Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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.