How do you find the derivative of 2^xln2?

1 Answer

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