What is the derivative of #2^x#?

1 Answer
Jul 28, 2015

#d/dx (2^x) = 2^x * ln2#

Explanation:

In order to be able to calculate the derivative of #2^x#, you're going to need to use two things

The idea here is that you can use the fact that you know what the derivative of #e^x# is to try and determine what the derivative of another constant raised to the power of #x#, in this case equal to #2#, is.

To do that, you need to write #2# as an exponential number that has the base equal to #e#.

Use the fact that

#color(blue)(e^(ln(a)) = a)#

to write

#e^(ln2) = 2#

This implies that #2^x# will be equivalent to

#2^x = (e^(ln2))^x = e^(x * ln2)#

Your derivative now looks like this

#d/dx(e^(x * ln2))#

This is where the chain rule comes into play. You know that the derivative of a function #y = f(u)# can be written as

#dy/dx = dy/(du) * (du)/dx#

In your case, #y = e^(x * ln2)#, and #u = x * ln2#, so that your derivative becomes

#d/dx(e^u) = underbrace(e^u/(du))_(color(blue)(=e^u)) * d/dx(u)#

#d/dx(e^u) = e^u * d/dx(u)#

Now replace #u# to calculate #d/dx(u)#

#d/dx(e^(x * ln2)) = e^(x * ln2) * d/dx(x * ln2)#

#d/dx(e^(x * ln2)) = e^(x * ln2) * ln2 d/dx(x)#

#d/dx(e^(x * ln2)) = e^(x * ln2) * ln2#

Therefore,

#d/dx(2^x) = color(green)(2^x * ln 2)#