Calculus Examples

$f (x) = 2 - 3 e^{- 2 x}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $2 - 3 e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [- 3 e^{- 2 x}]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [- 3 e^{- 2 x}]$

Step 1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 3 e^{- 2 x}]$

Step 2

Evaluate $\frac{d}{d x} [- 3 e^{- 2 x}]$ .

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Step 2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 e^{- 2 x}$ with respect to $x$ is $- 3 \frac{d}{d x} [e^{- 2 x}]$ .

$0 - 3 \frac{d}{d x} [e^{- 2 x}]$

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 $g (x) = - 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

$0 - 3 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 2 x])$

Step 2.2.2

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

$0 - 3 (e^{u} \frac{d}{d x} [- 2 x])$

Step 2.2.3

Replace all occurrences of $u$ with $- 2 x$ .

$0 - 3 (e^{- 2 x} \frac{d}{d x} [- 2 x])$

Step 2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 - 3 (e^{- 2 x} (- 2 \frac{d}{d x} [x]))$

Step 2.4

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

$0 - 3 (e^{- 2 x} (- 2 \cdot 1))$

Step 2.5

Multiply $- 2$ by $1$ .

$0 - 3 (e^{- 2 x} \cdot - 2)$

Step 2.6

Move $- 2$ to the left of $e^{- 2 x}$ .

$0 - 3 (- 2 \cdot e^{- 2 x})$

Step 2.7

Multiply $- 2$ by $- 3$ .

$0 + 6 e^{- 2 x}$

Step 3

Add $0$ and $6 e^{- 2 x}$ .

$6 e^{- 2 x}$