## josephT 3 years ago This question refers to lecture 6 of the Single Variable Calculus course, where Dr. Jerison is showing how to calculate d/dx a^x. (18-01sc-single-variable-calculus-fall-2010 Session 17: The Exponential Function, its Derivative, and its Inverse) Isn't there an error at 18:12 of the lecture? The equation d/dx f(kx) = k f'(kx) mixes Newton's and Leibnitz's notation, but, using Newton's, the equation is f'(kx) = k f'(kx). It seems this could only be correct for k=1 (or f'(kx) = 0). But k is "any number", not just 1; and f'(kx) = 2^(kx) ln(kx), so f'(kx) could only be 0 if kx = 1. Well, that rules out k being "any number - it has to be the reciprocal of x - and it couldn't be 0. Any help explaining this apparent difficulty with the proof would be appreciated. Thanks.

1. nmonson

d/dx f(kx) is not equivalent to f'(kx), but rather (f(kx))'. In this case f(kx) is a composition of functions where f( ) is the outer function and kx is the inner function. Apply the chain rule. Multiply the derivative of the outer function - f'(kx) - by the derivative of the inner function - k. So d/dx f(kx) = kf'(kx). Or (f(kx))' = kf'(kx).

2. josephT

I think I understand what you mean: When Dr Jerison writes f'(kx), that means the deriv with respect to kx of f(kx). [Would that be d/d(kx) f(kx) in Leibnitz notation?] And if u = kx, then, by the chain rule, d/dx f(u) = d/du f(u) * du/dx d/dx f(u) = d/du f(u) * d/dx kx d/dx f(u) = d/du f(u) * k which, replacing the rest of the u's with kx's gives: d/dx f(kx) = d/d(kx) f(kx) * k or as Dr Jerison correctly wrote: d/dx f(kx) = k * f'(kx) Please tell me if I got anything wrong here. And thanks so much for your taking the time to answer.

3. nmonson

Looks good to me. Don't know enough about notation to say if it's conventional to write d/d(kx). Conceptually though, you seem to have it as far as I can tell.