Question #df9f8

1 Answer
Andrew I.
Dec 29, 2017

#f'(x) = 2^xln(2)#

Explanation:

Let's say we have #y=2^x#

The trick with these kinds of functions is to take the natural log of both sides and use implicit differentiation like so:

#lny = ln 2^x#

#-> lny = xln2#

Now using implicit differentiation (remember we are differentiating with respect to #x# so the derivative of #y# will be #dy/dy# and if #y# is inside a function, as it is here, then we must use the chain rule accordingly):

#1/y dy/dx=ln(2)#

#-> dy/dx = yln(2)#

Now substitute in the original expression for #y# from the start and we get:

#dy/dx = 2^xln(2)#

In short the derivative of an exponential function where the base is not #e# is given as:

#f(x) = a^x -> f'(x)=ln(a)a^x#

so just multiply by the natural log of the base to get the derivative. When the base is #e# then it is not hard to see that it will cancel with the natural log leaving you with what you started.

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