# How do you differentiate y=e^-x?

May 15, 2018

$\frac{d}{\mathrm{dx}} {e}^{-} x = - {e}^{-} x$

#### Explanation:

We know that $\frac{d}{\mathrm{dx}} {a}^{x} = \ln \left(a\right) \cdot {a}^{x}$

${e}^{-} x = {\left(\frac{1}{e}\right)}^{x}$

Combining the two gives us

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

$\ln \left(\frac{1}{e}\right) = \ln \left({e}^{-} 1\right) = - 1$

${\left(\frac{1}{e}\right)}^{x} = {e}^{-} x$

Multiplying the two gives:

${e}^{-} x \cdot - 1 = - {e}^{-} x$

May 16, 2018

$- {e}^{-} x$

#### Explanation:

Given: $y = {e}^{-} x$.

Differentiate using the chain rule, which states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = - x , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = - 1$.

Then, $y = {e}^{u} , \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u}$.

Combine the results together to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \cdot - 1$

$= - {e}^{u}$

Substitute back $u = - x$ to get the final answer:

color(blue)(bar(ul|=-e^-x|)