Calculus Examples

$y = \cos (8 x) \ln (\cos^{2} (8 x))$

Step 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 $f (x) = \cos (8 x)$ and $g (x) = \ln (\cos^{2} (8 x))$ .

$\cos (8 x) \frac{d}{d x} [\ln (\cos^{2} (8 x))] + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \cos^{2} (8 x)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $\cos^{2} (8 x)$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $\cos^{2} (8 x)$ .

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

Step 3

Convert from $\frac{1}{\cos^{2} (8 x)}$ to $\sec^{2} (8 x)$ .

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

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \cos (8 x)$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $\cos (8 x)$ .

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

Step 4.2

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

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

Step 4.3

Replace all occurrences of $u_{2}$ with $\cos (8 x)$ .

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

Step 5

Raise $\cos (8 x)$ to the power of $1$ .

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

Step 6

Raise $\cos (8 x)$ to the power of $1$ .

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

Step 7

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

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

Step 8

Simplify the expression.

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Step 8.1

Add $1$ and $1$ .

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

Step 8.2

Move $2$ to the left of $\sec^{2} (8 x)$ .

$2 \cdot \sec^{2} (8 x) \cos^{2} (8 x) \frac{d}{d x} [\cos (8 x)] + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 9

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 8 x$ .

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Step 9.1

To apply the Chain Rule, set $u_{3}$ as $8 x$ .

$2 \sec^{2} (8 x) \cos^{2} (8 x) (\frac{d}{d u_{3}} [\cos (u_{3})] \frac{d}{d x} [8 x]) + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 9.2

The derivative of $\cos (u_{3})$ with respect to $u_{3}$ is $- \sin (u_{3})$ .

$2 \sec^{2} (8 x) \cos^{2} (8 x) (- \sin (u_{3}) \frac{d}{d x} [8 x]) + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 9.3

Replace all occurrences of $u_{3}$ with $8 x$ .

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

Step 10

Differentiate.

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Step 10.1

Multiply $- 1$ by $2$ .

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

Step 10.2

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

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

Step 10.3

Multiply $8$ by $- 2$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) \frac{d}{d x} [x] + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 10.4

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

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) \cdot 1 + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 10.5

Multiply $- 16$ by $1$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) \frac{d}{d x} [\cos (8 x)]$

Step 11

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 $g (x) = 8 x$ .

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Step 11.1

To apply the Chain Rule, set $u_{4}$ as $8 x$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (\frac{d}{d u_{4}} [\cos (u_{4})] \frac{d}{d x} [8 x])$

Step 11.2

The derivative of $\cos (u_{4})$ with respect to $u_{4}$ is $- \sin (u_{4})$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- \sin (u_{4}) \frac{d}{d x} [8 x])$

Step 11.3

Replace all occurrences of $u_{4}$ with $8 x$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- \sin (8 x) \frac{d}{d x} [8 x])$

Step 12

Differentiate.

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Step 12.1

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

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- \sin (8 x) (8 \frac{d}{d x} [x]))$

Step 12.2

Multiply $8$ by $- 1$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- 8 \sin (8 x) \frac{d}{d x} [x])$

Step 12.3

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

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- 8 \sin (8 x) \cdot 1)$

Step 12.4

Multiply $- 8$ by $1$ .

$- 16 \sec^{2} (8 x) \cos^{2} (8 x) \sin (8 x) + \ln (\cos^{2} (8 x)) (- 8 \sin (8 x))$

Step 13

Simplify.

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Step 13.1

Reorder terms.

$- 16 \cos^{2} (8 x) \sec^{2} (8 x) \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2

Simplify each term.

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Step 13.2.1

Rewrite $\sec (8 x)$ in terms of sines and cosines.

$- 16 \cos^{2} (8 x) {(\frac{1}{\cos (8 x)})}^{2} \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.2

Apply the product rule to $\frac{1}{\cos (8 x)}$ .

$- 16 \cos^{2} (8 x) \frac{1^{2}}{\cos^{2} (8 x)} \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.3

Cancel the common factor of $\cos^{2} (8 x)$ .

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Step 13.2.3.1

Factor $\cos^{2} (8 x)$ out of $- 16 \cos^{2} (8 x)$ .

$\cos^{2} (8 x) \cdot - 16 \frac{1^{2}}{\cos^{2} (8 x)} \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.3.2

Cancel the common factor.

$\cos^{2} (8 x) \cdot - 16 \frac{1^{2}}{\cos^{2} (8 x)} \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.3.3

Rewrite the expression.

$- 16 \cdot 1^{2} \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.4

One to any power is one.

$- 16 \cdot 1 \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$

Step 13.2.5

Multiply $- 16$ by $1$ .

$- 16 \sin (8 x) - 8 \sin (8 x) \ln (\cos^{2} (8 x))$