Question

How do you find the derivative of `2^xln2`?

Answer 1

We are going to use the formula `(u*v)'=u'v+uv'`

`u=2^x`
`u'=2^x ln2` (See below)
`v=ln2`
`v'=0` (Note that `v` is a constant.)

Thus `f'(x)=(2^x ln2)(ln2)+(2^x ln2)(0)`

`f'(x)=2^x (ln2)^2`

Note !:

If you have not memorized `d/dx(a^x) = a^x lna`, then you'll need to use:

`2^x = e^(xln2)`, whose derivative can be found by the chain rule:

`d/dx(2^x) = d/dx(e^(xln2)) = e^(xln2) * d/dx(xln2)`

`= e^(xln2) ln2 = 2^x ln2`

Note2:

Because `ln2` is a constant, we don't really need the product rule.

`d/dx(2^x ln2) =( ln2) d/dx(2^x) = (ln2) (2^x ln2) = 2^x (ln2)^2`