Calculus Examples

$18 + 18 e^{- x}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $18 + 18 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [18] + \frac{d}{d x} [18 e^{- x}]$ .

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

Step 1.2

Since $18$ is constant with respect to $x$ , the derivative of $18$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [18 e^{- x}]$

Step 2

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

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Step 2.1

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

$0 + 18 \frac{d}{d x} [e^{- x}]$

Step 2.2

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 $g (x) = - x$ .

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Step 2.2.1

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

$0 + 18 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 2.2.2

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

$0 + 18 (e^{u} \frac{d}{d x} [- x])$

Step 2.2.3

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

$0 + 18 (e^{- x} \frac{d}{d x} [- x])$

Step 2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$0 + 18 (e^{- x} (- \frac{d}{d x} [x]))$

Step 2.4

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

$0 + 18 (e^{- x} (- 1 \cdot 1))$

Step 2.5

Multiply $- 1$ by $1$ .

$0 + 18 (e^{- x} \cdot - 1)$

Step 2.6

Move $- 1$ to the left of $e^{- x}$ .

$0 + 18 (- 1 \cdot e^{- x})$

Step 2.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$0 + 18 (- e^{- x})$

Step 2.8

Multiply $- 1$ by $18$ .

$0 - 18 e^{- x}$

Step 3

Subtract $18 e^{- x}$ from $0$ .

$- 18 e^{- x}$