Calculus Examples

$f (x) = \sec (x) + 1 + \frac{2 \sqrt{3} π}{3}$ at $x = - \frac{π}{3}$

Step 1

Find the corresponding $y$ -value to $x = - \frac{π}{3}$ .

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Step 1.1

Substitute $- \frac{π}{3}$ in for $x$ .

$y = \sec (- \frac{π}{3}) + 1 + \frac{2 \sqrt{3} π}{3}$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = \sec (- \frac{π}{3}) + 1 + \frac{2 \sqrt{3} π}{3}$

Step 1.2.2

Simplify $\sec (- \frac{π}{3}) + 1 + \frac{2 \sqrt{3} π}{3}$ .

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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 $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$y = \sec (\frac{5 π}{3}) + 1 + \frac{2 \sqrt{3} π}{3}$

Step 1.2.2.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$y = \sec (\frac{π}{3}) + 1 + \frac{2 \sqrt{3} π}{3}$

Step 1.2.2.1.3

The exact value of $\sec (\frac{π}{3})$ is $2$ .

$y = 2 + 1 + \frac{2 \sqrt{3} π}{3}$

Step 1.2.2.2

Add $2$ and $1$ .

$y = 3 + \frac{2 \sqrt{3} π}{3}$

Step 2

Find the first derivative and evaluate at $x = - \frac{π}{3}$ and $y = 3 + \frac{2 \sqrt{3} π}{3}$ to find the slope of the tangent line.

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Step 2.1

By the Sum Rule, the derivative of $\sec (x) + 1 + \frac{2 \sqrt{3} π}{3}$ with respect to $x$ is $\frac{d}{d x} [\sec (x)] + \frac{d}{d x} [1] + \frac{d}{d x} [\frac{2 \sqrt{3} π}{3}]$ .

$\frac{d}{d x} [\sec (x)] + \frac{d}{d x} [1] + \frac{d}{d x} [\frac{2 \sqrt{3} π}{3}]$

Step 2.2

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$\sec (x) \tan (x) + \frac{d}{d x} [1] + \frac{d}{d x} [\frac{2 \sqrt{3} π}{3}]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\sec (x) \tan (x) + 0 + \frac{d}{d x} [\frac{2 \sqrt{3} π}{3}]$

Step 2.3.2

Since $\frac{2 \sqrt{3} π}{3}$ is constant with respect to $x$ , the derivative of $\frac{2 \sqrt{3} π}{3}$ with respect to $x$ is $0$ .

$\sec (x) \tan (x) + 0 + 0$

Step 2.4

Combine terms.

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Step 2.4.1

Add $\sec (x) \tan (x)$ and $0$ .

$\sec (x) \tan (x) + 0$

Step 2.4.2

Add $\sec (x) \tan (x)$ and $0$ .

$\sec (x) \tan (x)$

Step 2.5

Evaluate the derivative at $x = - \frac{π}{3}$ .

$\sec (- \frac{π}{3}) \tan (- \frac{π}{3})$

Step 2.6

Simplify.

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Step 2.6.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\sec (\frac{5 π}{3}) \tan (- \frac{π}{3})$

Step 2.6.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$\sec (\frac{π}{3}) \tan (- \frac{π}{3})$

Step 2.6.3

The exact value of $\sec (\frac{π}{3})$ is $2$ .

$2 \tan (- \frac{π}{3})$

Step 2.6.4

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$2 \tan (\frac{5 π}{3})$

Step 2.6.5

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

$2 (- \tan (\frac{π}{3}))$

Step 2.6.6

The exact value of $\tan (\frac{π}{3})$ is $\sqrt{3}$ .

$2 (- \sqrt{3})$

Step 2.6.7

Multiply $- 1$ by $2$ .

$- 2 \sqrt{3}$

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 $- 2 \sqrt{3}$ and a given point $(- \frac{π}{3}, 3 + \frac{2 \sqrt{3} π}{3})$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (3 + \frac{2 \sqrt{3} π}{3}) = - 2 \sqrt{3} \cdot (x - (- \frac{π}{3}))$

Step 3.2

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

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} \cdot (x + \frac{π}{3})$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $- 2 \sqrt{3} \cdot (x + \frac{π}{3})$ .

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Step 3.3.1.1

Rewrite.

$y - 3 - \frac{2 \sqrt{3} π}{3} = 0 + 0 - 2 \sqrt{3} \cdot (x + \frac{π}{3})$

Step 3.3.1.2

Simplify by adding zeros.

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} \cdot (x + \frac{π}{3})$

Step 3.3.1.3

Apply the distributive property.

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x - 2 \sqrt{3} \frac{π}{3}$

Step 3.3.1.4

Multiply $- 2 \sqrt{3} \frac{π}{3}$ .

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Step 3.3.1.4.1

Combine $\frac{π}{3}$ and $- 2$ .

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x + \frac{π \cdot - 2}{3} \sqrt{3}$

Step 3.3.1.4.2

Combine $\frac{π \cdot - 2}{3}$ and $\sqrt{3}$ .

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x + \frac{π \cdot - 2 \sqrt{3}}{3}$

Step 3.3.1.5

Simplify each term.

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Step 3.3.1.5.1

Move $- 2$ to the left of $π$ .

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x + \frac{- 2 \cdot π \sqrt{3}}{3}$

Step 3.3.1.5.2

Move the negative in front of the fraction.

$y - 3 - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x - \frac{2 π \sqrt{3}}{3}$

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 $3$ to both sides of the equation.

$y - \frac{2 \sqrt{3} π}{3} = - 2 \sqrt{3} x - \frac{2 π \sqrt{3}}{3} + 3$

Step 3.3.2.2

Add $\frac{2 \sqrt{3} π}{3}$ to both sides of the equation.

$y = - 2 \sqrt{3} x - \frac{2 π \sqrt{3}}{3} + 3 + \frac{2 \sqrt{3} π}{3}$

Step 3.3.2.3

Combine the opposite terms in $- 2 \sqrt{3} x - \frac{2 π \sqrt{3}}{3} + 3 + \frac{2 \sqrt{3} π}{3}$ .

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Step 3.3.2.3.1

Reorder the factors in the terms $- \frac{2 π \sqrt{3}}{3}$ and $\frac{2 \sqrt{3} π}{3}$ .

$y = - 2 \sqrt{3} x - \frac{2 π \sqrt{3}}{3} + 3 + \frac{2 π \sqrt{3}}{3}$

Step 3.3.2.3.2

Add $- \frac{2 π \sqrt{3}}{3}$ and $\frac{2 π \sqrt{3}}{3}$ .

$y = - 2 \sqrt{3} x + 0 + 3$

Step 3.3.2.3.3

Add $- 2 \sqrt{3} x$ and $0$ .

$y = - 2 \sqrt{3} x + 3$

Step 4