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

##### 3 Answers
Mar 26, 2017

$2 {e}^{2 x}$

#### Explanation:

Use the exponent rule ${\left({x}^{m}\right)}^{n} = {x}^{m \cdot n}$

$y = {\left({e}^{x}\right)}^{2} = {e}^{2 x}$

Use $\left({e}^{u}\right) ' = {e}^{u} \cdot u ' = u ' \cdot {e}^{u}$

Let $u = 2 x , \text{ } u ' = 2$

$y ' = 2 {e}^{2 x}$

Mar 26, 2017

$2 {e}^{2 x}$

#### Explanation:

$\textcolor{b l u e}{\text{Note that }} {\left({e}^{x}\right)}^{2} = {e}^{2 x}$

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

• d/dx(e^(f(x)))=e^(f(x))xxf'(x)

$\Rightarrow \frac{d}{\mathrm{dx}} \left({e}^{2 x}\right) = {e}^{2 x} \times \frac{d}{\mathrm{dx}} \left(2 x\right)$

$\textcolor{w h i t e}{\times \times \times \times} = 2 {e}^{2 x}$

Mar 26, 2017

For the heck of it, we can also treat ${\left({e}^{x}\right)}^{2}$ as a function squared.

Generally, we see that the derivative of ${\left(f \left(x\right)\right)}^{2}$ is $2 {\left(f \left(x\right)\right)}^{1} \cdot f ' \left(x\right)$.

The previous relation comes from the chain rule. The derivative of ${x}^{2}$ is $2 x$, which comes through the power rule. The chain rule says to follow that same logic, but also to multiply by the derivative of the inner function.

Here, we see that the derivative of ${\left({e}^{x}\right)}^{2}$ will be $2 {\left({e}^{x}\right)}^{1}$ then multiplied by the derivative of ${e}^{x}$.

The derivative of ${e}^{x}$ is ${e}^{x}$, so we see that:

$\frac{d}{\mathrm{dx}} {\left({e}^{x}\right)}^{2} = 2 {\left({e}^{x}\right)}^{1} \left(\frac{d}{\mathrm{dx}} {e}^{x}\right) = 2 {e}^{x} \left({e}^{x}\right)$

Which we can simplify using the rules of exponents:

d/dx(e^x)^2=2e^(x+x)=color(blue)(2e^(2x)