Calculus Examples

\csc (x)

$\csc (x)$

Step 1

Find the derivative.

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Step 1.1

The derivative of

\csc (x)

$\csc (x)$ with respect to

x

$x$ is

- \csc (x) \cot (x)

$- \csc (x) \cot (x)$ .

- \csc (x) \cot (x)

$- \csc (x) \cot (x)$

Step 1.2

Reorder the factors of

- \csc (x) \cot (x)

$- \csc (x) \cot (x)$ .

- \cot (x) \csc (x)

$- \cot (x) \csc (x)$

- \cot (x) \csc (x)

$- \cot (x) \csc (x)$

Step 2

Set the derivative equal to

0

$0$ then solve the equation

- \cot (x) \csc (x) = 0

$- \cot (x) \csc (x) = 0$ .

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Step 2.1

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

\cot (x) = 0

$\cot (x) = 0$

\csc (x) = 0

$\csc (x) = 0$

Step 2.2

Set

\cot (x)

$\cot (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 2.2.1

Set

\cot (x)

$\cot (x)$ equal to

0

$0$ .

\cot (x) = 0

$\cot (x) = 0$

Step 2.2.2

Solve

\cot (x) = 0

$\cot (x) = 0$ for

x

$x$ .

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Step 2.2.2.1

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

x

$x$ from inside the cotangent.

x = arccot (0)

$x = arccot (0)$

Step 2.2.2.2

Simplify the right side.

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Step 2.2.2.2.1

The exact value of

arccot (0)

$arccot (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 2.2.2.3

The cotangent 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 = π + \frac{π}{2}

$x = π + \frac{π}{2}$

Step 2.2.2.4

Simplify

π + \frac{π}{2}

$π + \frac{π}{2}$ .

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Step 2.2.2.4.1

To write

π

$π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

x = π \cdot \frac{2}{2} + \frac{π}{2}

$x = π \cdot \frac{2}{2} + \frac{π}{2}$

Step 2.2.2.4.2

Combine fractions.

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Step 2.2.2.4.2.1

Combine

π

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

x = \frac{π \cdot 2}{2} + \frac{π}{2}

$x = \frac{π \cdot 2}{2} + \frac{π}{2}$

Step 2.2.2.4.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 2 + π}{2}

$x = \frac{π \cdot 2 + π}{2}$

x = \frac{π \cdot 2 + π}{2}

$x = \frac{π \cdot 2 + π}{2}$

Step 2.2.2.4.3

Simplify the numerator.

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Step 2.2.2.4.3.1

Move

2

$2$ to the left of

π

$π$ .

x = \frac{2 \cdot π + π}{2}

$x = \frac{2 \cdot π + π}{2}$

Step 2.2.2.4.3.2

Add

2 π

$2 π$ and

π

$π$ .

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 2.2.2.5

Find the period of

\cot (x)

$\cot (x)$ .

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Step 2.2.2.5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 2.2.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 2.2.2.5.3

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 2.2.2.5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 2.2.2.6

The period of the

\cot (x)

$\cot (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = \frac{π}{2} + π n, \frac{3 π}{2} + π n

$x = \frac{π}{2} + π n, \frac{3 π}{2} + π n$ , for any integer

n

$n$

x = \frac{π}{2} + π n, \frac{3 π}{2} + π n

$x = \frac{π}{2} + π n, \frac{3 π}{2} + π n$ , for any integer

n

$n$

x = \frac{π}{2} + π n, \frac{3 π}{2} + π n

$x = \frac{π}{2} + π n, \frac{3 π}{2} + π n$ , for any integer

n

$n$

Step 2.3

Set

\csc (x)

$\csc (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 2.3.1

Set

\csc (x)

$\csc (x)$ equal to

0

$0$ .

\csc (x) = 0

$\csc (x) = 0$

Step 2.3.2

The range of cosecant is

y \leq - 1

$y \leq - 1$ and

y \geq 1

$y \geq 1$ . Since

0

$0$ does not fall in this range, there is no solution.

No solution

Step 2.4

The final solution is all the values that make

- \cot (x) \csc (x) = 0

$- \cot (x) \csc (x) = 0$ true.

x = \frac{π}{2} + π n, \frac{3 π}{2} + π n

$x = \frac{π}{2} + π n, \frac{3 π}{2} + π n$ , for any integer

n

$n$

Step 2.5

Consolidate the answers.

x = \frac{π}{2} + π n

$x = \frac{π}{2} + π n$ , for any integer

n

$n$

x = \frac{π}{2} + π n

$x = \frac{π}{2} + π n$ , for any integer

n

$n$

Step 3

Solve the original function

f (x) = \csc (x)

$f (x) = \csc (x)$ at

x = \frac{π}{2}

$x = \frac{π}{2}$ .

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Step 3.1

Replace the variable

x

$x$ with

\frac{π}{2}

$\frac{π}{2}$ in the expression.

f (\frac{π}{2}) = \csc (\frac{π}{2})

$f (\frac{π}{2}) = \csc (\frac{π}{2})$

Step 3.2

Simplify the result.

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Step 3.2.1

The exact value of

\csc (\frac{π}{2})

$\csc (\frac{π}{2})$ is

1

$1$ .

f (\frac{π}{2}) = 1

$f (\frac{π}{2}) = 1$

Step 3.2.2

The final answer is

1

$1$ .

1

$1$

1

$1$

1

$1$

Step 4

Solve the original function

f (x) = \csc (x)

$f (x) = \csc (x)$ at

x = \frac{π}{2} + π

$x = \frac{π}{2} + π$ .

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Step 4.1

Replace the variable

x

$x$ with

\frac{π}{2} + π

$\frac{π}{2} + π$ in the expression.

f (\frac{π}{2} + π) = \csc (\frac{π}{2} + π)

$f (\frac{π}{2} + π) = \csc (\frac{π}{2} + π)$

Step 4.2

Simplify the result.

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Step 4.2.1

To write

π

$π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

f (\frac{π}{2} + π) = \csc (\frac{π}{2} + π \cdot \frac{2}{2})

$f (\frac{π}{2} + π) = \csc (\frac{π}{2} + π \cdot \frac{2}{2})$

Step 4.2.2

Combine fractions.

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Step 4.2.2.1

Combine

π

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

f (\frac{π}{2} + π) = \csc (\frac{π}{2} + \frac{π \cdot 2}{2})

$f (\frac{π}{2} + π) = \csc (\frac{π}{2} + \frac{π \cdot 2}{2})$

Step 4.2.2.2

Combine the numerators over the common denominator.

f (\frac{π}{2} + π) = \csc (\frac{π + π \cdot 2}{2})

$f (\frac{π}{2} + π) = \csc (\frac{π + π \cdot 2}{2})$

f (\frac{π}{2} + π) = \csc (\frac{π + π \cdot 2}{2})

$f (\frac{π}{2} + π) = \csc (\frac{π + π \cdot 2}{2})$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Move

2

$2$ to the left of

π

$π$ .

f (\frac{π}{2} + π) = \csc (\frac{π + 2 \cdot π}{2})

$f (\frac{π}{2} + π) = \csc (\frac{π + 2 \cdot π}{2})$

Step 4.2.3.2

Add

π

$π$ and

2 π

$2 π$ .

f (\frac{π}{2} + π) = \csc (\frac{3 π}{2})

$f (\frac{π}{2} + π) = \csc (\frac{3 π}{2})$

f (\frac{π}{2} + π) = \csc (\frac{3 π}{2})

$f (\frac{π}{2} + π) = \csc (\frac{3 π}{2})$

Step 4.2.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosecant is negative in the fourth quadrant.

f (\frac{π}{2} + π) = - \csc (\frac{π}{2})

$f (\frac{π}{2} + π) = - \csc (\frac{π}{2})$

Step 4.2.5

The exact value of

\csc (\frac{π}{2})

$\csc (\frac{π}{2})$ is

1

$1$ .

f (\frac{π}{2} + π) = - 1 \cdot 1

$f (\frac{π}{2} + π) = - 1 \cdot 1$

Step 4.2.6

Multiply

- 1

$- 1$ by

1

$1$ .

f (\frac{π}{2} + π) = - 1

$f (\frac{π}{2} + π) = - 1$

Step 4.2.7

The final answer is

- 1

$- 1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 5

The horizontal tangent line on function

f (x) = \csc (x)

$f (x) = \csc (x)$ is

y = 1, y = - 1

$y = 1, y = - 1$ .

y = 1, y = - 1

$y = 1, y = - 1$

Step 6