Calculus Examples

\ln (\tan (x))

$\ln (\tan (x))$

Step 1

Find the asymptotes.

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Step 1.1

For any

y = \tan (x)

$y = \tan (x)$ , vertical asymptotes occur at

x = \frac{π}{2} + n π

$x = \frac{π}{2} + n π$ , where

n

$n$ is an integer. Use the basic period for

y = \tan (x)

$y = \tan (x)$ ,

(- \frac{π}{2}, \frac{π}{2})

$(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for

y = \ln (\tan (x))

$y = \ln (\tan (x))$ . Set the inside of the tangent function,

b x + c

$b x + c$ , for

y = a \tan (b x + c) + d

$y = a \tan (b x + c) + d$ equal to

- \frac{π}{2}

$- \frac{π}{2}$ to find where the vertical asymptote occurs for

y = \ln (\tan (x))

$y = \ln (\tan (x))$ .

\tan (x) = - \frac{π}{2}

$\tan (x) = - \frac{π}{2}$

Step 1.2

Solve for

x

$x$ .

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Step 1.2.1

Take the inverse tangent of both sides of the equation to extract

x

$x$ from inside the tangent.

x = \arctan (- \frac{π}{2})

$x = \arctan (- \frac{π}{2})$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Evaluate

\arctan (- \frac{π}{2})

$\arctan (- \frac{π}{2})$ .

x = - 1.00388482

$x = - 1.00388482$

x = - 1.00388482

$x = - 1.00388482$

Step 1.2.3

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from

π

$π$ to find the solution in the third quadrant.

x = - 1.00388482 - (3.14159265)

$x = - 1.00388482 - (3.14159265)$

Step 1.2.4

Simplify the expression to find the second solution.

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Step 1.2.4.1

Add

2 π

$2 π$ to

- 1.00388482 - (3.14159265)

$- 1.00388482 - (3.14159265)$ .

x = - 1.00388482 - (3.14159265) + 2 π

$x = - 1.00388482 - (3.14159265) + 2 π$

Step 1.2.4.2

The resulting angle of

2.13770783

$2.13770783$ is positive and coterminal with

- 1.00388482 - (3.14159265)

$- 1.00388482 - (3.14159265)$ .

x = 2.13770783

$x = 2.13770783$

x = 2.13770783

$x = 2.13770783$

Step 1.2.5

Find the period of

\tan (x)

$\tan (x)$ .

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Step 1.2.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 1.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 1.2.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 1.2.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 1.2.6

Add

π

$π$ to every negative angle to get positive angles.

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Step 1.2.6.1

Add

π

$π$ to

- 1.00388482

$- 1.00388482$ to find the positive angle.

- 1.00388482 + π

$- 1.00388482 + π$

Step 1.2.6.2

Replace with decimal approximation.

3.14159265 - 1.00388482

$3.14159265 - 1.00388482$

Step 1.2.6.3

Subtract

1.00388482

$1.00388482$ from

3.14159265

$3.14159265$ .

2.13770783

$2.13770783$

Step 1.2.6.4

List the new angles.

x = 2.13770783

$x = 2.13770783$

x = 2.13770783

$x = 2.13770783$

Step 1.2.7

The period of the

\tan (x)

$\tan (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = 2.13770783 + π n, 2.13770783 + π n

$x = 2.13770783 + π n, 2.13770783 + π n$ , for any integer

n

$n$

Step 1.2.8

Consolidate

2.13770783 + π n

$2.13770783 + π n$ and

2.13770783 + π n

$2.13770783 + π n$ to

2.13770783 + π n

$2.13770783 + π n$ .

x = 2.13770783 + π n

$x = 2.13770783 + π n$ , for any integer

n

$n$

x = 2.13770783 + π n

$x = 2.13770783 + π n$ , for any integer

n

$n$

Step 1.3

Set the inside of the tangent function

\tan (x)

$\tan (x)$ equal to

\frac{π}{2}

$\frac{π}{2}$ .

\tan (x) = \frac{π}{2}

$\tan (x) = \frac{π}{2}$

Step 1.4

Solve for

x

$x$ .

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Step 1.4.1

Take the inverse tangent of both sides of the equation to extract

x

$x$ from inside the tangent.

x = \arctan (\frac{π}{2})

$x = \arctan (\frac{π}{2})$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Evaluate

\arctan (\frac{π}{2})

$\arctan (\frac{π}{2})$ .

x = 1.00388482

$x = 1.00388482$

x = 1.00388482

$x = 1.00388482$

Step 1.4.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from

π

$π$ to find the solution in the fourth quadrant.

x = (3.14159265) + 1.00388482

$x = (3.14159265) + 1.00388482$

Step 1.4.4

Solve for

x

$x$ .

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Step 1.4.4.1

Remove parentheses.

x = 3.14159265 + 1.00388482

$x = 3.14159265 + 1.00388482$

Step 1.4.4.2

Remove parentheses.

x = (3.14159265) + 1.00388482

$x = (3.14159265) + 1.00388482$

Step 1.4.4.3

Add

3.14159265

$3.14159265$ and

1.00388482

$1.00388482$ .

x = 4.14547747

$x = 4.14547747$

x = 4.14547747

$x = 4.14547747$

Step 1.4.5

Find the period of

\tan (x)

$\tan (x)$ .

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Step 1.4.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 1.4.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 1.4.5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 1.4.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 1.4.6

The period of the

\tan (x)

$\tan (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = 1.00388482 + π n, 4.14547747 + π n

$x = 1.00388482 + π n, 4.14547747 + π n$ , for any integer

n

$n$

Step 1.4.7

Consolidate

1.00388482 + π n

$1.00388482 + π n$ and

4.14547747 + π n

$4.14547747 + π n$ to

1.00388482 + π n

$1.00388482 + π n$ .

x = 1.00388482 + π n

$x = 1.00388482 + π n$ , for any integer

n

$n$

x = 1.00388482 + π n

$x = 1.00388482 + π n$ , for any integer

n

$n$

Step 1.5

The basic period for

y = \ln (\tan (x))

$y = \ln (\tan (x))$ will occur at

(2.13770783 + π n, 1.00388482 + π n)

$(2.13770783 + π n, 1.00388482 + π n)$ , where

2.13770783 + π n

$2.13770783 + π n$ and

1.00388482 + π n

$1.00388482 + π n$ are vertical asymptotes.

(2.13770783 + π n, 1.00388482 + π n)

$(2.13770783 + π n, 1.00388482 + π n)$

Step 1.6

Find the period

\frac{π}{| b |}

$\frac{π}{| b |}$ to find where the vertical asymptotes exist.

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Step 1.6.1

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 1.6.2

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 1.7

The vertical asymptotes for

y = \ln (\tan (x))

$y = \ln (\tan (x))$ occur at

2.13770783 + π n

$2.13770783 + π n$ ,

1.00388482 + π n

$1.00388482 + π n$ , and every

π n

$π n$ , where

n

$n$ is an integer.

π n

$π n$

Step 1.8

There are only vertical asymptotes for tangent and cotangent functions.

Vertical Asymptotes:

x = 2.13770783 + π n + π n

$x = 2.13770783 + π n + π n$ for any integer

n

$n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes:

x = 2.13770783 + π n + π n

$x = 2.13770783 + π n + π n$ for any integer

n

$n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 2

Find the point at

x = 1

$x = 1$ .

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Step 2.1

Replace the variable

x

$x$ with

1

$1$ in the expression.

f (1) = \ln (\tan (1))

$f (1) = \ln (\tan (1))$

Step 2.2

Simplify the result.

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Step 2.2.1

Evaluate

\tan (1)

$\tan (1)$ .

f (1) = \ln (1.55740772)

$f (1) = \ln (1.55740772)$

Step 2.2.2

The final answer is

\ln (1.55740772)

$\ln (1.55740772)$ .

\ln (1.55740772)

$\ln (1.55740772)$

\ln (1.55740772)

$\ln (1.55740772)$

\ln (1.55740772)

$\ln (1.55740772)$

Step 3

Find the point at

x = 4

$x = 4$ .

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Step 3.1

Replace the variable

x

$x$ with

4

$4$ in the expression.

f (4) = \ln (\tan (4))

$f (4) = \ln (\tan (4))$

Step 3.2

Simplify the result.

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Step 3.2.1

Evaluate

\tan (4)

$\tan (4)$ .

f (4) = \ln (1.15782128)

$f (4) = \ln (1.15782128)$

Step 3.2.2

The final answer is

\ln (1.15782128)

$\ln (1.15782128)$ .

\ln (1.15782128)

$\ln (1.15782128)$

\ln (1.15782128)

$\ln (1.15782128)$

\ln (1.15782128)

$\ln (1.15782128)$

Step 4

Find the point at

x = 7

$x = 7$ .

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Step 4.1

Replace the variable

x

$x$ with

7

$7$ in the expression.

f (7) = \ln (\tan (7))

$f (7) = \ln (\tan (7))$

Step 4.2

Simplify the result.

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Step 4.2.1

Evaluate

\tan (7)

$\tan (7)$ .

f (7) = \ln (0.87144798)

$f (7) = \ln (0.87144798)$

Step 4.2.2

The final answer is

\ln (0.87144798)

$\ln (0.87144798)$ .

\ln (0.87144798)

$\ln (0.87144798)$

\ln (0.87144798)

$\ln (0.87144798)$

\ln (0.87144798)

$\ln (0.87144798)$

Step 5

The log function can be graphed using the vertical asymptote at

x = 2.13770783 + π n + π n (f o r) (a n y) (i n t e g e r) n

$x = 2.13770783 + π n + π n (f o r) (a n y) (i n t e g e r) n$ and the points

(1, 0.44302272), (4, 0.14654003), (7, - 0.1375991)

$(1, 0.44302272), (4, 0.14654003), (7, - 0.1375991)$ .

Vertical Asymptote:

x = 2.13770783 + π n + π n (f o r) (a n y) (i n t e g e r) n

$x = 2.13770783 + π n + π n (f o r) (a n y) (i n t e g e r) n$

\begin{array}{c} x & y \\ 1 & 0.443 \\ 4 & 0.147 \\ 7 & - 0.138 \end{array}

$\begin{array}{c} x & y \\ 1 & 0.443 \\ 4 & 0.147 \\ 7 & - 0.138 \end{array}$

Step 6