rule, if we consider that f(x) = e^x and g(x) = x ln a, and that we are really looking for d/dx f(g(x)), shouldn't the solution be f'(g(x))(g'(x)), which would be [e^(x ln a)][1(ln a) + x (1/a)(da/dx)]?

anonymous · Answer

[f′(g(x)][g′(x)] is correct, but I went about differentiating [g'(x)] incorrectly. Because a is a constant, da/dx = 0, so...

$$[e ^{xlna}][1(\ln a)+x(1/a)(da/dx)] = [e ^{xlna}][1(\ln a)]$$