Calculus Examples

$f (x) = e^{x^{2} - 4}$

Step 1

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 $g (x) = x^{2} - 4$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $x^{2} - 4$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{2} - 4]$

Step 1.2

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

$e^{u} \frac{d}{d x} [x^{2} - 4]$

Step 1.3

Replace all occurrences of $u$ with $x^{2} - 4$ .

$e^{x^{2} - 4} \frac{d}{d x} [x^{2} - 4]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

$e^{x^{2} - 4} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4])$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$e^{x^{2} - 4} (2 x + \frac{d}{d x} [- 4])$

Step 2.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$e^{x^{2} - 4} (2 x + 0)$

Step 2.4

Add $2 x$ and $0$ .

$e^{x^{2} - 4} (2 x)$

Step 3

Simplify.

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Step 3.1

Reorder the factors of $e^{x^{2} - 4} (2 x)$ .

$2 e^{x^{2} - 4} x$

Step 3.2

Reorder factors in $2 e^{x^{2} - 4} x$ .

$2 x e^{x^{2} - 4}$