Calculus Examples

$y = \ln (\sec (9 x) + \tan (9 x))$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \sec (9 x) + \tan (9 x)$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\sec (9 x) + \tan (9 x)$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [\sec (9 x) + \tan (9 x)]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [\sec (9 x) + \tan (9 x)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\sec (9 x) + \tan (9 x)$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} \frac{d}{d x} [\sec (9 x) + \tan (9 x)]$

Step 2

By the Sum Rule, the derivative of $\sec (9 x) + \tan (9 x)$ with respect to $x$ is $\frac{d}{d x} [\sec (9 x)] + \frac{d}{d x} [\tan (9 x)]$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (\frac{d}{d x} [\sec (9 x)] + \frac{d}{d x} [\tan (9 x)])$

Step 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 $g (x) = 9 x$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $9 x$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (\frac{d}{d u_{2}} [\sec (u_{2})] \frac{d}{d x} [9 x] + \frac{d}{d x} [\tan (9 x)])$

Step 3.2

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

$\frac{1}{\sec (9 x) + \tan (9 x)} (\sec (u_{2}) \tan (u_{2}) \frac{d}{d x} [9 x] + \frac{d}{d x} [\tan (9 x)])$

Step 3.3

Replace all occurrences of $u_{2}$ with $9 x$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (\sec (9 x) \tan (9 x) \frac{d}{d x} [9 x] + \frac{d}{d x} [\tan (9 x)])$

Step 4

Differentiate.

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Step 4.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (\sec (9 x) \tan (9 x) (9 \frac{d}{d x} [x]) + \frac{d}{d x} [\tan (9 x)])$

Step 4.2

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

$\frac{1}{\sec (9 x) + \tan (9 x)} (\sec (9 x) \tan (9 x) (9 \cdot 1) + \frac{d}{d x} [\tan (9 x)])$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $9$ by $1$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (\sec (9 x) \tan (9 x) \cdot 9 + \frac{d}{d x} [\tan (9 x)])$

Step 4.3.2

Move $9$ to the left of $\sec (9 x) \tan (9 x)$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \cdot (\sec (9 x) \tan (9 x)) + \frac{d}{d x} [\tan (9 x)])$

Step 5

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 $g (x) = 9 x$ .

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Step 5.1

To apply the Chain Rule, set $u_{3}$ as $9 x$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \frac{d}{d u_{3}} [\tan (u_{3})] \frac{d}{d x} [9 x])$

Step 5.2

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

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \sec^{2} (u_{3}) \frac{d}{d x} [9 x])$

Step 5.3

Replace all occurrences of $u_{3}$ with $9 x$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \sec^{2} (9 x) \frac{d}{d x} [9 x])$

Step 6

Differentiate.

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Step 6.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \sec^{2} (9 x) (9 \frac{d}{d x} [x]))$

Step 6.2

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

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \sec^{2} (9 x) (9 \cdot 1))$

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $9$ by $1$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + \sec^{2} (9 x) \cdot 9)$

Step 6.3.2

Move $9$ to the left of $\sec^{2} (9 x)$ .

$\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + 9 \cdot \sec^{2} (9 x))$

Step 7

Simplify.

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Step 7.1

Reorder the factors of $\frac{1}{\sec (9 x) + \tan (9 x)} (9 \sec (9 x) \tan (9 x) + 9 \sec^{2} (9 x))$ .

$(9 \sec (9 x) \tan (9 x) + 9 \sec^{2} (9 x)) \frac{1}{\sec (9 x) + \tan (9 x)}$

Step 7.2

Apply the distributive property.

$9 \sec (9 x) \tan (9 x) \frac{1}{\sec (9 x) + \tan (9 x)} + 9 \sec^{2} (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$

Step 7.3

Multiply $9 \sec (9 x) \tan (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$ .

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Step 7.3.1

Combine $\frac{1}{\sec (9 x) + \tan (9 x)}$ and $9$ .

$\frac{9}{\sec (9 x) + \tan (9 x)} \sec (9 x) \tan (9 x) + 9 \sec^{2} (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$

Step 7.3.2

Combine $\frac{9}{\sec (9 x) + \tan (9 x)}$ and $\sec (9 x)$ .

$\frac{9 \sec (9 x)}{\sec (9 x) + \tan (9 x)} \tan (9 x) + 9 \sec^{2} (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$

Step 7.3.3

Combine $\frac{9 \sec (9 x)}{\sec (9 x) + \tan (9 x)}$ and $\tan (9 x)$ .

$\frac{9 \sec (9 x) \tan (9 x)}{\sec (9 x) + \tan (9 x)} + 9 \sec^{2} (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$

Step 7.4

Multiply $9 \sec^{2} (9 x) \frac{1}{\sec (9 x) + \tan (9 x)}$ .

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Step 7.4.1

Combine $\frac{1}{\sec (9 x) + \tan (9 x)}$ and $9$ .

$\frac{9 \sec (9 x) \tan (9 x)}{\sec (9 x) + \tan (9 x)} + \frac{9}{\sec (9 x) + \tan (9 x)} \sec^{2} (9 x)$

Step 7.4.2

Combine $\frac{9}{\sec (9 x) + \tan (9 x)}$ and $\sec^{2} (9 x)$ .

$\frac{9 \sec (9 x) \tan (9 x)}{\sec (9 x) + \tan (9 x)} + \frac{9 \sec^{2} (9 x)}{\sec (9 x) + \tan (9 x)}$

Step 7.5

Combine the numerators over the common denominator.

$\frac{9 \sec (9 x) \tan (9 x) + 9 \sec^{2} (9 x)}{\sec (9 x) + \tan (9 x)}$

Step 7.6

Factor $9 \sec (9 x)$ out of $9 \sec (9 x) \tan (9 x) + 9 \sec^{2} (9 x)$ .

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Step 7.6.1

Factor $9 \sec (9 x)$ out of $9 \sec (9 x) \tan (9 x)$ .

$\frac{9 \sec (9 x) (\tan (9 x)) + 9 \sec^{2} (9 x)}{\sec (9 x) + \tan (9 x)}$

Step 7.6.2

Factor $9 \sec (9 x)$ out of $9 \sec^{2} (9 x)$ .

$\frac{9 \sec (9 x) (\tan (9 x)) + 9 \sec (9 x) (\sec (9 x))}{\sec (9 x) + \tan (9 x)}$

Step 7.6.3

Factor $9 \sec (9 x)$ out of $9 \sec (9 x) (\tan (9 x)) + 9 \sec (9 x) (\sec (9 x))$ .

$\frac{9 \sec (9 x) (\tan (9 x) + \sec (9 x))}{\sec (9 x) + \tan (9 x)}$