Calculus Examples

$\ln (\tan (x))$

Step 1

Find the asymptotes.

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Step 1.1

For any $y = \tan (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = \tan (x)$ , $(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for $y = \ln (\tan (x))$ . Set the inside of the tangent function, $b x + c$ , for $y = a \tan (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = \ln (\tan (x))$ .

$\tan (x) = - \frac{π}{2}$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (- \frac{π}{2})$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Evaluate $\arctan (- \frac{π}{2})$ .

$x = - 1.00388482$

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

$x = - 1.00388482 - (3.14159265)$

Step 1.2.4

Simplify the expression to find the second solution.

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Step 1.2.4.1

Add $2 π$ to $- 1.00388482 - (3.14159265)$ .

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

Step 1.2.4.2

The resulting angle of $2.13770783$ is positive and coterminal with $- 1.00388482 - (3.14159265)$ .

$x = 2.13770783$

Step 1.2.5

Find the period of $\tan (x)$ .

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Step 1.2.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 1.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 1.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.2.5.4

Divide $π$ by $1$ .

$π$

Step 1.2.6

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

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Step 1.2.6.1

Add $π$ to $- 1.00388482$ to find the positive angle.

$- 1.00388482 + π$

Step 1.2.6.2

Replace with decimal approximation.

$3.14159265 - 1.00388482$

Step 1.2.6.3

Subtract $1.00388482$ from $3.14159265$ .

$2.13770783$

Step 1.2.6.4

List the new angles.

$x = 2.13770783$

Step 1.2.7

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

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

Step 1.2.8

Consolidate $2.13770783 + π n$ and $2.13770783 + π n$ to $2.13770783 + π n$ .

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

Step 1.3

Set the inside of the tangent function $\tan (x)$ equal to $\frac{π}{2}$ .

$\tan (x) = \frac{π}{2}$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (\frac{π}{2})$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Evaluate $\arctan (\frac{π}{2})$ .

$x = 1.00388482$

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

$x = (3.14159265) + 1.00388482$

Step 1.4.4

Solve for $x$ .

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Step 1.4.4.1

Remove parentheses.

$x = 3.14159265 + 1.00388482$

Step 1.4.4.2

Remove parentheses.

$x = (3.14159265) + 1.00388482$

Step 1.4.4.3

Add $3.14159265$ and $1.00388482$ .

$x = 4.14547747$

Step 1.4.5

Find the period of $\tan (x)$ .

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Step 1.4.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 1.4.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 1.4.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.4.5.4

Divide $π$ by $1$ .

$π$

Step 1.4.6

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

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

Step 1.4.7

Consolidate $1.00388482 + π n$ and $4.14547747 + π n$ to $1.00388482 + π n$ .

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

Step 1.5

The basic period for $y = \ln (\tan (x))$ will occur at $(2.13770783 + π n, 1.00388482 + π n)$ , where $2.13770783 + π n$ and $1.00388482 + π n$ are vertical asymptotes.

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

Step 1.6

Find the period $\frac{π}{| b |}$ 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 $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.6.2

Divide $π$ by $1$ .

$π$

Step 1.7

The vertical asymptotes for $y = \ln (\tan (x))$ occur at $2.13770783 + π n$ , $1.00388482 + π n$ , and every $π n$ , where $n$ is an integer.

$π n$

Step 1.8

There are only vertical asymptotes for tangent and cotangent functions.

Vertical Asymptotes: $x = 2.13770783 + π n + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 2.13770783 + π n + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 2

Find the point at $x = 1$ .

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Step 2.1

Replace the variable $x$ with $1$ in the expression.

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

Step 2.2

Simplify the result.

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Step 2.2.1

Evaluate $\tan (1)$ .

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

Step 2.2.2

The final answer is $\ln (1.55740772)$ .

$\ln (1.55740772)$

Step 3

Find the point at $x = 4$ .

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Step 3.1

Replace the variable $x$ with $4$ in the expression.

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

Step 3.2

Simplify the result.

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Step 3.2.1

Evaluate $\tan (4)$ .

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

Step 3.2.2

The final answer is $\ln (1.15782128)$ .

$\ln (1.15782128)$

Step 4

Find the point at $x = 7$ .

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Step 4.1

Replace the variable $x$ with $7$ in the expression.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Evaluate $\tan (7)$ .

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

Step 4.2.2

The final answer is $\ln (0.87144798)$ .

$\ln (0.87144798)$

Step 5

The log function can be graphed using the vertical asymptote at $x = 2.13770783 + π n + π n (f o r) (a n y) (i n t e g e r) n$ and the points $(1, 0.44302272), (4, 0.14654003), (7, - 0.1375991)$ .

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

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

Step 6