Question

If f(kx) = 2^(kx), where k is a constant. How can the derivative of that function be equal to k*f'(kx)?

Answer 1

I've been watching lecture 7 of MIT's OCW program of Calculus (introducing the derivative of \(a^x\)) and I'm stuck trying to understand what you can see in the image I've attached below (If someone is interested, it's at 17:12 of the lecture).

\(f(kx) = 2^{kx}\) and \(f(kx) = b^x\), where \(k\) is a constant and \(b= 2^x\).

What I don't understand is that:

\(\frac{d}{dx}f(kx) = kf'(kx)\)

Where is that \(k\) coming from? By the chain rule I thought it should be:

\(\frac{d}{dx}f(kx) = k2^{k-1}f'(kx)\)

I'm puzzled |:

Answer 2

Really? For me, with chain rule, we should have \(kf'(kx)\)

See, let \(g(x)=kx\), we have \(\frac{d}{dx}f(g(x)) = f'(g(x))g'(x) = \boxed{kf'(kx)}\)

Answer 3

Pff. I don't know I was doing.

Answer 4

I get it. Thanks !!! (:

Answer 5

Ok, glad I helped.

Answer 6

(: