Calculus Examples

$y = x e^{x}$

Step 1

Find the first derivative.

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x)$

Step 1.2

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

$f' (x) = x e^{x} + e^{x} \frac{d}{d x} (x)$

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

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

$f' (x) = x e^{x} + e^{x} \cdot 1$

Step 1.3.2

Multiply $e^{x}$ by $1$ .

$f' (x) = x e^{x} + e^{x}$

Step 2

Find the second derivative.

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Step 2.1

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

$f'' (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (e^{x})$

Step 2.2

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

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Step 2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f'' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (e^{x})$

Step 2.2.2

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

$f'' (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (e^{x})$

Step 2.2.3

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

$f'' (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (e^{x})$

Step 2.2.4

Multiply $e^{x}$ by $1$ .

$f'' (x) = x e^{x} + e^{x} + \frac{d}{d x} (e^{x})$

Step 2.3

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

$f'' (x) = x e^{x} + e^{x} + e^{x}$

Step 2.4

Simplify.

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Step 2.4.1

Add $e^{x}$ and $e^{x}$ .

$f'' (x) = x e^{x} + 2 e^{x}$

Step 2.4.2

Reorder terms.

$f'' (x) = e^{x} x + 2 e^{x}$

Step 2.4.3

Reorder factors in $e^{x} x + 2 e^{x}$ .

$f'' (x) = x e^{x} + 2 e^{x}$

Step 3

Find the third derivative.

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Step 3.1

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

$f''' (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (2 e^{x})$

Step 3.2

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

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Step 3.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f''' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (2 e^{x})$

Step 3.2.2

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

$f''' (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (2 e^{x})$

Step 3.2.3

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

$f''' (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (2 e^{x})$

Step 3.2.4

Multiply $e^{x}$ by $1$ .

$f''' (x) = x e^{x} + e^{x} + \frac{d}{d x} (2 e^{x})$

Step 3.3

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

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Step 3.3.1

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

$f''' (x) = x e^{x} + e^{x} + 2 \frac{d}{d x} (e^{x})$

Step 3.3.2

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

$f''' (x) = x e^{x} + e^{x} + 2 e^{x}$

Step 3.4

Simplify.

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Step 3.4.1

Add $e^{x}$ and $2 e^{x}$ .

$f''' (x) = x e^{x} + 3 e^{x}$

Step 3.4.2

Reorder terms.

$f''' (x) = e^{x} x + 3 e^{x}$

Step 3.4.3

Reorder factors in $e^{x} x + 3 e^{x}$ .

$f''' (x) = x e^{x} + 3 e^{x}$

Step 4

Find the fourth derivative.

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Step 4.1

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

$f^{4} (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (3 e^{x})$

Step 4.2

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

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Step 4.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f^{4} (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (3 e^{x})$

Step 4.2.2

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

$f^{4} (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (3 e^{x})$

Step 4.2.3

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

$f^{4} (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (3 e^{x})$

Step 4.2.4

Multiply $e^{x}$ by $1$ .

$f^{4} (x) = x e^{x} + e^{x} + \frac{d}{d x} (3 e^{x})$

Step 4.3

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

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Step 4.3.1

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

$f^{4} (x) = x e^{x} + e^{x} + 3 \frac{d}{d x} (e^{x})$

Step 4.3.2

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

$f^{4} (x) = x e^{x} + e^{x} + 3 e^{x}$

Step 4.4

Simplify.

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Step 4.4.1

Add $e^{x}$ and $3 e^{x}$ .

$f^{4} (x) = x e^{x} + 4 e^{x}$

Step 4.4.2

Reorder terms.

$f^{4} (x) = e^{x} x + 4 e^{x}$

Step 4.4.3

Reorder factors in $e^{x} x + 4 e^{x}$ .

$f^{4} (x) = x e^{x} + 4 e^{x}$