Calculus Examples

$y = \sec (2 x)$

Step 1

Find the first derivative.

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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 $g (x) = 2 x$ .

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

$f' (x) = \frac{d}{d u} (\sec (u)) \frac{d}{d x} (2 x)$

The derivative of $\sec (u)$ with respect to $u$ is $\sec (u) \tan (u)$ .

$f' (x) = \sec (u) \tan (u) \frac{d}{d x} (2 x)$

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

$f' (x) = \sec (2 x) \tan (2 x) \frac{d}{d x} (2 x)$

Differentiate.

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

$f' (x) = \sec (2 x) \tan (2 x) (2 \frac{d}{d x} (x))$

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

$f' (x) = \sec (2 x) \tan (2 x) (2 \cdot 1)$

Simplify the expression.

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Multiply $2$ by $1$ .

$f' (x) = \sec (2 x) \tan (2 x) \cdot 2$

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

$f' (x) = 2 \sec (2 x) \tan (2 x)$

Step 2

Find the second derivative.

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

$f'' (x) = 2 \frac{d}{d x} (\sec (2 x) \tan (2 x))$

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

$f'' (x) = 2 (\sec (2 x) \frac{d}{d x} (\tan (2 x)) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

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 $g (x) = 2 x$ .

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

$f'' (x) = 2 (\sec (2 x) (\frac{d}{d u (1)} (\tan (u_{1})) \frac{d}{d x} (2 x)) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

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

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

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

$f'' (x) = 2 (\sec (2 x) (\sec^{2} (2 x) \frac{d}{d x} (2 x)) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

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

$f'' (x) = 2 (\sec^{2} (2 x) \sec (2 x) \frac{d}{d x} (2 x) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

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

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

Differentiate.

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Add $2$ and $1$ .

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

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

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

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

$f'' (x) = 2 (\sec^{3} (2 x) (2 \cdot 1) + \tan (2 x) \frac{d}{d x} (\sec (2 x)))$

Simplify the expression.

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Multiply $2$ by $1$ .

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

Move $2$ to the left of $\sec^{3} (2 x)$ .

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

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 $g (x) = 2 x$ .

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

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

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

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

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

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

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

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

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

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

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

$f'' (x) = 2 (2 \sec^{3} (2 x) + {\tan (2 x)}^{1 + 1} (\sec (2 x) \frac{d}{d x} (2 x)))$

Add $1$ and $1$ .

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

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

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

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

$f'' (x) = 2 (2 \sec^{3} (2 x) + \tan^{2} (2 x) \sec (2 x) (2 \cdot 1))$

Simplify the expression.

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Multiply $2$ by $1$ .

$f'' (x) = 2 (2 \sec^{3} (2 x) + \tan^{2} (2 x) \sec (2 x) \cdot 2)$

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

$f'' (x) = 2 (2 \sec^{3} (2 x) + 2 \cdot (\tan^{2} (2 x) \sec (2 x)))$

Simplify.

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Apply the distributive property.

$f'' (x) = 2 (2 \sec^{3} (2 x)) + 2 (2 \tan^{2} (2 x) \sec (2 x))$

Combine terms.

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Multiply $2$ by $2$ .

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

Multiply $2$ by $2$ .

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

Reorder terms.

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

Step 3

Find the third derivative.

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By the Sum Rule, the derivative of $4 \tan^{2} (2 x) \sec (2 x) + 4 \sec^{3} (2 x)$ with respect to $x$ is $\frac{d}{d x} [4 \tan^{2} (2 x) \sec (2 x)] + \frac{d}{d x} [4 \sec^{3} (2 x)]$ .

$f''' (x) = \frac{d}{d x} (4 \tan^{2} (2 x) \sec (2 x)) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

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

$f''' (x) = 4 \frac{d}{d x} (\tan^{2} (2 x) \sec (2 x)) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) \frac{d}{d x} (\sec (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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 $g (x) = 2 x$ .

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

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

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (u_{1}) \tan (u_{1}) \frac{d}{d x} (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \frac{d}{d x} (x))) + \sec (2 x) \frac{d}{d x} (\tan^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) \frac{d}{d x} (\tan^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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 $g (x) = \tan (2 x)$ .

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (\frac{d}{d u (2)} ({u_{2}}^{2}) \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 u_{2} \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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 $g (x) = 2 x$ .

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) (\frac{d}{d u (3)} (\tan (u_{3})) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

The derivative of $\tan (u_{3})$ with respect to $u_{3}$ is $\sec^{2} (u_{3})$ .

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (u_{3}) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \frac{d}{d x} (x))))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Multiply $2$ by $1$ .

$f''' (x) = 4 (\tan^{2} (2 x) (\sec (2 x) \tan (2 x) \cdot 2) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (\tan^{2} (2 x) (2 \cdot (\sec (2 x) \tan (2 x))) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

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Move $\tan (2 x)$ .

$f''' (x) = 4 (\tan (2 x) \tan^{2} (2 x) (2 \sec (2 x)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

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

$f''' (x) = 4 (\tan (2 x) \tan^{2} (2 x) (2 \sec (2 x)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 ({\tan (2 x)}^{1 + 2} (2 \sec (2 x)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Add $1$ and $2$ .

$f''' (x) = 4 (\tan^{3} (2 x) (2 \sec (2 x)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (2 \cdot (\tan^{3} (2 x) \sec (2 x)) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Multiply $2$ by $1$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + \sec (2 x) (2 \tan (2 x) (\sec^{2} (2 x) \cdot 2))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + \sec (2 x) (2 \tan (2 x) (2 \cdot \sec^{2} (2 x)))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Multiply $2$ by $2$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + \sec (2 x) (4 \tan (2 x) \sec^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) (\sec (2 x) \sec^{2} (2 x))) + \frac{d}{d x} (4 \sec^{3} (2 x))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) {\sec (2 x)}^{1 + 2}) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Add $1$ and $2$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + \frac{d}{d x} (4 \sec^{3} (2 x))$

Evaluate $\frac{d}{d x} [4 \sec^{3} (2 x)]$ .

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 \frac{d}{d x} (\sec^{3} (2 x))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = \sec (2 x)$ .

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To apply the Chain Rule, set $u_{4}$ as $\sec (2 x)$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (\frac{d}{d u (4)} ({u_{4}}^{3}) \frac{d}{d x} (\sec (2 x)))$

Differentiate using the Power Rule which states that $\frac{d}{d u_{4}} [{u_{4}}^{n}]$ is $n {u_{4}}^{n - 1}$ where $n = 3$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 {u_{4}}^{2} \frac{d}{d x} (\sec (2 x)))$

Replace all occurrences of $u_{4}$ with $\sec (2 x)$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) \frac{d}{d x} (\sec (2 x)))$

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 $g (x) = 2 x$ .

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To apply the Chain Rule, set $u_{5}$ as $2 x$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (\frac{d}{d u (5)} (\sec (u_{5})) \frac{d}{d x} (2 x)))$

The derivative of $\sec (u_{5})$ with respect to $u_{5}$ is $\sec (u_{5}) \tan (u_{5})$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (\sec (u_{5}) \tan (u_{5}) \frac{d}{d x} (2 x)))$

Replace all occurrences of $u_{5}$ with $2 x$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} (2 x)))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \frac{d}{d x} (x))))$

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

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

Multiply $2$ by $1$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) \cdot 2))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (3 \sec^{2} (2 x) (2 \cdot (\sec (2 x) \tan (2 x))))$

Multiply $2$ by $3$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (6 \sec^{2} (2 x) (\sec (2 x) \tan (2 x)))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (6 (\sec (2 x) \sec^{2} (2 x)) \tan (2 x))$

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

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (6 {\sec (2 x)}^{1 + 2} \tan (2 x))$

Add $1$ and $2$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 4 (6 \sec^{3} (2 x) \tan (2 x))$

Multiply $6$ by $4$ .

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x) + 4 \tan (2 x) \sec^{3} (2 x)) + 24 \sec^{3} (2 x) \tan (2 x)$

Simplify.

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Apply the distributive property.

$f''' (x) = 4 (2 \tan^{3} (2 x) \sec (2 x)) + 4 (4 \tan (2 x) \sec^{3} (2 x)) + 24 \sec^{3} (2 x) \tan (2 x)$

Combine terms.

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Multiply $2$ by $4$ .

$f''' (x) = 8 (\tan^{3} (2 x) \sec (2 x)) + 4 (4 \tan (2 x) \sec^{3} (2 x)) + 24 \sec^{3} (2 x) \tan (2 x)$

Multiply $4$ by $4$ .

$f''' (x) = 8 \tan^{3} (2 x) \sec (2 x) + 16 (\tan (2 x) \sec^{3} (2 x)) + 24 \sec^{3} (2 x) \tan (2 x)$

Reorder the factors of $16 \tan (2 x) \sec^{3} (2 x)$ .

$f''' (x) = 8 \tan^{3} (2 x) \sec (2 x) + 16 \sec^{3} (2 x) \tan (2 x) + 24 \sec^{3} (2 x) \tan (2 x)$

Add $16 \sec^{3} (2 x) \tan (2 x)$ and $24 \sec^{3} (2 x) \tan (2 x)$ .

$f''' (x) = 8 \tan^{3} (2 x) \sec (2 x) + 40 \sec^{3} (2 x) \tan (2 x)$

Step 4

Find the fourth derivative.

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By the Sum Rule, the derivative of $8 \tan^{3} (2 x) \sec (2 x) + 40 \sec^{3} (2 x) \tan (2 x)$ with respect to $x$ is $\frac{d}{d x} [8 \tan^{3} (2 x) \sec (2 x)] + \frac{d}{d x} [40 \sec^{3} (2 x) \tan (2 x)]$ .

$f^{4} (x) = \frac{d}{d x} (8 \tan^{3} (2 x) \sec (2 x)) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Evaluate $\frac{d}{d x} [8 \tan^{3} (2 x) \sec (2 x)]$ .

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Since $8$ is constant with respect to $x$ , the derivative of $8 \tan^{3} (2 x) \sec (2 x)$ with respect to $x$ is $8 \frac{d}{d x} [\tan^{3} (2 x) \sec (2 x)]$ .

$f^{4} (x) = 8 \frac{d}{d x} (\tan^{3} (2 x) \sec (2 x)) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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 $f (x) = \tan^{3} (2 x)$ and $g (x) = \sec (2 x)$ .

$f^{4} (x) = 8 (\tan^{3} (2 x) \frac{d}{d x} (\sec (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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 $g (x) = 2 x$ .

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\frac{d}{d u (1)} (\sec (u_{1})) \frac{d}{d x} (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (u_{1}) \tan (u_{1}) \frac{d}{d x} (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} (2 x)) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \frac{d}{d x} (x))) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) \frac{d}{d x} (\tan^{3} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = \tan (2 x)$ .

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (\frac{d}{d u (2)} ({u_{2}}^{3}) \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 {u_{2}}^{2} \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) \frac{d}{d x} (\tan (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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 $g (x) = 2 x$ .

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) (\frac{d}{d u (3)} (\tan (u_{3})) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

The derivative of $\tan (u_{3})$ with respect to $u_{3}$ is $\sec^{2} (u_{3})$ .

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (u_{3}) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) \frac{d}{d x} (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \frac{d}{d x} (x))))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Multiply $2$ by $1$ .

$f^{4} (x) = 8 (\tan^{3} (2 x) (\sec (2 x) \tan (2 x) \cdot 2) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (\tan^{3} (2 x) (2 \cdot (\sec (2 x) \tan (2 x))) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Multiply $\tan^{3} (2 x)$ by $\tan (2 x)$ by adding the exponents.

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Move $\tan (2 x)$ .

$f^{4} (x) = 8 (\tan (2 x) \tan^{3} (2 x) (2 \sec (2 x)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Multiply $\tan (2 x)$ by $\tan^{3} (2 x)$ .

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

$f^{4} (x) = 8 (\tan (2 x) \tan^{3} (2 x) (2 \sec (2 x)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 ({\tan (2 x)}^{1 + 3} (2 \sec (2 x)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Add $1$ and $3$ .

$f^{4} (x) = 8 (\tan^{4} (2 x) (2 \sec (2 x)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (2 \cdot (\tan^{4} (2 x) \sec (2 x)) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) (2 \cdot 1)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Multiply $2$ by $1$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + \sec (2 x) (3 \tan^{2} (2 x) (\sec^{2} (2 x) \cdot 2))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + \sec (2 x) (3 \tan^{2} (2 x) (2 \cdot \sec^{2} (2 x)))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Multiply $2$ by $3$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + \sec (2 x) (6 \tan^{2} (2 x) \sec^{2} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) (\sec (2 x) \sec^{2} (2 x))) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) {\sec (2 x)}^{1 + 2}) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Add $1$ and $2$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + \frac{d}{d x} (40 \sec^{3} (2 x) \tan (2 x))$

Evaluate $\frac{d}{d x} [40 \sec^{3} (2 x) \tan (2 x)]$ .

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Since $40$ is constant with respect to $x$ , the derivative of $40 \sec^{3} (2 x) \tan (2 x)$ with respect to $x$ is $40 \frac{d}{d x} [\sec^{3} (2 x) \tan (2 x)]$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 \frac{d}{d x} (\sec^{3} (2 x) \tan (2 x))$

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 $f (x) = \sec^{3} (2 x)$ and $g (x) = \tan (2 x)$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) \frac{d}{d x} (\tan (2 x)) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

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 $g (x) = 2 x$ .

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\frac{d}{d u (4)} (\tan (u_{4})) \frac{d}{d x} (2 x)) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

The derivative of $\tan (u_{4})$ with respect to $u_{4}$ is $\sec^{2} (u_{4})$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (u_{4}) \frac{d}{d x} (2 x)) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) \frac{d}{d x} (2 x)) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \frac{d}{d x} (x))) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) \frac{d}{d x} (\sec^{3} (2 x)))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = \sec (2 x)$ .

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (\frac{d}{d u (5)} ({u_{5}}^{3}) \frac{d}{d x} (\sec (2 x))))$

Differentiate using the Power Rule which states that $\frac{d}{d u_{5}} [{u_{5}}^{n}]$ is $n {u_{5}}^{n - 1}$ where $n = 3$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 {u_{5}}^{2} \frac{d}{d x} (\sec (2 x))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) \frac{d}{d x} (\sec (2 x))))$

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 $g (x) = 2 x$ .

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) (\frac{d}{d u (6)} (\sec (u_{6})) \frac{d}{d x} (2 x))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (u_{6}) \tan (u_{6}) \frac{d}{d x} (2 x))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} (2 x))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \frac{d}{d x} (x)))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) (2 \cdot 1)) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

Multiply $2$ by $1$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (\sec^{2} (2 x) \cdot 2) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{3} (2 x) (2 \cdot \sec^{2} (2 x)) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

Multiply $\sec^{3} (2 x)$ by $\sec^{2} (2 x)$ by adding the exponents.

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{2} (2 x) \sec^{3} (2 x) \cdot 2 + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 ({\sec (2 x)}^{2 + 3} \cdot 2 + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

Add $2$ and $3$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (\sec^{5} (2 x) \cdot 2 + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

Move $2$ to the left of $\sec^{5} (2 x)$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \cdot \sec^{5} (2 x) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) (2 \cdot 1))))$

Multiply $2$ by $1$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (3 \sec^{2} (2 x) (\sec (2 x) \tan (2 x) \cdot 2)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (3 \sec^{2} (2 x) (2 \cdot (\sec (2 x) \tan (2 x)))))$

Multiply $2$ by $3$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (6 \sec^{2} (2 x) (\sec (2 x) \tan (2 x))))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (6 (\sec (2 x) \sec^{2} (2 x)) \tan (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (6 {\sec (2 x)}^{1 + 2} \tan (2 x)))$

Add $1$ and $2$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + \tan (2 x) (6 \sec^{3} (2 x) \tan (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + 6 \sec^{3} (2 x) (\tan (2 x) \tan (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + 6 \sec^{3} (2 x) (\tan (2 x) \tan (2 x)))$

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

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + 6 \sec^{3} (2 x) {\tan (2 x)}^{1 + 1})$

Add $1$ and $1$ .

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x) + 6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + 6 \sec^{3} (2 x) \tan^{2} (2 x))$

Simplify.

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Apply the distributive property.

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x)) + 8 (6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x) + 6 \sec^{3} (2 x) \tan^{2} (2 x))$

Apply the distributive property.

$f^{4} (x) = 8 (2 \tan^{4} (2 x) \sec (2 x)) + 8 (6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x)) + 40 (6 \sec^{3} (2 x) \tan^{2} (2 x))$

Combine terms.

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Multiply $2$ by $8$ .

$f^{4} (x) = 16 (\tan^{4} (2 x) \sec (2 x)) + 8 (6 \tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x)) + 40 (6 \sec^{3} (2 x) \tan^{2} (2 x))$

Multiply $6$ by $8$ .

$f^{4} (x) = 16 \tan^{4} (2 x) \sec (2 x) + 48 (\tan^{2} (2 x) \sec^{3} (2 x)) + 40 (2 \sec^{5} (2 x)) + 40 (6 \sec^{3} (2 x) \tan^{2} (2 x))$

Multiply $2$ by $40$ .

$f^{4} (x) = 16 \tan^{4} (2 x) \sec (2 x) + 48 \tan^{2} (2 x) \sec^{3} (2 x) + 80 \sec^{5} (2 x) + 40 (6 \sec^{3} (2 x) \tan^{2} (2 x))$

Multiply $6$ by $40$ .

$f^{4} (x) = 16 \tan^{4} (2 x) \sec (2 x) + 48 \tan^{2} (2 x) \sec^{3} (2 x) + 80 \sec^{5} (2 x) + 240 (\sec^{3} (2 x) \tan^{2} (2 x))$

Reorder the factors of $48 \tan^{2} (2 x) \sec^{3} (2 x)$ .

$f^{4} (x) = 16 \tan^{4} (2 x) \sec (2 x) + 48 \sec^{3} (2 x) \tan^{2} (2 x) + 80 \sec^{5} (2 x) + 240 \sec^{3} (2 x) \tan^{2} (2 x)$

Add $48 \sec^{3} (2 x) \tan^{2} (2 x)$ and $240 \sec^{3} (2 x) \tan^{2} (2 x)$ .

$f^{4} (x) = 16 \tan^{4} (2 x) \sec (2 x) + 288 \sec^{3} (2 x) \tan^{2} (2 x) + 80 \sec^{5} (2 x)$