Calculus Examples

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

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $- 1$ .

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

Step 1.4

Raise $x$ to the power of $1$ .

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

Step 1.5

Raise $x$ to the power of $1$ .

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

Step 1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.7

Simplify the expression.

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Step 1.7.1

Add $1$ and $1$ .

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

Step 1.7.2

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

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

Step 1.8

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

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

Step 1.9

Multiply $e^{- x^{2}}$ by $1$ .

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

Step 1.10

Simplify.

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Step 1.10.1

Reorder terms.

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

Step 1.10.2

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

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

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 $f (x) = x^{2}$ and $g (x) = e^{- x^{2}}$ .

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

Step 2.2.3

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

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x^{2}$ .

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

Step 2.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $- x^{2}$ .

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

Step 2.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Multiply $2$ by $- 1$ .

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

Step 2.2.8

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.2.8.1

Move $x$ .

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

Step 2.2.8.2

Multiply $x$ by $x^{2}$ .

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Step 2.2.8.2.1

Raise $x$ to the power of $1$ .

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

Step 2.2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.2.8.3

Add $1$ and $2$ .

$f'' (x) = - 2 (x^{3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (e^{- x^{2}})$

Step 2.2.9

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

$f'' (x) = - 2 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (e^{- x^{2}})$

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

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

$f'' (x) = - 2 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + \frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x^{2})$

Step 2.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - 2 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + e^{u_{2}} \frac{d}{d x} (- x^{2})$

Step 2.3.1.3

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

$f'' (x) = - 2 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + e^{- x^{2}} \frac{d}{d x} (- x^{2})$

Step 2.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 2 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + e^{- x^{2}} (- \frac{d}{d x} x^{2})$

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $- 1$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $- 2$ by $- 2$ .

$f'' (x) = 4 (x^{3} (e^{- x^{2}})) - 2 (e^{- x^{2}} (2 x)) + e^{- x^{2}} (- 2 x)$

Step 2.4.2.2

Multiply $2$ by $- 2$ .

$f'' (x) = 4 x^{3} e^{- x^{2}} - 4 (e^{- x^{2}} (x)) + e^{- x^{2}} (- 2 x)$

Step 2.4.2.3

Move $e^{- x^{2}}$ .

$f'' (x) = 4 x^{3} e^{- x^{2}} - 4 x e^{- x^{2}} - 2 x e^{- x^{2}}$

Step 2.4.2.4

Subtract $2 x e^{- x^{2}}$ from $- 4 x e^{- x^{2}}$ .

$f'' (x) = 4 x^{3} e^{- x^{2}} - 6 x e^{- x^{2}}$

Step 2.4.3

Reorder terms.

$f'' (x) = 4 e^{- x^{2}} x^{3} - 6 e^{- x^{2}} x$

Step 2.4.4

Reorder factors in $4 e^{- x^{2}} x^{3} - 6 e^{- x^{2}} x$ .

$f'' (x) = 4 x^{3} e^{- x^{2}} - 6 x e^{- x^{2}}$

Step 3

Find the third derivative.

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Step 3.1

By the Sum Rule, the derivative of $4 x^{3} e^{- x^{2}} - 6 x e^{- x^{2}}$ with respect to $x$ is $\frac{d}{d x} [4 x^{3} e^{- x^{2}}] + \frac{d}{d x} [- 6 x e^{- x^{2}}]$ .

$f''' (x) = \frac{d}{d x} (4 x^{3} e^{- x^{2}}) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2

Evaluate $\frac{d}{d x} [4 x^{3} e^{- x^{2}}]$ .

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Step 3.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3} e^{- x^{2}}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3} e^{- x^{2}}]$ .

$f''' (x) = 4 \frac{d}{d x} (x^{3} e^{- x^{2}}) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.2

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

$f''' (x) = 4 (x^{3} \frac{d}{d x} (e^{- x^{2}}) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.3

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

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Step 3.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x^{2}$ .

$f''' (x) = 4 (x^{3} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f''' (x) = 4 (x^{3} (e^{u_{1}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.3.3

Replace all occurrences of $u_{1}$ with $- x^{2}$ .

$f''' (x) = 4 (x^{3} (e^{- x^{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f''' (x) = 4 (x^{3} (e^{- x^{2}} (- \frac{d}{d x} x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.5

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

$f''' (x) = 4 (x^{3} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \frac{d}{d x} (x^{3})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f''' (x) = 4 (x^{3} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.7

Multiply $2$ by $- 1$ .

$f''' (x) = 4 (x^{3} (e^{- x^{2}} (- 2 x)) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.8

Multiply $x^{3}$ by $x$ by adding the exponents.

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Step 3.2.8.1

Move $x$ .

$f''' (x) = 4 (x \cdot x^{3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.8.2

Multiply $x$ by $x^{3}$ .

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Step 3.2.8.2.1

Raise $x$ to the power of $1$ .

$f''' (x) = 4 (x \cdot x^{3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f''' (x) = 4 (x^{1 + 3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.8.3

Add $1$ and $3$ .

$f''' (x) = 4 (x^{4} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.2.9

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) + \frac{d}{d x} (- 6 x e^{- x^{2}})$

Step 3.3

Evaluate $\frac{d}{d x} [- 6 x e^{- x^{2}}]$ .

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Step 3.3.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x e^{- x^{2}}$ with respect to $x$ is $- 6 \frac{d}{d x} [x e^{- x^{2}}]$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 \frac{d}{d x} (x e^{- x^{2}})$

Step 3.3.2

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x \frac{d}{d x} (e^{- x^{2}}) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.3

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

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Step 3.3.3.1

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{u_{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.3.3

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{- x^{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{- x^{2}} (- \frac{d}{d x} x^{2})) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.5

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \frac{d}{d x} (x))$

Step 3.3.6

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \cdot 1)$

Step 3.3.7

Multiply $2$ by $- 1$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x (e^{- x^{2}} (- 2 x)) + e^{- x^{2}} \cdot 1)$

Step 3.3.8

Raise $x$ to the power of $1$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x \cdot x (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} \cdot 1)$

Step 3.3.9

Raise $x$ to the power of $1$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x \cdot x (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} \cdot 1)$

Step 3.3.10

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x^{1 + 1} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} \cdot 1)$

Step 3.3.11

Add $1$ and $1$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} \cdot 1)$

Step 3.3.12

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

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (- 2 \cdot e^{- x^{2}}) + e^{- x^{2}} \cdot 1)$

Step 3.3.13

Multiply $e^{- x^{2}}$ by $1$ .

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}}) + e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (- 2 e^{- x^{2}}) + e^{- x^{2}})$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}})) + 4 (e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (- 2 e^{- x^{2}}) + e^{- x^{2}})$

Step 3.4.2

Apply the distributive property.

$f''' (x) = 4 (x^{4} (- 2 e^{- x^{2}})) + 4 (e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (- 2 e^{- x^{2}})) - 6 e^{- x^{2}}$

Step 3.4.3

Combine terms.

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Step 3.4.3.1

Multiply $- 2$ by $4$ .

$f''' (x) = - 8 (x^{4} (e^{- x^{2}})) + 4 (e^{- x^{2}} (3 x^{2})) - 6 (x^{2} (- 2 e^{- x^{2}})) - 6 e^{- x^{2}}$

Step 3.4.3.2

Multiply $3$ by $4$ .

$f''' (x) = - 8 x^{4} e^{- x^{2}} + 12 (e^{- x^{2}} (x^{2})) - 6 (x^{2} (- 2 e^{- x^{2}})) - 6 e^{- x^{2}}$

Step 3.4.3.3

Multiply $- 2$ by $- 6$ .

$f''' (x) = - 8 x^{4} e^{- x^{2}} + 12 e^{- x^{2}} x^{2} + 12 (x^{2} (e^{- x^{2}})) - 6 e^{- x^{2}}$

Step 3.4.3.4

Add $12 e^{- x^{2}} x^{2}$ and $12 x^{2} e^{- x^{2}}$ .

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Step 3.4.3.4.1

Move $e^{- x^{2}}$ .

$f''' (x) = - 8 x^{4} e^{- x^{2}} + 12 x^{2} e^{- x^{2}} + 12 x^{2} e^{- x^{2}} - 6 e^{- x^{2}}$

Step 3.4.3.4.2

Add $12 x^{2} e^{- x^{2}}$ and $12 x^{2} e^{- x^{2}}$ .

$f''' (x) = - 8 x^{4} e^{- x^{2}} + 24 x^{2} e^{- x^{2}} - 6 e^{- x^{2}}$

Step 3.4.4

Reorder terms.

$f''' (x) = - 8 e^{- x^{2}} x^{4} + 24 e^{- x^{2}} x^{2} - 6 e^{- x^{2}}$

Step 3.4.5

Reorder factors in $- 8 e^{- x^{2}} x^{4} + 24 e^{- x^{2}} x^{2} - 6 e^{- x^{2}}$ .

$f''' (x) = - 8 x^{4} e^{- x^{2}} + 24 x^{2} e^{- x^{2}} - 6 e^{- x^{2}}$

Step 4

Find the fourth derivative.

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Step 4.1

By the Sum Rule, the derivative of $- 8 x^{4} e^{- x^{2}} + 24 x^{2} e^{- x^{2}} - 6 e^{- x^{2}}$ with respect to $x$ is $\frac{d}{d x} [- 8 x^{4} e^{- x^{2}}] + \frac{d}{d x} [24 x^{2} e^{- x^{2}}] + \frac{d}{d x} [- 6 e^{- x^{2}}]$ .

$f^{4} (x) = \frac{d}{d x} (- 8 x^{4} e^{- x^{2}}) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2

Evaluate $\frac{d}{d x} [- 8 x^{4} e^{- x^{2}}]$ .

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Step 4.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x^{4} e^{- x^{2}}$ with respect to $x$ is $- 8 \frac{d}{d x} [x^{4} e^{- x^{2}}]$ .

$f^{4} (x) = - 8 \frac{d}{d x} (x^{4} e^{- x^{2}}) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.2

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

$f^{4} (x) = - 8 (x^{4} \frac{d}{d x} (e^{- x^{2}}) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.3

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

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Step 4.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x^{2}$ .

$f^{4} (x) = - 8 (x^{4} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 8 (x^{4} (e^{u_{1}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.3.3

Replace all occurrences of $u_{1}$ with $- x^{2}$ .

$f^{4} (x) = - 8 (x^{4} (e^{- x^{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f^{4} (x) = - 8 (x^{4} (e^{- x^{2}} (- \frac{d}{d x} x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.5

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

$f^{4} (x) = - 8 (x^{4} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f^{4} (x) = - 8 (x^{4} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.7

Multiply $2$ by $- 1$ .

$f^{4} (x) = - 8 (x^{4} (e^{- x^{2}} (- 2 x)) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.8

Multiply $x^{4}$ by $x$ by adding the exponents.

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Step 4.2.8.1

Move $x$ .

$f^{4} (x) = - 8 (x \cdot x^{4} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.8.2

Multiply $x$ by $x^{4}$ .

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Step 4.2.8.2.1

Raise $x$ to the power of $1$ .

$f^{4} (x) = - 8 (x \cdot x^{4} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{4} (x) = - 8 (x^{1 + 4} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.8.3

Add $1$ and $4$ .

$f^{4} (x) = - 8 (x^{5} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.2.9

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + \frac{d}{d x} (24 x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3

Evaluate $\frac{d}{d x} [24 x^{2} e^{- x^{2}}]$ .

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Step 4.3.1

Since $24$ is constant with respect to $x$ , the derivative of $24 x^{2} e^{- x^{2}}$ with respect to $x$ is $24 \frac{d}{d x} [x^{2} e^{- x^{2}}]$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 \frac{d}{d x} (x^{2} e^{- x^{2}}) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.2

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 $f (x) = x^{2}$ and $g (x) = e^{- x^{2}}$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} \frac{d}{d x} (e^{- x^{2}}) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.3

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

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Step 4.3.3.1

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{u_{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.3.3

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{- x^{2}} \frac{d}{d x} (- x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{- x^{2}} (- \frac{d}{d x} x^{2})) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.5

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.6

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.7

Multiply $2$ by $- 1$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{2} (e^{- x^{2}} (- 2 x)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.8

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 4.3.8.1

Move $x$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x \cdot x^{2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.8.2

Multiply $x$ by $x^{2}$ .

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Step 4.3.8.2.1

Raise $x$ to the power of $1$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x \cdot x^{2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{1 + 2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.8.3

Add $1$ and $2$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.3.9

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + \frac{d}{d x} (- 6 e^{- x^{2}})$

Step 4.4

Evaluate $\frac{d}{d x} [- 6 e^{- x^{2}}]$ .

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Step 4.4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 e^{- x^{2}}$ with respect to $x$ is $- 6 \frac{d}{d x} [e^{- x^{2}}]$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 \frac{d}{d x} e^{- x^{2}}$

Step 4.4.2

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

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Step 4.4.2.1

To apply the Chain Rule, set $u_{3}$ as $- x^{2}$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (\frac{d}{d u (3)} (e^{u_{3}}) \frac{d}{d x} (- x^{2}))$

Step 4.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (e^{u_{3}} \frac{d}{d x} (- x^{2}))$

Step 4.4.2.3

Replace all occurrences of $u_{3}$ with $- x^{2}$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (e^{- x^{2}} \frac{d}{d x} (- x^{2}))$

Step 4.4.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (e^{- x^{2}} (- \frac{d}{d x} x^{2}))$

Step 4.4.4

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

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (e^{- x^{2}} (- (2 x)))$

Step 4.4.5

Multiply $2$ by $- 1$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 6 (e^{- x^{2}} (- 2 x))$

Step 4.4.6

Multiply $- 2$ by $- 6$ .

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}}) + e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5

Simplify.

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Step 4.5.1

Apply the distributive property.

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}})) - 8 (e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5.2

Apply the distributive property.

$f^{4} (x) = - 8 (x^{5} (- 2 e^{- x^{2}})) - 8 (e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}})) + 24 (e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5.3

Combine terms.

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Step 4.5.3.1

Multiply $- 2$ by $- 8$ .

$f^{4} (x) = 16 (x^{5} (e^{- x^{2}})) - 8 (e^{- x^{2}} (4 x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}})) + 24 (e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5.3.2

Multiply $4$ by $- 8$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 32 (e^{- x^{2}} (x^{3})) + 24 (x^{3} (- 2 e^{- x^{2}})) + 24 (e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5.3.3

Multiply $- 2$ by $24$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 32 e^{- x^{2}} x^{3} - 48 (x^{3} (e^{- x^{2}})) + 24 (e^{- x^{2}} (2 x)) + 12 e^{- x^{2}} x$

Step 4.5.3.4

Multiply $2$ by $24$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 32 e^{- x^{2}} x^{3} - 48 x^{3} e^{- x^{2}} + 48 (e^{- x^{2}} (x)) + 12 e^{- x^{2}} x$

Step 4.5.3.5

Subtract $48 x^{3} e^{- x^{2}}$ from $- 32 e^{- x^{2}} x^{3}$ .

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Step 4.5.3.5.1

Move $e^{- x^{2}}$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 32 x^{3} e^{- x^{2}} - 48 x^{3} e^{- x^{2}} + 48 e^{- x^{2}} x + 12 e^{- x^{2}} x$

Step 4.5.3.5.2

Subtract $48 x^{3} e^{- x^{2}}$ from $- 32 x^{3} e^{- x^{2}}$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 80 x^{3} e^{- x^{2}} + 48 e^{- x^{2}} x + 12 e^{- x^{2}} x$

Step 4.5.3.6

Add $48 e^{- x^{2}} x$ and $12 e^{- x^{2}} x$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 80 x^{3} e^{- x^{2}} + 60 e^{- x^{2}} x$

Step 4.5.4

Reorder terms.

$f^{4} (x) = 16 e^{- x^{2}} x^{5} - 80 e^{- x^{2}} x^{3} + 60 e^{- x^{2}} x$

Step 4.5.5

Reorder factors in $16 e^{- x^{2}} x^{5} - 80 e^{- x^{2}} x^{3} + 60 e^{- x^{2}} x$ .

$f^{4} (x) = 16 x^{5} e^{- x^{2}} - 80 x^{3} e^{- x^{2}} + 60 x e^{- x^{2}}$

Step 5

The fourth derivative of $f (x)$ with respect to $x$ is $16 x^{5} e^{- x^{2}} - 80 x^{3} e^{- x^{2}} + 60 x e^{- x^{2}}$ .

$16 x^{5} e^{- x^{2}} - 80 x^{3} e^{- x^{2}} + 60 x e^{- x^{2}}$