# Question #df9f8

Dec 29, 2017

$f ' \left(x\right) = {2}^{x} \ln \left(2\right)$

#### 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:

$\ln y = \ln {2}^{x}$

$\to \ln y = x \ln 2$

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

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(2\right)$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = y \ln \left(2\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {2}^{x} \ln \left(2\right)$

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

$f \left(x\right) = {a}^{x} \to f ' \left(x\right) = \ln \left(a\right) {a}^{x}$

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.