Calculus Examples

$y = \tan (2 x + 1)$

Step 1

Find the first derivative.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x + 1$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$\frac{d}{d u} [\tan (u)] \frac{d}{d x} [2 x + 1]$

Step 1.1.2

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

$\sec^{2} (u) \frac{d}{d x} [2 x + 1]$

Step 1.1.3

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $2$ by $1$ .

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

Step 1.2.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\sec^{2} (2 x + 1) (2 + 0)$

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Add $2$ and $0$ .

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

Step 1.2.6.2

Move $2$ to the left of $\sec^{2} (2 x + 1)$ .

$f' (x) = 2 \sec^{2} (2 x + 1)$

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sec (2 x + 1)$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $\sec (2 x + 1)$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u_{1}$ with $\sec (2 x + 1)$ .

$2 (2 \sec (2 x + 1) \frac{d}{d x} [\sec (2 x + 1)])$

Step 2.3

Multiply $2$ by $2$ .

$4 (\sec (2 x + 1) \frac{d}{d x} [\sec (2 x + 1)])$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x + 1$ .

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Step 2.4.1

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

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

Step 2.4.2

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

$4 \sec (2 x + 1) (\sec (u_{2}) \tan (u_{2}) \frac{d}{d x} [2 x + 1])$

Step 2.4.3

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

$4 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 2.5

Raise $\sec (2 x + 1)$ to the power of $1$ .

$4 (\sec^{1} (2 x + 1) \sec (2 x + 1)) (\tan (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 2.6

Raise $\sec (2 x + 1)$ to the power of $1$ .

$4 (\sec^{1} (2 x + 1) \sec^{1} (2 x + 1)) (\tan (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 2.7

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

$4 {\sec (2 x + 1)}^{1 + 1} (\tan (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 2.8

Add $1$ and $1$ .

$4 \sec^{2} (2 x + 1) (\tan (2 x + 1) \frac{d}{d x} [2 x + 1])$

Step 2.9

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$4 \sec^{2} (2 x + 1) \tan (2 x + 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])$

Step 2.10

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

$4 \sec^{2} (2 x + 1) \tan (2 x + 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 2.11

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

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

Step 2.12

Multiply $2$ by $1$ .

$4 \sec^{2} (2 x + 1) \tan (2 x + 1) (2 + \frac{d}{d x} [1])$

Step 2.13

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$4 \sec^{2} (2 x + 1) \tan (2 x + 1) (2 + 0)$

Step 2.14

Simplify the expression.

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Step 2.14.1

Add $2$ and $0$ .

$4 \sec^{2} (2 x + 1) \tan (2 x + 1) \cdot 2$

Step 2.14.2

Multiply $2$ by $4$ .

$f'' (x) = 8 \sec^{2} (2 x + 1) \tan (2 x + 1)$

Step 3

Find the third derivative.

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Step 3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 \sec^{2} (2 x + 1) \tan (2 x + 1)$ with respect to $x$ is $8 \frac{d}{d x} [\sec^{2} (2 x + 1) \tan (2 x + 1)]$ .

$8 \frac{d}{d x} [\sec^{2} (2 x + 1) \tan (2 x + 1)]$

Step 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 $f (x) = \sec^{2} (2 x + 1)$ and $g (x) = \tan (2 x + 1)$ .

$8 (\sec^{2} (2 x + 1) \frac{d}{d x} [\tan (2 x + 1)] + \tan (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)])$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x + 1$ .

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Step 3.3.1

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

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

Step 3.3.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

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

Step 3.3.3

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

$8 (\sec^{2} (2 x + 1) (\sec^{2} (2 x + 1) \frac{d}{d x} [2 x + 1]) + \tan (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)])$

Step 3.4

Multiply $\sec^{2} (2 x + 1)$ by $\sec^{2} (2 x + 1)$ by adding the exponents.

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Step 3.4.1

Move $\sec^{2} (2 x + 1)$ .

$8 (\sec^{2} (2 x + 1) \sec^{2} (2 x + 1) \frac{d}{d x} [2 x + 1] + \tan (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)])$

Step 3.4.2

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

$8 ({\sec (2 x + 1)}^{2 + 2} \frac{d}{d x} [2 x + 1] + \tan (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)])$

Step 3.4.3

Add $2$ and $2$ .

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

Step 3.5

Differentiate.

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Step 3.5.1

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 3.5.2

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

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

Step 3.5.3

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

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

Step 3.5.4

Multiply $2$ by $1$ .

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

Step 3.5.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$8 (\sec^{4} (2 x + 1) (2 + 0) + \tan (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)])$

Step 3.5.6

Simplify the expression.

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Step 3.5.6.1

Add $2$ and $0$ .

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

Step 3.5.6.2

Move $2$ to the left of $\sec^{4} (2 x + 1)$ .

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

Step 3.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sec (2 x + 1)$ .

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Step 3.6.1

To apply the Chain Rule, set $u_{2}$ as $\sec (2 x + 1)$ .

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

Step 3.6.2

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

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

Step 3.6.3

Replace all occurrences of $u_{2}$ with $\sec (2 x + 1)$ .

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

Step 3.7

Move $2$ to the left of $\tan (2 x + 1)$ .

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

Step 3.8

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x + 1$ .

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Step 3.8.1

To apply the Chain Rule, set $u_{3}$ as $2 x + 1$ .

$8 (2 \sec^{4} (2 x + 1) + 2 \tan (2 x + 1) \sec (2 x + 1) (\frac{d}{d u_{3}} [\sec (u_{3})] \frac{d}{d x} [2 x + 1]))$

Step 3.8.2

The derivative of $\sec (u_{3})$ with respect to $u_{3}$ is $\sec (u_{3}) \tan (u_{3})$ .

$8 (2 \sec^{4} (2 x + 1) + 2 \tan (2 x + 1) \sec (2 x + 1) (\sec (u_{3}) \tan (u_{3}) \frac{d}{d x} [2 x + 1]))$

Step 3.8.3

Replace all occurrences of $u_{3}$ with $2 x + 1$ .

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

Step 3.9

Raise $\tan (2 x + 1)$ to the power of $1$ .

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

Step 3.10

Raise $\tan (2 x + 1)$ to the power of $1$ .

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

Step 3.11

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

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

Step 3.12

Add $1$ and $1$ .

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

Step 3.13

Raise $\sec (2 x + 1)$ to the power of $1$ .

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

Step 3.14

Raise $\sec (2 x + 1)$ to the power of $1$ .

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

Step 3.15

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

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

Step 3.16

Add $1$ and $1$ .

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

Step 3.17

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 3.18

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

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

Step 3.19

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

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

Step 3.20

Multiply $2$ by $1$ .

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

Step 3.21

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$8 (2 \sec^{4} (2 x + 1) + 2 \tan^{2} (2 x + 1) \sec^{2} (2 x + 1) (2 + 0))$

Step 3.22

Simplify the expression.

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Step 3.22.1

Add $2$ and $0$ .

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

Step 3.22.2

Multiply $2$ by $2$ .

$8 (2 \sec^{4} (2 x + 1) + 4 \tan^{2} (2 x + 1) \sec^{2} (2 x + 1))$

Step 3.23

Simplify.

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Step 3.23.1

Apply the distributive property.

$8 (2 \sec^{4} (2 x + 1)) + 8 (4 \tan^{2} (2 x + 1) \sec^{2} (2 x + 1))$

Step 3.23.2

Combine terms.

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Step 3.23.2.1

Multiply $2$ by $8$ .

$16 \sec^{4} (2 x + 1) + 8 (4 \tan^{2} (2 x + 1) \sec^{2} (2 x + 1))$

Step 3.23.2.2

Multiply $4$ by $8$ .

$16 \sec^{4} (2 x + 1) + 32 \tan^{2} (2 x + 1) \sec^{2} (2 x + 1)$

Step 3.23.3

Reorder terms.

$f''' (x) = 32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1) + 16 \sec^{4} (2 x + 1)$

Step 4

Find the fourth derivative.

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Step 4.1

By the Sum Rule, the derivative of $32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1) + 16 \sec^{4} (2 x + 1)$ with respect to $x$ is $\frac{d}{d x} [32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1)] + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$ .

$\frac{d}{d x} [32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1)] + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2

Evaluate $\frac{d}{d x} [32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1)]$ .

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Step 4.2.1

Since $32$ is constant with respect to $x$ , the derivative of $32 \sec^{2} (2 x + 1) \tan^{2} (2 x + 1)$ with respect to $x$ is $32 \frac{d}{d x} [\sec^{2} (2 x + 1) \tan^{2} (2 x + 1)]$ .

$32 \frac{d}{d x} [\sec^{2} (2 x + 1) \tan^{2} (2 x + 1)] + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.2

$32 (\sec^{2} (2 x + 1) \frac{d}{d x} [\tan^{2} (2 x + 1)] + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

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 $f (x) = x^{2}$ and $g (x) = \tan (2 x + 1)$ .

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Step 4.2.3.1

To apply the Chain Rule, set $u_{1}$ as $\tan (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tan (2 x + 1)]) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.3.2

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

$32 (\sec^{2} (2 x + 1) (2 u_{1} \frac{d}{d x} [\tan (2 x + 1)]) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.3.3

Replace all occurrences of $u_{1}$ with $\tan (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) \frac{d}{d x} [\tan (2 x + 1)]) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x + 1$ .

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Step 4.2.4.1

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [2 x + 1])) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.4.2

The derivative of $\tan (u_{2})$ with respect to $u_{2}$ is $\sec^{2} (u_{2})$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (u_{2}) \frac{d}{d x} [2 x + 1])) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.4.3

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) \frac{d}{d x} [2 x + 1])) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.5

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [1]))) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.6

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [1]))) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.7

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + \frac{d}{d x} [1]))) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.8

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) \frac{d}{d x} [\sec^{2} (2 x + 1)]) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.9

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sec (2 x + 1)$ .

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Step 4.2.9.1

To apply the Chain Rule, set $u_{3}$ as $\sec (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (\frac{d}{d u_{3}} [{u_{3}}^{2}] \frac{d}{d x} [\sec (2 x + 1)])) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.9.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 2$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 u_{3} \frac{d}{d x} [\sec (2 x + 1)])) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.9.3

Replace all occurrences of $u_{3}$ with $\sec (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) \frac{d}{d x} [\sec (2 x + 1)])) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.10

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x + 1$ .

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Step 4.2.10.1

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\frac{d}{d u_{4}} [\sec (u_{4})] \frac{d}{d x} [2 x + 1]))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.10.2

The derivative of $\sec (u_{4})$ with respect to $u_{4}$ is $\sec (u_{4}) \tan (u_{4})$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (u_{4}) \tan (u_{4}) \frac{d}{d x} [2 x + 1]))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.10.3

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) \frac{d}{d x} [2 x + 1]))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.11

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.12

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.13

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

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + \frac{d}{d x} [1])))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.14

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 \cdot 1 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.15

Multiply $2$ by $1$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) (2 + 0))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.16

Add $2$ and $0$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (\sec^{2} (2 x + 1) \cdot 2)) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.17

Move $2$ to the left of $\sec^{2} (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) (2 \tan (2 x + 1) (2 \cdot \sec^{2} (2 x + 1))) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.18

Multiply $2$ by $2$ .

$32 (\sec^{2} (2 x + 1) (4 \tan (2 x + 1) \sec^{2} (2 x + 1)) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.19

Multiply $\sec^{2} (2 x + 1)$ by $\sec^{2} (2 x + 1)$ by adding the exponents.

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Step 4.2.19.1

Move $\sec^{2} (2 x + 1)$ .

$32 (\sec^{2} (2 x + 1) \sec^{2} (2 x + 1) (4 \tan (2 x + 1)) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.19.2

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

$32 ({\sec (2 x + 1)}^{2 + 2} (4 \tan (2 x + 1)) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.19.3

Add $2$ and $2$ .

$32 (\sec^{4} (2 x + 1) (4 \tan (2 x + 1)) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.20

Move $4$ to the left of $\sec^{4} (2 x + 1)$ .

$32 (4 \cdot \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.21

Multiply $2$ by $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 + 0)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.22

Add $2$ and $0$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) \cdot 2))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.23

Move $2$ to the left of $\sec (2 x + 1) \tan (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (2 \sec (2 x + 1) (2 \cdot (\sec (2 x + 1) \tan (2 x + 1))))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.24

Multiply $2$ by $2$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (4 \sec (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1)))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.25

Raise $\sec (2 x + 1)$ to the power of $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (4 (\sec^{1} (2 x + 1) \sec (2 x + 1)) \tan (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.26

Raise $\sec (2 x + 1)$ to the power of $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (4 (\sec^{1} (2 x + 1) \sec^{1} (2 x + 1)) \tan (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.27

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

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (4 {\sec (2 x + 1)}^{1 + 1} \tan (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.28

Add $1$ and $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{2} (2 x + 1) (4 \sec^{2} (2 x + 1) \tan (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.29

Multiply $\tan^{2} (2 x + 1)$ by $\tan (2 x + 1)$ by adding the exponents.

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Step 4.2.29.1

Move $\tan (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan (2 x + 1) \tan^{2} (2 x + 1) (4 \sec^{2} (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.29.2

Multiply $\tan (2 x + 1)$ by $\tan^{2} (2 x + 1)$ .

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Step 4.2.29.2.1

Raise $\tan (2 x + 1)$ to the power of $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{1} (2 x + 1) \tan^{2} (2 x + 1) (4 \sec^{2} (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.29.2.2

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

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + {\tan (2 x + 1)}^{1 + 2} (4 \sec^{2} (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.29.3

Add $1$ and $2$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + \tan^{3} (2 x + 1) (4 \sec^{2} (2 x + 1))) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.2.30

Move $4$ to the left of $\tan^{3} (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + \frac{d}{d x} [16 \sec^{4} (2 x + 1)]$

Step 4.3

Evaluate $\frac{d}{d x} [16 \sec^{4} (2 x + 1)]$ .

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Step 4.3.1

Since $16$ is constant with respect to $x$ , the derivative of $16 \sec^{4} (2 x + 1)$ with respect to $x$ is $16 \frac{d}{d x} [\sec^{4} (2 x + 1)]$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 \frac{d}{d x} [\sec^{4} (2 x + 1)]$

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 $f (x) = x^{4}$ and $g (x) = \sec (2 x + 1)$ .

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Step 4.3.2.1

To apply the Chain Rule, set $u_{5}$ as $\sec (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (\frac{d}{d u_{5}} [{u_{5}}^{4}] \frac{d}{d x} [\sec (2 x + 1)])$

Step 4.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{5}} [{u_{5}}^{n}]$ is $n {u_{5}}^{n - 1}$ where $n = 4$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 {u_{5}}^{3} \frac{d}{d x} [\sec (2 x + 1)])$

Step 4.3.2.3

Replace all occurrences of $u_{5}$ with $\sec (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) \frac{d}{d x} [\sec (2 x + 1)])$

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 $f (x) = \sec (x)$ and $g (x) = 2 x + 1$ .

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Step 4.3.3.1

To apply the Chain Rule, set $u_{6}$ as $2 x + 1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\frac{d}{d u_{6}} [\sec (u_{6})] \frac{d}{d x} [2 x + 1]))$

Step 4.3.3.2

The derivative of $\sec (u_{6})$ with respect to $u_{6}$ is $\sec (u_{6}) \tan (u_{6})$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (u_{6}) \tan (u_{6}) \frac{d}{d x} [2 x + 1]))$

Step 4.3.3.3

Replace all occurrences of $u_{6}$ with $2 x + 1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) \frac{d}{d x} [2 x + 1]))$

Step 4.3.4

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])))$

Step 4.3.5

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

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])))$

Step 4.3.6

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

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + \frac{d}{d x} [1])))$

Step 4.3.7

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 \cdot 1 + 0)))$

Step 4.3.8

Multiply $2$ by $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) (2 + 0)))$

Step 4.3.9

Add $2$ and $0$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1) \cdot 2))$

Step 4.3.10

Move $2$ to the left of $\sec (2 x + 1) \tan (2 x + 1)$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (4 \sec^{3} (2 x + 1) (2 \cdot (\sec (2 x + 1) \tan (2 x + 1))))$

Step 4.3.11

Multiply $2$ by $4$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (8 \sec^{3} (2 x + 1) (\sec (2 x + 1) \tan (2 x + 1)))$

Step 4.3.12

Raise $\sec (2 x + 1)$ to the power of $1$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (8 (\sec^{1} (2 x + 1) \sec^{3} (2 x + 1)) \tan (2 x + 1))$

Step 4.3.13

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

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (8 {\sec (2 x + 1)}^{1 + 3} \tan (2 x + 1))$

Step 4.3.14

Add $1$ and $3$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 16 (8 \sec^{4} (2 x + 1) \tan (2 x + 1))$

Step 4.3.15

Multiply $8$ by $16$ .

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1) + 4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 128 \sec^{4} (2 x + 1) \tan (2 x + 1)$

Step 4.4

Simplify.

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Step 4.4.1

Apply the distributive property.

$32 (4 \sec^{4} (2 x + 1) \tan (2 x + 1)) + 32 (4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 128 \sec^{4} (2 x + 1) \tan (2 x + 1)$

Step 4.4.2

Combine terms.

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Step 4.4.2.1

Multiply $4$ by $32$ .

$128 (\sec^{4} (2 x + 1) \tan (2 x + 1)) + 32 (4 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 128 \sec^{4} (2 x + 1) \tan (2 x + 1)$

Step 4.4.2.2

Multiply $4$ by $32$ .

$128 \sec^{4} (2 x + 1) \tan (2 x + 1) + 128 (\tan^{3} (2 x + 1) \sec^{2} (2 x + 1)) + 128 \sec^{4} (2 x + 1) \tan (2 x + 1)$

Step 4.4.2.3

Add $128 \sec^{4} (2 x + 1) \tan (2 x + 1)$ and $128 \sec^{4} (2 x + 1) \tan (2 x + 1)$ .

$f^{4} (x) = 256 \sec^{4} (2 x + 1) \tan (2 x + 1) + 128 \tan^{3} (2 x + 1) \sec^{2} (2 x + 1)$