Calculus Examples

$x (\cos (2 x))$

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 $f (x) = x$ and $g (x) = \cos (2 x)$ .

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

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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 1.2.1

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

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

Step 1.2.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$x (- \sin (u) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]$

Step 1.2.3

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

$x (- \sin (2 x) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]$

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 $2 \frac{d}{d x} [x]$ .

$x (- \sin (2 x) (2 \frac{d}{d x} [x])) + \cos (2 x) \frac{d}{d x} [x]$

Step 1.3.2

Multiply $2$ by $- 1$ .

$x (- 2 \sin (2 x) \frac{d}{d x} [x]) + \cos (2 x) \frac{d}{d x} [x]$

Step 1.3.3

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

$x (- 2 \sin (2 x) \cdot 1) + \cos (2 x) \frac{d}{d x} [x]$

Step 1.3.4

Multiply $- 2$ by $1$ .

$x (- 2 \sin (2 x)) + \cos (2 x) \frac{d}{d x} [x]$

Step 1.3.5

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

$x (- 2 \sin (2 x)) + \cos (2 x) \cdot 1$

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Multiply $\cos (2 x)$ by $1$ .

$x (- 2 \sin (2 x)) + \cos (2 x)$

Step 1.3.6.2

Reorder terms.

$f' (x) = - 2 x \sin (2 x) + \cos (2 x)$

Step 2

Find the second derivative.

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Step 2.1

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

$\frac{d}{d x} [- 2 x \sin (2 x)] + \frac{d}{d x} [\cos (2 x)]$

Step 2.2

Evaluate $\frac{d}{d x} [- 2 x \sin (2 x)]$ .

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Step 2.2.1

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

$- 2 \frac{d}{d x} [x \sin (2 x)] + \frac{d}{d x} [\cos (2 x)]$

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 $f (x) = x$ and $g (x) = \sin (2 x)$ .

$- 2 (x \frac{d}{d x} [\sin (2 x)] + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 2.2.3.1

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

$- 2 (x (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.3.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$- 2 (x (\cos (u_{1}) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.3.3

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

$- 2 (x (\cos (2 x) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.4

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

$- 2 (x (\cos (2 x) (2 \frac{d}{d x} [x])) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.5

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

$- 2 (x (\cos (2 x) (2 \cdot 1)) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.6

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

$- 2 (x (\cos (2 x) (2 \cdot 1)) + \sin (2 x) \cdot 1) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.7

Multiply $2$ by $1$ .

$- 2 (x (\cos (2 x) \cdot 2) + \sin (2 x) \cdot 1) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.8

Move $2$ to the left of $\cos (2 x)$ .

$- 2 (x (2 \cdot \cos (2 x)) + \sin (2 x) \cdot 1) + \frac{d}{d x} [\cos (2 x)]$

Step 2.2.9

Multiply $\sin (2 x)$ by $1$ .

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) + \frac{d}{d x} [\cos (2 x)]$

Step 2.3

Evaluate $\frac{d}{d x} [\cos (2 x)]$ .

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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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 2.3.1.1

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

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) + \frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d x} [2 x]$

Step 2.3.1.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - \sin (u_{2}) \frac{d}{d x} [2 x]$

Step 2.3.1.3

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

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - \sin (2 x) \frac{d}{d x} [2 x]$

Step 2.3.2

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

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - \sin (2 x) (2 \frac{d}{d x} [x])$

Step 2.3.3

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

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - \sin (2 x) (2 \cdot 1)$

Step 2.3.4

Multiply $2$ by $1$ .

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - \sin (2 x) \cdot 2$

Step 2.3.5

Multiply $2$ by $- 1$ .

$- 2 (x (2 \cos (2 x)) + \sin (2 x)) - 2 \sin (2 x)$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$- 2 (x (2 \cos (2 x))) - 2 \sin (2 x) - 2 \sin (2 x)$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $2$ by $- 2$ .

$- 4 (x (\cos (2 x))) - 2 \sin (2 x) - 2 \sin (2 x)$

Step 2.4.2.2

Subtract $2 \sin (2 x)$ from $- 2 \sin (2 x)$ .

$f'' (x) = - 4 x \cos (2 x) - 4 \sin (2 x)$

Step 3

Find the third derivative.

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Step 3.1

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

$\frac{d}{d x} [- 4 x \cos (2 x)] + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2

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

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Step 3.2.1

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

$- 4 \frac{d}{d x} [x \cos (2 x)] + \frac{d}{d x} [- 4 \sin (2 x)]$

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 $f (x) = x$ and $g (x) = \cos (2 x)$ .

$- 4 (x \frac{d}{d x} [\cos (2 x)] + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 3.2.3.1

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

$- 4 (x (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.3.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$- 4 (x (- \sin (u_{1}) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.3.3

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

$- 4 (x (- \sin (2 x) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.4

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

$- 4 (x (- \sin (2 x) (2 \frac{d}{d x} [x])) + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.5

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

$- 4 (x (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.6

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

$- 4 (x (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) \cdot 1) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.7

Multiply $2$ by $1$ .

$- 4 (x (- \sin (2 x) \cdot 2) + \cos (2 x) \cdot 1) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.8

Multiply $2$ by $- 1$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x) \cdot 1) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.2.9

Multiply $\cos (2 x)$ by $1$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) + \frac{d}{d x} [- 4 \sin (2 x)]$

Step 3.3

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

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Step 3.3.1

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

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 \frac{d}{d x} [\sin (2 x)]$

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 3.3.2.1

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

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [2 x])$

Step 3.3.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\cos (u_{2}) \frac{d}{d x} [2 x])$

Step 3.3.2.3

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

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\cos (2 x) \frac{d}{d x} [2 x])$

Step 3.3.3

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

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\cos (2 x) (2 \frac{d}{d x} [x]))$

Step 3.3.4

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

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\cos (2 x) (2 \cdot 1))$

Step 3.3.5

Multiply $2$ by $1$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (\cos (2 x) \cdot 2)$

Step 3.3.6

Move $2$ to the left of $\cos (2 x)$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 4 (2 \cdot \cos (2 x))$

Step 3.3.7

Multiply $2$ by $- 4$ .

$- 4 (x (- 2 \sin (2 x)) + \cos (2 x)) - 8 \cos (2 x)$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$- 4 (x (- 2 \sin (2 x))) - 4 \cos (2 x) - 8 \cos (2 x)$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Multiply $- 2$ by $- 4$ .

$8 (x (\sin (2 x))) - 4 \cos (2 x) - 8 \cos (2 x)$

Step 3.4.2.2

Subtract $8 \cos (2 x)$ from $- 4 \cos (2 x)$ .

$f''' (x) = 8 x \sin (2 x) - 12 \cos (2 x)$

Step 4

Find the fourth derivative.

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Step 4.1

By the Sum Rule, the derivative of $8 x \sin (2 x) - 12 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [8 x \sin (2 x)] + \frac{d}{d x} [- 12 \cos (2 x)]$ .

$\frac{d}{d x} [8 x \sin (2 x)] + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2

Evaluate $\frac{d}{d x} [8 x \sin (2 x)]$ .

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Step 4.2.1

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

$8 \frac{d}{d x} [x \sin (2 x)] + \frac{d}{d x} [- 12 \cos (2 x)]$

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 $f (x) = x$ and $g (x) = \sin (2 x)$ .

$8 (x \frac{d}{d x} [\sin (2 x)] + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 4.2.3.1

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

$8 (x (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.3.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$8 (x (\cos (u_{1}) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.3.3

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

$8 (x (\cos (2 x) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.4

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

$8 (x (\cos (2 x) (2 \frac{d}{d x} [x])) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.5

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

$8 (x (\cos (2 x) (2 \cdot 1)) + \sin (2 x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.6

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

$8 (x (\cos (2 x) (2 \cdot 1)) + \sin (2 x) \cdot 1) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.7

Multiply $2$ by $1$ .

$8 (x (\cos (2 x) \cdot 2) + \sin (2 x) \cdot 1) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.8

Move $2$ to the left of $\cos (2 x)$ .

$8 (x (2 \cdot \cos (2 x)) + \sin (2 x) \cdot 1) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.2.9

Multiply $\sin (2 x)$ by $1$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) + \frac{d}{d x} [- 12 \cos (2 x)]$

Step 4.3

Evaluate $\frac{d}{d x} [- 12 \cos (2 x)]$ .

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Step 4.3.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 \cos (2 x)$ with respect to $x$ is $- 12 \frac{d}{d x} [\cos (2 x)]$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 \frac{d}{d x} [\cos (2 x)]$

Step 4.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 4.3.2.1

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

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (\frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d x} [2 x])$

Step 4.3.2.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- \sin (u_{2}) \frac{d}{d x} [2 x])$

Step 4.3.2.3

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

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 4.3.3

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

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- \sin (2 x) (2 \frac{d}{d x} [x]))$

Step 4.3.4

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

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- \sin (2 x) (2 \cdot 1))$

Step 4.3.5

Multiply $2$ by $1$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- \sin (2 x) \cdot 2)$

Step 4.3.6

Multiply $2$ by $- 1$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) - 12 (- 2 \sin (2 x))$

Step 4.3.7

Multiply $- 2$ by $- 12$ .

$8 (x (2 \cos (2 x)) + \sin (2 x)) + 24 \sin (2 x)$

Step 4.4

Simplify.

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Step 4.4.1

Apply the distributive property.

$8 (x (2 \cos (2 x))) + 8 \sin (2 x) + 24 \sin (2 x)$

Step 4.4.2

Combine terms.

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Step 4.4.2.1

Multiply $2$ by $8$ .

$16 (x (\cos (2 x))) + 8 \sin (2 x) + 24 \sin (2 x)$

Step 4.4.2.2

Add $8 \sin (2 x)$ and $24 \sin (2 x)$ .

$f^{4} (x) = 16 x \cos (2 x) + 32 \sin (2 x)$