Calculus Examples

$\sec (x)$

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

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

Find the second derivative.

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

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

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

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

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

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Reorder terms.

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

Find the third derivative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Move $\tan (x)$ .

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

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $2$ .

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

Combine terms.

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

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

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

$f''' (x) = \tan^{3} (x) \sec (x) + 5 \sec^{3} (x) \tan (x)$

Find the fourth derivative.

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By the Sum Rule, the derivative of $\tan^{3} (x) \sec (x) + 5 \sec^{3} (x) \tan (x)$ with respect to $x$ is $\frac{d}{d x} [\tan^{3} (x) \sec (x)] + \frac{d}{d x} [5 \sec^{3} (x) \tan (x)]$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Move $\tan (x)$ .

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

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

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

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

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

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

Add $1$ and $3$ .

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

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

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

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

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

Add $1$ and $2$ .

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

Evaluate $\frac{d}{d x} [5 \sec^{3} (x) \tan (x)]$ .

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Since $5$ is constant with respect to $x$ , the derivative of $5 \sec^{3} (x) \tan (x)$ with respect to $x$ is $5 \frac{d}{d x} [\sec^{3} (x) \tan (x)]$ .

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

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

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

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

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

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 $g (x) = \sec (x)$ .

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

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

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

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

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

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

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

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

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

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

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

Add $3$ and $2$ .

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Multiply $3$ by $5$ .

$f^{4} (x) = \tan^{4} (x) \sec (x) + 3 \tan^{2} (x) \sec^{3} (x) + 5 \sec^{5} (x) + 15 (\sec^{3} (x) \tan^{2} (x))$

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

$f^{4} (x) = \tan^{4} (x) \sec (x) + 3 \sec^{3} (x) \tan^{2} (x) + 5 \sec^{5} (x) + 15 \sec^{3} (x) \tan^{2} (x)$

Add $3 \sec^{3} (x) \tan^{2} (x)$ and $15 \sec^{3} (x) \tan^{2} (x)$ .

$f^{4} (x) = \tan^{4} (x) \sec (x) + 18 \sec^{3} (x) \tan^{2} (x) + 5 \sec^{5} (x)$