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)#