# How do you find the derivative of  2e^(x+3)?

Jan 17, 2017

$f \left(x\right) = 2 {e}^{x + 3} \implies f ' \left(x\right) = 2 {e}^{x + 3}$

#### Explanation:

First remember a derivative times a constant equals a constant times a derivative

$\left(c f \left(x\right)\right) ' = c f ' \left(x\right)$ in this case $c = 2$

So

$\left(2 {e}^{x + 3}\right) ' = 2 \left({e}^{x + 3}\right) '$

Then we use the chain rule (f(g(x))'=f'(g(x)g'(x)

$2 \left({e}^{x + 3}\right) ' = 2 \left({e}^{x + 3}\right) ' \left(x + 3\right) '$

Since $\left({e}^{x}\right) ' = {e}^{x}$

$2 \left({e}^{x + 3}\right) ' \left(x + 3\right) ' = 2 \left({e}^{x + 3}\right) \left(x + 3\right) '$

Since $\left(f \left(x\right) + g \left(x\right)\right) ' = f ' \left(x\right) + g ' \left(x\right)$

$2 \left({e}^{x + 3}\right) \left(x + 3\right) ' = 2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + \left(3\right) '\right)$

Since 3 is a constant its derivative is zero

$2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + \left(3\right) '\right) = 2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + 0\right) = 2 \left({e}^{x + 3}\right) \left(x\right) '$

and since (f(x)=x => f'(x)=1)

$2 \left({e}^{x + 3}\right) \left(x\right) ' = 2 \left({e}^{x + 3}\right) \left(1\right) = \underline{2 \left({e}^{x + 3}\right)}$