Calculus Examples

$f (x) = \log_{5} (\tan (2 x))$

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 $f (x) = \log_{5} (x)$ and $g (x) = \tan (2 x)$ .

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Step 1.1.1

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

$\frac{d}{d u_{1}} [\log_{5} (u_{1})] \frac{d}{d x} [\tan (2 x)]$

Step 1.1.2

The derivative of $\log_{5} (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1} \ln (5)}$ .

$\frac{1}{u_{1} \ln (5)} \frac{d}{d x} [\tan (2 x)]$

Step 1.1.3

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

$\frac{1}{\tan (2 x) \ln (5)} \frac{d}{d x} [\tan (2 x)]$

Step 1.2

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

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Step 1.2.1

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

$\frac{1}{\tan (2 x) \ln (5)} (\frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [2 x])$

Step 1.2.2

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

$\frac{1}{\tan (2 x) \ln (5)} (\sec^{2} (u_{2}) \frac{d}{d x} [2 x])$

Step 1.2.3

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

$\frac{1}{\tan (2 x) \ln (5)} (\sec^{2} (2 x) \frac{d}{d x} [2 x])$

Step 1.3

Differentiate.

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Step 1.3.1

Combine $\sec^{2} (2 x)$ and $\frac{1}{\tan (2 x) \ln (5)}$ .

$\frac{\sec^{2} (2 x)}{\tan (2 x) \ln (5)} \frac{d}{d x} [2 x]$

Step 1.3.2

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

$\frac{\sec^{2} (2 x)}{\tan (2 x) \ln (5)} (2 \frac{d}{d x} [x])$

Step 1.3.3

Combine $2$ and $\frac{\sec^{2} (2 x)}{\tan (2 x) \ln (5)}$ .

$\frac{2 \sec^{2} (2 x)}{\tan (2 x) \ln (5)} \frac{d}{d x} [x]$

Step 1.3.4

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

$\frac{2 \sec^{2} (2 x)}{\tan (2 x) \ln (5)} \cdot 1$

Step 1.3.5

Multiply $\frac{2 \sec^{2} (2 x)}{\tan (2 x) \ln (5)}$ by $1$ .

$f' (x) = \frac{2 \sec^{2} (2 x)}{\tan (2 x) \ln (5)}$

Step 2

Find the second derivative.

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Step 2.1

Since $\frac{2}{\ln (5)}$ is constant with respect to $x$ , the derivative of $\frac{2 \sec^{2} (2 x)}{\tan (2 x) \ln (5)}$ with respect to $x$ is $\frac{2}{\ln (5)} \frac{d}{d x} [\frac{\sec^{2} (2 x)}{\tan (2 x)}]$ .

$\frac{2}{\ln (5)} \frac{d}{d x} [\frac{\sec^{2} (2 x)}{\tan (2 x)}]$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = \sec^{2} (2 x)$ and $g (x) = \tan (2 x)$ .

$\frac{2}{\ln (5)} \cdot \frac{\tan (2 x) \frac{d}{d x} [\sec^{2} (2 x)] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 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 $g (x) = \sec (2 x)$ .

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Step 2.3.1

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

$\frac{2}{\ln (5)} \cdot \frac{\tan (2 x) (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\sec (2 x)]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 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 $n = 2$ .

$\frac{2}{\ln (5)} \cdot \frac{\tan (2 x) (2 u_{1} \frac{d}{d x} [\sec (2 x)]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.3.3

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

$\frac{2}{\ln (5)} \cdot \frac{\tan (2 x) (2 \sec (2 x) \frac{d}{d x} [\sec (2 x)]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.4

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

$\frac{2}{\ln (5)} \cdot \frac{2 \cdot \tan (2 x) \sec (2 x) \frac{d}{d x} [\sec (2 x)] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.5

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

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Step 2.5.1

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan (2 x) \sec (2 x) (\frac{d}{d u_{2}} [\sec (u_{2})] \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.5.2

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan (2 x) \sec (2 x) (\sec (u_{2}) \tan (u_{2}) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.5.3

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan (2 x) \sec (2 x) (\sec (2 x) \tan (2 x) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.6

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

$\frac{2}{\ln (5)} \cdot \frac{2 (\tan^{1} (2 x) \tan (2 x)) \sec (2 x) (\sec (2 x) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.7

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

$\frac{2}{\ln (5)} \cdot \frac{2 (\tan^{1} (2 x) \tan^{1} (2 x)) \sec (2 x) (\sec (2 x) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.8

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

$\frac{2}{\ln (5)} \cdot \frac{2 {\tan (2 x)}^{1 + 1} \sec (2 x) (\sec (2 x) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.9

Add $1$ and $1$ .

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) \sec (2 x) (\sec (2 x) \frac{d}{d x} [2 x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.10

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) (\sec^{1} (2 x) \sec (2 x)) \frac{d}{d x} [2 x] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.11

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) (\sec^{1} (2 x) \sec^{1} (2 x)) \frac{d}{d x} [2 x] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.12

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) {\sec (2 x)}^{1 + 1} \frac{d}{d x} [2 x] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.13

Differentiate.

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Step 2.13.1

Add $1$ and $1$ .

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) \sec^{2} (2 x) \frac{d}{d x} [2 x] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.13.2

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

$\frac{2}{\ln (5)} \cdot \frac{2 \tan^{2} (2 x) \sec^{2} (2 x) (2 \frac{d}{d x} [x]) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.13.3

Multiply $2$ by $2$ .

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) \frac{d}{d x} [x] - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.13.4

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) \cdot 1 - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.13.5

Multiply $4$ by $1$ .

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{2} (2 x) \frac{d}{d x} [\tan (2 x)]}{\tan^{2} (2 x)}$

Step 2.14

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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Step 2.14.1

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{2} (2 x) (\frac{d}{d u_{3}} [\tan (u_{3})] \frac{d}{d x} [2 x])}{\tan^{2} (2 x)}$

Step 2.14.2

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{2} (2 x) (\sec^{2} (u_{3}) \frac{d}{d x} [2 x])}{\tan^{2} (2 x)}$

Step 2.14.3

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{2} (2 x) (\sec^{2} (2 x) \frac{d}{d x} [2 x])}{\tan^{2} (2 x)}$

Step 2.15

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - {\sec (2 x)}^{2 + 2} \frac{d}{d x} [2 x]}{\tan^{2} (2 x)}$

Step 2.16

Add $2$ and $2$ .

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{4} (2 x) \frac{d}{d x} [2 x]}{\tan^{2} (2 x)}$

Step 2.17

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - \sec^{4} (2 x) (2 \frac{d}{d x} [x])}{\tan^{2} (2 x)}$

Step 2.18

Multiply $2$ by $- 1$ .

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - 2 \sec^{4} (2 x) \frac{d}{d x} [x]}{\tan^{2} (2 x)}$

Step 2.19

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

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - 2 \sec^{4} (2 x) \cdot 1}{\tan^{2} (2 x)}$

Step 2.20

Combine fractions.

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Step 2.20.1

Multiply $- 2$ by $1$ .

$\frac{2}{\ln (5)} \cdot \frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - 2 \sec^{4} (2 x)}{\tan^{2} (2 x)}$

Step 2.20.2

Multiply $\frac{2}{\ln (5)}$ by $\frac{4 \tan^{2} (2 x) \sec^{2} (2 x) - 2 \sec^{4} (2 x)}{\tan^{2} (2 x)}$ .

$\frac{2 (4 \tan^{2} (2 x) \sec^{2} (2 x) - 2 \sec^{4} (2 x))}{\ln (5) \tan^{2} (2 x)}$

Step 2.21

Simplify.

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Step 2.21.1

Apply the distributive property.

$\frac{2 (4 \tan^{2} (2 x) \sec^{2} (2 x)) + 2 (- 2 \sec^{4} (2 x))}{\ln (5) \tan^{2} (2 x)}$

Step 2.21.2

Simplify each term.

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Step 2.21.2.1

Multiply $4$ by $2$ .

$\frac{8 (\tan^{2} (2 x) \sec^{2} (2 x)) + 2 (- 2 \sec^{4} (2 x))}{\ln (5) \tan^{2} (2 x)}$

Step 2.21.2.2

Multiply $- 2$ by $2$ .

$\frac{8 \tan^{2} (2 x) \sec^{2} (2 x) - 4 \sec^{4} (2 x)}{\ln (5) \tan^{2} (2 x)}$

Step 2.21.3

Reorder terms.

$\frac{8 \sec^{2} (2 x) \tan^{2} (2 x) - 4 \sec^{4} (2 x)}{\tan^{2} (2 x) \ln (5)}$

Step 2.21.4

Factor $4 \sec^{2} (2 x)$ out of $8 \sec^{2} (2 x) \tan^{2} (2 x) - 4 \sec^{4} (2 x)$ .

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Step 2.21.4.1

Factor $4 \sec^{2} (2 x)$ out of $8 \sec^{2} (2 x) \tan^{2} (2 x)$ .

$\frac{4 \sec^{2} (2 x) (2 \tan^{2} (2 x)) - 4 \sec^{4} (2 x)}{\tan^{2} (2 x) \ln (5)}$

Step 2.21.4.2

Factor $4 \sec^{2} (2 x)$ out of $- 4 \sec^{4} (2 x)$ .

$\frac{4 \sec^{2} (2 x) (2 \tan^{2} (2 x)) + 4 \sec^{2} (2 x) (- \sec^{2} (2 x))}{\tan^{2} (2 x) \ln (5)}$

Step 2.21.4.3

Factor $4 \sec^{2} (2 x)$ out of $4 \sec^{2} (2 x) (2 \tan^{2} (2 x)) + 4 \sec^{2} (2 x) (- \sec^{2} (2 x))$ .

$f'' (x) = \frac{4 \sec^{2} (2 x) (2 \tan^{2} (2 x) - \sec^{2} (2 x))}{\tan^{2} (2 x) \ln (5)}$

Step 3

The second derivative of $f (x)$ with respect to $x$ is $\frac{4 \sec^{2} (2 x) (2 \tan^{2} (2 x) - \sec^{2} (2 x))}{\tan^{2} (2 x) \ln (5)}$ .

$\frac{4 \sec^{2} (2 x) (2 \tan^{2} (2 x) - \sec^{2} (2 x))}{\tan^{2} (2 x) \ln (5)}$