# How do you find the derivative of 2^x?

##### 2 Answers
Apr 6, 2018

$\frac{d}{\mathrm{dx}} {2}^{x} = {2}^{x} \ln 2$

#### Explanation:

In general, the derivative of an exponential with some constant base is

$\frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \ln a$.

A proof of this will be shown.

So, $\frac{d}{\mathrm{dx}} {2}^{x} = {2}^{x} \ln 2$

Proof:

Rewrite ${a}^{x}$ as ${e}^{\ln} \left({a}^{x}\right) = {e}^{x \ln a}$.

Now,

$\frac{d}{\mathrm{dx}} {a}^{x} = \frac{d}{\mathrm{dx}} {e}^{x \ln a} = {e}^{x \ln a} \cdot \frac{d}{\mathrm{dx}} \left(x \ln a\right) ,$ as per the Chain Rule.

We then have

$\frac{d}{\mathrm{dx}} {a}^{x} = {e}^{x \ln a} \ln \left(a\right) ,$ and, recalling that ${e}^{x \ln a} = {a}^{x} ,$ we finally have

$\frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \ln a$.

Apr 6, 2018

Let $y = {2}^{x}$

Taking log on both sides,

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

$\implies \log y = x \log 2$

Applying derivative,

$\implies \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \log 2$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {2}^{x} \log 2$