Calculus Examples

f (x) = - 5 cot (x) - \frac{5 π}{2} - 4

$f (x) = - 5 \cot (x) - \frac{5 π}{2} - 4$ at

x = - \frac{π}{4}

$x = - \frac{π}{4}$

Step 1

Find the corresponding

y

$y$ -value to

x = - \frac{π}{4}

$x = - \frac{π}{4}$ .

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Step 1.1

Substitute

- \frac{π}{4}

$- \frac{π}{4}$ in for

x

$x$ .

y = - 5 cot (- \frac{π}{4}) - \frac{5 π}{2} - 4

$y = - 5 \cot (- \frac{π}{4}) - \frac{5 π}{2} - 4$

Step 1.2

Solve for

y

$y$ .

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Step 1.2.1

Remove parentheses.

y = - 5 cot (- \frac{π}{4}) - \frac{5 π}{2} - 4

$y = - 5 \cot (- \frac{π}{4}) - \frac{5 π}{2} - 4$

Step 1.2.2

Simplify

- 5 cot (- \frac{π}{4}) - \frac{5 π}{2} - 4

$- 5 \cot (- \frac{π}{4}) - \frac{5 π}{2} - 4$ .

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Add full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

y = - 5 cot (\frac{7 π}{4}) - \frac{5 π}{2} - 4

$y = - 5 \cot (\frac{7 π}{4}) - \frac{5 π}{2} - 4$

Step 1.2.2.1.2

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

y = - 5 (- cot (\frac{π}{4})) - \frac{5 π}{2} - 4

$y = - 5 (- \cot (\frac{π}{4})) - \frac{5 π}{2} - 4$

Step 1.2.2.1.3

The exact value of

cot (\frac{π}{4})

$\cot (\frac{π}{4})$ is

1

$1$ .

y = - 5 (- 1 \cdot 1) - \frac{5 π}{2} - 4

$y = - 5 (- 1 \cdot 1) - \frac{5 π}{2} - 4$

Step 1.2.2.1.4

Multiply

- 5 (- 1 \cdot 1)

$- 5 (- 1 \cdot 1)$ .

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Step 1.2.2.1.4.1

Multiply

- 1

$- 1$ by

1

$1$ .

y = - 5 \cdot - 1 - \frac{5 π}{2} - 4

$y = - 5 \cdot - 1 - \frac{5 π}{2} - 4$

Step 1.2.2.1.4.2

Multiply

- 5

$- 5$ by

- 1

$- 1$ .

y = 5 - \frac{5 π}{2} - 4

$y = 5 - \frac{5 π}{2} - 4$

y = 5 - \frac{5 π}{2} - 4

$y = 5 - \frac{5 π}{2} - 4$

y = 5 - \frac{5 π}{2} - 4

$y = 5 - \frac{5 π}{2} - 4$

Step 1.2.2.2

Subtract

4

$4$ from

5

$5$ .

y = 1 - \frac{5 π}{2}

$y = 1 - \frac{5 π}{2}$

y = 1 - \frac{5 π}{2}

$y = 1 - \frac{5 π}{2}$

y = 1 - \frac{5 π}{2}

$y = 1 - \frac{5 π}{2}$

y = 1 - \frac{5 π}{2}

$y = 1 - \frac{5 π}{2}$

Step 2

Find the first derivative and evaluate at

x = - \frac{π}{4}

$x = - \frac{π}{4}$ and

y = 1 - \frac{5 π}{2}

$y = 1 - \frac{5 π}{2}$ to find the slope of the tangent line.

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Step 2.1

By the Sum Rule, the derivative of

- 5 cot (x) - \frac{5 π}{2} - 4

$- 5 \cot (x) - \frac{5 π}{2} - 4$ with respect to

x

$x$ is

\frac{d}{d x} [- 5 cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$\frac{d}{d x} [- 5 \cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$ .

\frac{d}{d x} [- 5 cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$\frac{d}{d x} [- 5 \cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$

Step 2.2

Evaluate

\frac{d}{d x} [- 5 cot (x)]

$\frac{d}{d x} [- 5 \cot (x)]$ .

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Step 2.2.1

Since

- 5

$- 5$ is constant with respect to

x

$x$ , the derivative of

- 5 cot (x)

$- 5 \cot (x)$ with respect to

x

$x$ is

- 5 \frac{d}{d x} [cot (x)]

$- 5 \frac{d}{d x} [\cot (x)]$ .

- 5 \frac{d}{d x} [cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$- 5 \frac{d}{d x} [\cot (x)] + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$

Step 2.2.2

The derivative of

cot (x)

$\cot (x)$ with respect to

x

$x$ is

- {csc}^{2} (x)

$- \csc^{2} (x)$ .

- 5 (- {csc}^{2} (x)) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$- 5 (- \csc^{2} (x)) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$

Step 2.2.3

Multiply

- 1

$- 1$ by

- 5

$- 5$ .

5 {csc}^{2} (x) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$5 \csc^{2} (x) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$

5 {csc}^{2} (x) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]

$5 \csc^{2} (x) + \frac{d}{d x} [- \frac{5 π}{2}] + \frac{d}{d x} [- 4]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since

- \frac{5 π}{2}

$- \frac{5 π}{2}$ is constant with respect to

x

$x$ , the derivative of

- \frac{5 π}{2}

$- \frac{5 π}{2}$ with respect to

x

$x$ is

0

$0$ .

5 {csc}^{2} (x) + 0 + \frac{d}{d x} [- 4]

$5 \csc^{2} (x) + 0 + \frac{d}{d x} [- 4]$

Step 2.3.2

Since

- 4

$- 4$ is constant with respect to

x

$x$ , the derivative of

- 4

$- 4$ with respect to

x

$x$ is

0

$0$ .

5 {csc}^{2} (x) + 0 + 0

$5 \csc^{2} (x) + 0 + 0$

5 {csc}^{2} (x) + 0 + 0

$5 \csc^{2} (x) + 0 + 0$

Step 2.4

Combine terms.

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Step 2.4.1

Add

5 {csc}^{2} (x)

$5 \csc^{2} (x)$ and

0

$0$ .

5 {csc}^{2} (x) + 0

$5 \csc^{2} (x) + 0$

Step 2.4.2

Add

5 {csc}^{2} (x)

$5 \csc^{2} (x)$ and

0

$0$ .

5 {csc}^{2} (x)

$5 \csc^{2} (x)$

5 {csc}^{2} (x)

$5 \csc^{2} (x)$

Step 2.5

Evaluate the derivative at

x = - \frac{π}{4}

$x = - \frac{π}{4}$ .

5 {csc}^{2} (- \frac{π}{4})

$5 \csc^{2} (- \frac{π}{4})$

Step 2.6

Simplify.

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Step 2.6.1

Add full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

5 {csc}^{2} (\frac{7 π}{4})

$5 \csc^{2} (\frac{7 π}{4})$

Step 2.6.2

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.

5 {(- csc (\frac{π}{4}))}^{2}

$5 {(- \csc (\frac{π}{4}))}^{2}$

Step 2.6.3

The exact value of

csc (\frac{π}{4})

$\csc (\frac{π}{4})$ is

\sqrt{2}

$\sqrt{2}$ .

5 {(- \sqrt{2})}^{2}

$5 {(- \sqrt{2})}^{2}$

Step 2.6.4

Simplify the expression.

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Step 2.6.4.1

Apply the product rule to

- \sqrt{2}

$- \sqrt{2}$ .

5 ({(- 1)}^{2} {\sqrt{2}}^{2})

$5 ({(- 1)}^{2} {\sqrt{2}}^{2})$

Step 2.6.4.2

Raise

- 1

$- 1$ to the power of

2

$2$ .

5 (1 {\sqrt{2}}^{2})

$5 (1 {\sqrt{2}}^{2})$

Step 2.6.4.3

Multiply

{\sqrt{2}}^{2}

${\sqrt{2}}^{2}$ by

1

$1$ .

5 {\sqrt{2}}^{2}

$5 {\sqrt{2}}^{2}$

5 {\sqrt{2}}^{2}

$5 {\sqrt{2}}^{2}$

Step 2.6.5

Rewrite

{\sqrt{2}}^{2}

${\sqrt{2}}^{2}$ as

2

$2$ .

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Step 2.6.5.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{2}

$\sqrt{2}$ as

2^{\frac{1}{2}}

$2^{\frac{1}{2}}$ .

5 {(2^{\frac{1}{2}})}^{2}

$5 {(2^{\frac{1}{2}})}^{2}$

Step 2.6.5.2

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

5 \cdot 2^{\frac{1}{2} \cdot 2}

$5 \cdot 2^{\frac{1}{2} \cdot 2}$

Step 2.6.5.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

5 \cdot 2^{\frac{2}{2}}

$5 \cdot 2^{\frac{2}{2}}$

Step 2.6.5.4

Cancel the common factor of

2

$2$ .

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Step 2.6.5.4.1

Cancel the common factor.

$5 \cdot 2^{\frac{2}{2}}$

Step 2.6.5.4.2

Rewrite the expression.

$5 \cdot 2^{1}$

Step 2.6.5.5

Evaluate the exponent.

$5 \cdot 2$

Step 2.6.6

Multiply

$5$ by

$2$ .

$10$

Step 3

Plug the slope and point values into the point-slope formula and solve for

$y$ .

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Step 3.1

Use the slope

$10$ and a given point

$(- \frac{π}{4}, 1 - \frac{5 π}{2})$ to substitute for

$x_{1}$ and

$y_{1}$ in the point-slope form

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1 - \frac{5 π}{2}) = 10 \cdot (x - (- \frac{π}{4}))$

Step 3.2

Simplify the equation and keep it in point-slope form.

$y - 1 + \frac{5 π}{2} = 10 \cdot (x + \frac{π}{4})$

Step 3.3

Solve for

$y$ .

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Step 3.3.1

Simplify

$10 \cdot (x + \frac{π}{4})$ .

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Step 3.3.1.1

Rewrite.

$y - 1 + \frac{5 π}{2} = 0 + 0 + 10 \cdot (x + \frac{π}{4})$

Step 3.3.1.2

Simplify by adding zeros.

$y - 1 + \frac{5 π}{2} = 10 \cdot (x + \frac{π}{4})$

Step 3.3.1.3

Apply the distributive property.

$y - 1 + \frac{5 π}{2} = 10 x + 10 \frac{π}{4}$

Step 3.3.1.4

Cancel the common factor of

$2$ .

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Step 3.3.1.4.1

Factor

$2$ out of

$10$ .

$y - 1 + \frac{5 π}{2} = 10 x + 2 (5) \frac{π}{4}$

Step 3.3.1.4.2

Factor

$2$ out of

$4$ .

$y - 1 + \frac{5 π}{2} = 10 x + 2 \cdot 5 \frac{π}{2 \cdot 2}$

Step 3.3.1.4.3

Cancel the common factor.

$y - 1 + \frac{5 π}{2} = 10 x + 2 \cdot 5 \frac{π}{2 \cdot 2}$

Step 3.3.1.4.4

Rewrite the expression.

$y - 1 + \frac{5 π}{2} = 10 x + 5 \frac{π}{2}$

Step 3.3.1.5

Combine

$5$ and

$\frac{π}{2}$ .

$y - 1 + \frac{5 π}{2} = 10 x + \frac{5 π}{2}$

Step 3.3.2

Move all terms not containing

$y$ to the right side of the equation.

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Step 3.3.2.1

Add

$1$ to both sides of the equation.

$y + \frac{5 π}{2} = 10 x + \frac{5 π}{2} + 1$

Step 3.3.2.2

Subtract

$\frac{5 π}{2}$ from both sides of the equation.

$y = 10 x + \frac{5 π}{2} + 1 - \frac{5 π}{2}$

Step 3.3.2.3

Combine the opposite terms in

$10 x + \frac{5 π}{2} + 1 - \frac{5 π}{2}$ .

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Step 3.3.2.3.1

Combine the numerators over the common denominator.

$y = 10 x + \frac{5 π - 5 π}{2} + 1$

Step 3.3.2.3.2

Subtract

$5 π$ from

$5 π$ .

$y = 10 x + \frac{0}{2} + 1$

Step 3.3.2.4

Divide

$0$ by

$2$ .

$y = 10 x + 0 + 1$

Step 3.3.2.5

Add

$10 x$ and

$0$ .

$y = 10 x + 1$

Step 4