Question

Question #df9f8

Answer 1

`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.