Proof for exponential derivatives

[nobahar]

f(x) = 2^x \left \left
f(kx) = 2^(kx) \left \left
b = 2^k \left \left
b^x = 2^(kx) \left \left
b^x = f(kx)
\frac{d}{dx}(b^x) = \frac{d}{dx}(f(kx)) = \frac{d}{dx}(2^(kx)) (1)
\frac{d}{dx}(f(kx)) = k.f'(kx) (2)
I can't see how step (1) gets to step (2).
Because I thought:
\frac{d}{dx}(f(kx)) = k.\frac{d}{dx}(f(x))

 

[nobahar]

I think it's:
\frac{d}{d(kx)}(f(kx))*\frac{d}{dx}(kx) = k*f'(kx)

 

[Mark44]

d/dx(f(u)) = d/du(f(u))*du/dx

 

[dextercioby]

f(x) = 2^x \left \left
f(kx) = 2^(kx) \left \left
b = 2^k \left \left
b^x = 2^(kx) \left \left
b^x = f(kx)
\frac{d}{dx}(b^x) = \frac{d}{dx}(f(kx)) = \frac{d}{dx}(2^(kx)) (1)
\frac{d}{dx}(f(kx)) = k.f'(kx) (2)
I can't see how step (1) gets to step (2).
Because I thought:
\frac{d}{dx}(f(kx)) = k.\frac{d}{dx}(f(x))

But isn't 2=e^{\ln 2} ? Or am i missing something ?

 

[Mark44]

Yes, 2 = e^{ln 2}

d/dx(f(kx)) = d/dx[(e^{ln 2})^kx] = e^{kxln 2} * k*ln2 = 2^kx *k*ln2. The OP's formatting and organization made it a bit difficult to follow.