Calculus Examples

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

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

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

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 $g (x) = 2 x$ .

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

$x^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [x^{2}]$

Step 1.3.3

Simplify the expression.

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Step 1.3.3.1

Multiply $2$ by $1$ .

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

Step 1.3.3.2

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

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

Step 1.3.4

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

$x^{2} (2 e^{2 x}) + e^{2 x} (2 x)$

Step 1.4

Simplify.

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Step 1.4.1

Reorder terms.

$2 e^{2 x} x^{2} + 2 e^{2 x} x$

Step 1.4.2

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

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

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

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

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

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 $g (x) = 2 x$ .

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Step 2.2.3.1

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

$2 (x^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [2 x e^{2 x}]$

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 $a$ = $e$ .

$2 (x^{2} (e^{u_{1}} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [2 x e^{2 x}]$

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

$2 (x^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [2 x e^{2 x}]$

Step 2.2.6

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

$2 (x^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (2 x)) + \frac{d}{d x} [2 x e^{2 x}]$

Step 2.2.7

Multiply $2$ by $1$ .

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

Step 2.2.8

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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 $g (x) = 2 x$ .

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

$2 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 2 (x (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [x])$

Step 2.3.6

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

$2 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 2 (x (e^{2 x} (2 \cdot 1)) + e^{2 x} \cdot 1)$

Step 2.3.7

Multiply $2$ by $1$ .

$2 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 2 (x (e^{2 x} \cdot 2) + e^{2 x} \cdot 1)$

Step 2.3.8

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

$2 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 2 (x (2 \cdot e^{2 x}) + e^{2 x} \cdot 1)$

Step 2.3.9

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

$2 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 2 (x (2 e^{2 x}) + e^{2 x})$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$2 (x^{2} (2 e^{2 x})) + 2 (e^{2 x} (2 x)) + 2 (x (2 e^{2 x}) + e^{2 x})$

Step 2.4.2

Apply the distributive property.

$2 (x^{2} (2 e^{2 x})) + 2 (e^{2 x} (2 x)) + 2 (x (2 e^{2 x})) + 2 e^{2 x}$

Step 2.4.3

Combine terms.

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Step 2.4.3.1

Multiply $2$ by $2$ .

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

Step 2.4.3.2

Multiply $2$ by $2$ .

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

Step 2.4.3.3

Multiply $2$ by $2$ .

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

Step 2.4.3.4

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

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Step 2.4.3.4.1

Move $e^{2 x}$ .

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

Step 2.4.3.4.2

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

$4 x^{2} e^{2 x} + 8 x e^{2 x} + 2 e^{2 x}$

Step 2.4.4

Reorder terms.

$4 e^{2 x} x^{2} + 8 e^{2 x} x + 2 e^{2 x}$

Step 2.4.5

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

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

Step 3

Find the third derivative.

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Step 3.1

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

$\frac{d}{d x} [4 x^{2} e^{2 x}] + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2

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

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Step 3.2.1

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

$4 \frac{d}{d x} [x^{2} e^{2 x}] + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

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

$4 (x^{2} \frac{d}{d x} [e^{2 x}] + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

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 $g (x) = 2 x$ .

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Step 3.2.3.1

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

$4 (x^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

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 $a$ = $e$ .

$4 (x^{2} (e^{u_{1}} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.3.3

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

$4 (x^{2} (e^{2 x} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.4

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

$4 (x^{2} (e^{2 x} (2 \frac{d}{d x} [x])) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.5

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

$4 (x^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.6

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

$4 (x^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (2 x)) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.7

Multiply $2$ by $1$ .

$4 (x^{2} (e^{2 x} \cdot 2) + e^{2 x} (2 x)) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.2.8

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + \frac{d}{d x} [8 x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3

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

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Step 3.3.1

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 \frac{d}{d x} [x e^{2 x}] + \frac{d}{d x} [2 e^{2 x}]$

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x \frac{d}{d x} [e^{2 x}] + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

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 $g (x) = 2 x$ .

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Step 3.3.3.1

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

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 $a$ = $e$ .

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{u_{2}} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.3.3

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{2 x} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.4

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{2 x} (2 \frac{d}{d x} [x])) + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.5

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [x]) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.6

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{2 x} (2 \cdot 1)) + e^{2 x} \cdot 1) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.7

Multiply $2$ by $1$ .

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (e^{2 x} \cdot 2) + e^{2 x} \cdot 1) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.8

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 \cdot e^{2 x}) + e^{2 x} \cdot 1) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.3.9

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + \frac{d}{d x} [2 e^{2 x}]$

Step 3.4

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

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Step 3.4.1

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 \frac{d}{d x} [e^{2 x}]$

Step 3.4.2

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 $g (x) = 2 x$ .

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Step 3.4.2.1

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d x} [2 x])$

Step 3.4.2.2

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (e^{u_{3}} \frac{d}{d x} [2 x])$

Step 3.4.2.3

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (e^{2 x} \frac{d}{d x} [2 x])$

Step 3.4.3

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (e^{2 x} (2 \frac{d}{d x} [x]))$

Step 3.4.4

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (e^{2 x} (2 \cdot 1))$

Step 3.4.5

Multiply $2$ by $1$ .

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (e^{2 x} \cdot 2)$

Step 3.4.6

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

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 2 (2 \cdot e^{2 x})$

Step 3.4.7

Multiply $2$ by $2$ .

$4 (x^{2} (2 e^{2 x}) + e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 4 e^{2 x}$

Step 3.5

Simplify.

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Step 3.5.1

Apply the distributive property.

$4 (x^{2} (2 e^{2 x})) + 4 (e^{2 x} (2 x)) + 8 (x (2 e^{2 x}) + e^{2 x}) + 4 e^{2 x}$

Step 3.5.2

Apply the distributive property.

$4 (x^{2} (2 e^{2 x})) + 4 (e^{2 x} (2 x)) + 8 (x (2 e^{2 x})) + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3

Combine terms.

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Step 3.5.3.1

Multiply $2$ by $4$ .

$8 (x^{2} (e^{2 x})) + 4 (e^{2 x} (2 x)) + 8 (x (2 e^{2 x})) + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3.2

Multiply $2$ by $4$ .

$8 x^{2} e^{2 x} + 8 (e^{2 x} (x)) + 8 (x (2 e^{2 x})) + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3.3

Multiply $2$ by $8$ .

$8 x^{2} e^{2 x} + 8 e^{2 x} x + 16 (x (e^{2 x})) + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3.4

Add $8 e^{2 x} x$ and $16 x e^{2 x}$ .

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Step 3.5.3.4.1

Move $e^{2 x}$ .

$8 x^{2} e^{2 x} + 8 x e^{2 x} + 16 x e^{2 x} + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3.4.2

Add $8 x e^{2 x}$ and $16 x e^{2 x}$ .

$8 x^{2} e^{2 x} + 24 x e^{2 x} + 8 e^{2 x} + 4 e^{2 x}$

Step 3.5.3.5

Add $8 e^{2 x}$ and $4 e^{2 x}$ .

$8 x^{2} e^{2 x} + 24 x e^{2 x} + 12 e^{2 x}$

Step 3.5.4

Reorder terms.

$8 e^{2 x} x^{2} + 24 e^{2 x} x + 12 e^{2 x}$

Step 3.5.5

Reorder factors in $8 e^{2 x} x^{2} + 24 e^{2 x} x + 12 e^{2 x}$ .

$f''' (x) = 8 x^{2} e^{2 x} + 24 x e^{2 x} + 12 e^{2 x}$

Step 4

The third derivative of $f (x)$ with respect to $x$ is $8 x^{2} e^{2 x} + 24 x e^{2 x} + 12 e^{2 x}$ .

$8 x^{2} e^{2 x} + 24 x e^{2 x} + 12 e^{2 x}$