Home > Calculus > Differentiating Exponential Functions with Other Bases

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

Answer 1

Emilia Aguilar

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Alvarez

To find the derivative of (2^{x\ln(2)}), you can use the chain rule.

Let (f(x) = 2^{x\ln(2)}).

Then, (f'(x) = \ln(2) \cdot 2^{x\ln(2)}).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.