Question

How do you differentiate `f(x)= 2^(3x)`?

Answer 1

`f^'(x) = 3ln(2)2^(3x)`

Explanation:

First we put this in terms of the base `e` exponential

`f(x) = e^(3ln(2)x)`

We know that `d/dx(e^(kx)) = ke^(kx)`, so for `k = 3ln(2)` we have

`f^'(x) = 3ln(2)2^(3x)`