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#