- #1

nobahar

- 497

- 2

[tex]f(kx) = 2^(kx) \left \left[/tex]

[tex]b = 2^k \left \left[/tex]

[tex]b^x = 2^(kx) \left \left[/tex]

[tex]b^x = f(kx)[/tex]

[tex]\frac{d}{dx}(b^x) = \frac{d}{dx}(f(kx)) = \frac{d}{dx}(2^(kx)) (1)[/tex]

[tex]\frac{d}{dx}(f(kx)) = k.f'(kx) (2)[/tex]

I can't see how step (1) gets to step (2).

Because I thought:

[tex]\frac{d}{dx}(f(kx)) = k.\frac{d}{dx}(f(x))[/tex]