Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.4

Differentiate.

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

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

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

Step 1.4.4

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

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

Step 1.4.5

Multiply $- 4$ by $1$ .

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

Step 1.4.6

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

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

Step 1.4.7

Multiply $e^{x^{2} - 4 x}$ by $1$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

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

Step 1.5.2

Reorder terms.

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

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) \frac{d}{d x} (e^{x^{2} - 4 x} x)) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.7

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \frac{d}{d x} (x) + x \frac{d}{d x} (e^{x^{2} - 4 x}))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.8

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x \frac{d}{d x} (e^{x^{2} - 4 x}))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.9

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

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Step 2.2.9.1

To apply the Chain Rule, set $u_{1}$ as $x^{2} - 4 x$ .

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (x^{2} - 4 x)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.9.2

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{u_{1}} \frac{d}{d x} (x^{2} - 4 x)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.9.3

Replace all occurrences of $u_{1}$ with $x^{2} - 4 x$ .

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} \frac{d}{d x} (x^{2} - 4 x)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.10

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 4 x))))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.11

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} (2 x + \frac{d}{d x} (- 4 x))))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.12

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} (2 x - 4 \frac{d}{d x} x)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.13

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

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 \cdot 1 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} (2 x - 4 \cdot 1)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.14

Multiply $2$ by $1$ .

$f'' (x) = 2 (e^{x^{2} - 4 x} x (2 + 0) + (2 x - 4) (e^{x^{2} - 4 x} \cdot 1 + x (e^{x^{2} - 4 x} (2 x - 4 \cdot 1)))) + \frac{d}{d x} (2 e^{x^{2} - 4 x})$

Step 2.2.15

Add $2$ and $0$ .

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

Step 2.2.16

Move $2$ to the left of $e^{x^{2} - 4 x} x$ .

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

Step 2.2.17

Multiply $e^{x^{2} - 4 x}$ by $1$ .

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

Step 2.2.18

Multiply $- 4$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

Multiply $- 4$ by $1$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Multiply $2$ by $2$ .

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

Step 2.4.3

Simplify each term.

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Step 2.4.3.1

Apply the distributive property.

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

Step 2.4.3.2

Multiply $2$ by $2$ .

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

Step 2.4.3.3

Multiply $2$ by $- 4$ .

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

Step 2.4.3.4

Simplify each term.

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Step 2.4.3.4.1

Apply the distributive property.

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

Step 2.4.3.4.2

Rewrite using the commutative property of multiplication.

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

Step 2.4.3.4.3

Move $- 4$ to the left of $e^{x^{2} - 4 x}$ .

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

Step 2.4.3.4.4

Apply the distributive property.

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

Step 2.4.3.4.5

Rewrite using the commutative property of multiplication.

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

Step 2.4.3.4.6

Rewrite using the commutative property of multiplication.

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

Step 2.4.3.4.7

Multiply $x$ by $x$ by adding the exponents.

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Step 2.4.3.4.7.1

Move $x$ .

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

Step 2.4.3.4.7.2

Multiply $x$ by $x$ .

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

Step 2.4.3.5

Expand $(4 x - 8) (e^{x^{2} - 4 x} + 2 x^{2} e^{x^{2} - 4 x} - 4 x e^{x^{2} - 4 x})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 2.4.3.6

Simplify each term.

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Step 2.4.3.6.1

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

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Step 2.4.3.6.1.1

Move $x^{2}$ .

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

Step 2.4.3.6.1.2

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

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Step 2.4.3.6.1.2.1

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

Step 2.4.3.6.1.2.2

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

$f'' (x) = 4 e^{x^{2} - 4 x} x + 4 x e^{x^{2} - 4 x} + 4 x^{2 + 1} (2 e^{x^{2} - 4 x}) + 4 x (- 4 x e^{x^{2} - 4 x}) - 8 e^{x^{2} - 4 x} - 8 (2 x^{2} e^{x^{2} - 4 x}) - 8 (- 4 x e^{x^{2} - 4 x}) + 2 e^{x^{2} - 4 x} (2 x - 4)$

Step 2.4.3.6.1.3

Add $2$ and $1$ .

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

Step 2.4.3.6.2

Rewrite using the commutative property of multiplication.

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

Step 2.4.3.6.3

Multiply $4$ by $2$ .

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

Step 2.4.3.6.4

Multiply $x$ by $x$ by adding the exponents.

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Step 2.4.3.6.4.1

Move $x$ .

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

Step 2.4.3.6.4.2

Multiply $x$ by $x$ .

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

Step 2.4.3.6.5

Rewrite using the commutative property of multiplication.

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

Step 2.4.3.6.6

Multiply $4$ by $- 4$ .

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

Step 2.4.3.6.7

Multiply $2$ by $- 8$ .

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

Step 2.4.3.6.8

Multiply $- 4$ by $- 8$ .

$f'' (x) = 4 e^{x^{2} - 4 x} x + 4 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 16 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} - 16 x^{2} e^{x^{2} - 4 x} + 32 x e^{x^{2} - 4 x} + 2 e^{x^{2} - 4 x} (2 x - 4)$

Step 2.4.3.7

Add $4 x e^{x^{2} - 4 x}$ and $32 x e^{x^{2} - 4 x}$ .

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 16 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} - 16 x^{2} e^{x^{2} - 4 x} + 2 e^{x^{2} - 4 x} (2 x - 4)$

Step 2.4.3.8

Subtract $16 x^{2} e^{x^{2} - 4 x}$ from $- 16 x^{2} e^{x^{2} - 4 x}$ .

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 2 e^{x^{2} - 4 x} (2 x - 4)$

Step 2.4.3.9

Apply the distributive property.

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 2 e^{x^{2} - 4 x} (2 x) + 2 e^{x^{2} - 4 x} \cdot - 4$

Step 2.4.3.10

Rewrite using the commutative property of multiplication.

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 2 \cdot (2 e^{x^{2} - 4 x} x) + 2 e^{x^{2} - 4 x} \cdot - 4$

Step 2.4.3.11

Multiply $- 4$ by $2$ .

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 2 \cdot (2 e^{x^{2} - 4 x} x) - 8 e^{x^{2} - 4 x}$

Step 2.4.3.12

Multiply $2$ by $2$ .

$f'' (x) = 4 e^{x^{2} - 4 x} x + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 4 e^{x^{2} - 4 x} x - 8 e^{x^{2} - 4 x}$

Step 2.4.4

Add $4 e^{x^{2} - 4 x} x$ and $36 x e^{x^{2} - 4 x}$ .

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Step 2.4.4.1

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

$f'' (x) = 4 x e^{x^{2} - 4 x} + 36 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 4 e^{x^{2} - 4 x} x - 8 e^{x^{2} - 4 x}$

Step 2.4.4.2

Add $4 x e^{x^{2} - 4 x}$ and $36 x e^{x^{2} - 4 x}$ .

$f'' (x) = 40 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} + 4 e^{x^{2} - 4 x} x - 8 e^{x^{2} - 4 x}$

Step 2.4.5

Add $40 x e^{x^{2} - 4 x}$ and $4 e^{x^{2} - 4 x} x$ .

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Step 2.4.5.1

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

$f'' (x) = 40 x e^{x^{2} - 4 x} + 4 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x}$

Step 2.4.5.2

Add $40 x e^{x^{2} - 4 x}$ and $4 x e^{x^{2} - 4 x}$ .

$f'' (x) = 44 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x} - 8 e^{x^{2} - 4 x}$

Step 2.4.6

Subtract $8 e^{x^{2} - 4 x}$ from $- 8 e^{x^{2} - 4 x}$ .

$f'' (x) = 44 x e^{x^{2} - 4 x} + 8 x^{3} e^{x^{2} - 4 x} - 32 x^{2} e^{x^{2} - 4 x} - 16 e^{x^{2} - 4 x}$