# How do you differentiate e^(-10x)?

Jun 11, 2018

$- 10 {e}^{- 10 x}$

#### Explanation:

Given: ${e}^{- 10 x}$.

Use the chain rule, which states that,

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

Let $u = - 10 x , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = - 10$.

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

Combining, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \cdot - 10$

$= - 10 {e}^{u}$

Substitute back $u = - 10 x$ to get the final answer:

$= - 10 {e}^{- 10 x}$

Note:

A common fact in derivatives is that $\frac{d}{\mathrm{dx}} \left({e}^{f \left(x\right)}\right) = f ' \left(x\right) {e}^{f \left(x\right)}$.

$\frac{d}{\mathrm{dx}} \left({e}^{- 10 x}\right) = - 10 \cdot {e}^{- 10 x}$

#### Explanation:

By the formula $\frac{d \left({e}^{u}\right)}{\mathrm{dx}} = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left({e}^{- 10 x}\right) = {e}^{- 10 x} \cdot \frac{d \left(- 10 x\right)}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left({e}^{- 10 x}\right) = {e}^{- 10 x} \cdot \left(- 10\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{- 10 x}\right) = - 10 \cdot {e}^{- 10 x}$

I hope the explanation is useful....God bless...