Find dy/dx y=e^(-18x)

$y = e^{- 18 x}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (e^{- 18 x})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 18 x$ .

To apply the Chain Rule, set $u$ as $- 18 x$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 18 x]$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d x} [- 18 x]$

Replace all occurrences of $u$ with $- 18 x$ .

$e^{- 18 x} \frac{d}{d x} [- 18 x]$

Differentiate.

Since $- 18$ is constant with respect to $x$ , the derivative of $- 18 x$ with respect to $x$ is $- 18 \frac{d}{d x} [x]$ .

$e^{- 18 x} (- 18 \frac{d}{d x} [x])$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$e^{- 18 x} (- 18 \cdot 1)$

Simplify the expression.

Multiply $- 18$ by $1$ .

$e^{- 18 x} \cdot - 18$

Move $- 18$ to the left of $e^{- 18 x}$ .

$- 18 e^{- 18 x}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = - 18 e^{- 18 x}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - 18 e^{- 18 x}$