Calculus Examples

$f (x) = 5 \tan (x)$ , $x = \frac{π}{3}$

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 $x$ .

$y = 5 \tan (\frac{π}{3})$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = 5 \tan (\frac{π}{3})$

Step 1.2.2

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

$y = 5 \sqrt{3}$

Step 2

Find the first derivative and evaluate at $x = \frac{π}{3}$ and $y = 5 \sqrt{3}$ to find the slope of the tangent line.

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Step 2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 \tan (x)$ with respect to $x$ is $5 \frac{d}{d x} [\tan (x)]$ .

$5 \frac{d}{d x} [\tan (x)]$

Step 2.2

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

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

Step 2.3

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

$5 \sec^{2} (\frac{π}{3})$

Step 2.4

Simplify.

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Step 2.4.1

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

$5 \cdot 2^{2}$

Step 2.4.2

Raise $2$ to the power of $2$ .

$5 \cdot 4$

Step 2.4.3

Multiply $5$ by $4$ .

$20$

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 $20$ and a given point $(\frac{π}{3}, 5 \sqrt{3})$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5 \sqrt{3}) = 20 \cdot (x - (\frac{π}{3}))$

Step 3.2

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

$y - 5 \sqrt{3} = 20 \cdot (x - \frac{π}{3})$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $20 \cdot (x - \frac{π}{3})$ .

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Step 3.3.1.1

Rewrite.

$y - 5 \sqrt{3} = 0 + 0 + 20 \cdot (x - \frac{π}{3})$

Step 3.3.1.2

Simplify by adding zeros.

$y - 5 \sqrt{3} = 20 \cdot (x - \frac{π}{3})$

Step 3.3.1.3

Apply the distributive property.

$y - 5 \sqrt{3} = 20 x + 20 (- \frac{π}{3})$

Step 3.3.1.4

Multiply $20 (- \frac{π}{3})$ .

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Step 3.3.1.4.1

Multiply $- 1$ by $20$ .

$y - 5 \sqrt{3} = 20 x - 20 \frac{π}{3}$

Step 3.3.1.4.2

Combine $- 20$ and $\frac{π}{3}$ .

$y - 5 \sqrt{3} = 20 x + \frac{- 20 π}{3}$

Step 3.3.1.5

Move the negative in front of the fraction.

$y - 5 \sqrt{3} = 20 x - \frac{20 π}{3}$

Step 3.3.2

Add $5 \sqrt{3}$ to both sides of the equation.

$y = 20 x - \frac{20 π}{3} + 5 \sqrt{3}$

Step 3.3.3

Write in $y = m x + b$ form.

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Step 3.3.3.1

To write $5 \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = 20 x - \frac{20 π}{3} + 5 \sqrt{3} \cdot \frac{3}{3}$

Step 3.3.3.2

Combine $5 \sqrt{3}$ and $\frac{3}{3}$ .

$y = 20 x - \frac{20 π}{3} + \frac{5 \sqrt{3} \cdot 3}{3}$

Step 3.3.3.3

Combine the numerators over the common denominator.

$y = 20 x + \frac{- 20 π + 5 \sqrt{3} \cdot 3}{3}$

Step 3.3.3.4

Multiply $3$ by $5$ .

$y = 20 x + \frac{- 20 π + 15 \sqrt{3}}{3}$

Step 3.3.3.5

Factor $- 1$ out of $- 20 π$ .

$y = 20 x + \frac{- (20 π) + 15 \sqrt{3}}{3}$

Step 3.3.3.6

Factor $- 1$ out of $15 \sqrt{3}$ .

$y = 20 x + \frac{- (20 π) - (- 15 \sqrt{3})}{3}$

Step 3.3.3.7

Factor $- 1$ out of $- (20 π) - (- 15 \sqrt{3})$ .

$y = 20 x + \frac{- (20 π - 15 \sqrt{3})}{3}$

Step 3.3.3.8

Rewrite $- (20 π - 15 \sqrt{3})$ as $- 1 (20 π - 15 \sqrt{3})$ .

$y = 20 x + \frac{- 1 (20 π - 15 \sqrt{3})}{3}$

Step 3.3.3.9

Move the negative in front of the fraction.

$y = 20 x - \frac{20 π - 15 \sqrt{3}}{3}$

Step 4