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.