Calculus Examples

$y = \sqrt{3} x + 2 \cos (x)$

Step 1

Set $y$ as a function of $x$ .

$f (x) = \sqrt{3} x + 2 \cos (x)$

Step 2

Find the derivative.

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Step 2.1

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

$\frac{d}{d x} [\sqrt{3} x] + \frac{d}{d x} [2 \cos (x)]$

Step 2.2

Evaluate $\frac{d}{d x} [\sqrt{3} x]$ .

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Step 2.2.1

Since $\sqrt{3}$ is constant with respect to $x$ , the derivative of $\sqrt{3} x$ with respect to $x$ is $\sqrt{3} \frac{d}{d x} [x]$ .

$\sqrt{3} \frac{d}{d x} [x] + \frac{d}{d x} [2 \cos (x)]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

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

Step 2.2.3

Multiply $\sqrt{3}$ by $1$ .

$\sqrt{3} + \frac{d}{d x} [2 \cos (x)]$

Step 2.3

Evaluate $\frac{d}{d x} [2 \cos (x)]$ .

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Step 2.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \cos (x)$ with respect to $x$ is $2 \frac{d}{d x} [\cos (x)]$ .

$\sqrt{3} + 2 \frac{d}{d x} [\cos (x)]$

Step 2.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\sqrt{3} + 2 (- \sin (x))$

Step 2.3.3

Multiply $- 1$ by $2$ .

$\sqrt{3} - 2 \sin (x)$

Step 2.4

Reorder terms.

$- 2 \sin (x) + \sqrt{3}$

Step 3

Set the derivative equal to $0$ then solve the equation $- 2 \sin (x) + \sqrt{3} = 0$ .

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Step 3.1

Subtract $\sqrt{3}$ from both sides of the equation.

$- 2 \sin (x) = - \sqrt{3}$

Step 3.2

Divide each term in $- 2 \sin (x) = - \sqrt{3}$ by $- 2$ and simplify.

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Step 3.2.1

Divide each term in $- 2 \sin (x) = - \sqrt{3}$ by $- 2$ .

$\frac{- 2 \sin (x)}{- 2} = \frac{- \sqrt{3}}{- 2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (x)}{- 2} = \frac{- \sqrt{3}}{- 2}$

Step 3.2.2.1.2

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{- \sqrt{3}}{- 2}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Dividing two negative values results in a positive value.

$\sin (x) = \frac{\sqrt{3}}{2}$

Step 3.3

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

$x = \arcsin (\frac{\sqrt{3}}{2})$

Step 3.4

Simplify the right side.

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Step 3.4.1

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

$x = \frac{π}{3}$

Step 3.5

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{3}$

Step 3.6

Simplify $π - \frac{π}{3}$ .

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Step 3.6.1

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

$x = π \cdot \frac{3}{3} - \frac{π}{3}$

Step 3.6.2

Combine fractions.

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Step 3.6.2.1

Combine $π$ and $\frac{3}{3}$ .

$x = \frac{π \cdot 3}{3} - \frac{π}{3}$

Step 3.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 - π}{3}$

Step 3.6.3

Simplify the numerator.

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Step 3.6.3.1

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π - π}{3}$

Step 3.6.3.2

Subtract $π$ from $3 π$ .

$x = \frac{2 π}{3}$

Step 3.7

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

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Step 3.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.7.2

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

$\frac{2 π}{| 1 |}$

Step 3.7.3

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

$\frac{2 π}{1}$

Step 3.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 4

Solve the original function $f (x) = \sqrt{3} x + 2 \cos (x)$ at $x = \frac{π}{3}$ .

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Step 4.1

Replace the variable $x$ with $\frac{π}{3}$ in the expression.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

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

Step 4.2.1.2

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

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

Step 4.2.1.3

Cancel the common factor of $2$ .

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Step 4.2.1.3.1

Cancel the common factor.

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

Step 4.2.1.3.2

Rewrite the expression.

$f (\frac{π}{3}) = \frac{\sqrt{3} π}{3} + 1$

Step 4.2.2

The final answer is $\frac{\sqrt{3} π}{3} + 1$ .

$\frac{\sqrt{3} π}{3} + 1$

Step 5

Solve the original function $f (x) = \sqrt{3} x + 2 \cos (x)$ at $x = \frac{2 π}{3}$ .

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Step 5.1

Replace the variable $x$ with $\frac{2 π}{3}$ in the expression.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

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

Step 5.2.1.2

Move $2$ to the left of $\sqrt{3}$ .

$f (\frac{2 π}{3}) = \frac{2 \cdot (\sqrt{3} π)}{3} + 2 \cos (\frac{2 π}{3})$

Step 5.2.1.3

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

$f (\frac{2 π}{3}) = \frac{2 \sqrt{3} π}{3} + 2 (- \cos (\frac{π}{3}))$

Step 5.2.1.4

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

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

Step 5.2.1.5

Cancel the common factor of $2$ .

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Step 5.2.1.5.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 5.2.1.5.2

Cancel the common factor.

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

Step 5.2.1.5.3

Rewrite the expression.

$f (\frac{2 π}{3}) = \frac{2 \sqrt{3} π}{3} - 1$

Step 5.2.2

The final answer is $\frac{2 \sqrt{3} π}{3} - 1$ .

$\frac{2 \sqrt{3} π}{3} - 1$

Step 6

The horizontal tangent lines on function $f (x) = \sqrt{3} x + 2 \cos (x)$ are $y = \frac{\sqrt{3} π}{3} + 1, y = \frac{2 \sqrt{3} π}{3} - 1$ .

$y = \frac{\sqrt{3} π}{3} + 1, y = \frac{2 \sqrt{3} π}{3} - 1$

Step 7