Calculus Examples

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

Step 1

Write $y = \cos (x) + \frac{x}{2}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

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

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

Step 2.2

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

$- \sin (x) + \frac{d}{d x} [\frac{x}{2}]$

Step 2.3

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

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Step 2.3.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

$- \sin (x) + \frac{1}{2} \frac{d}{d x} [x]$

Step 2.3.2

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

$- \sin (x) + \frac{1}{2} \cdot 1$

Step 2.3.3

Multiply $\frac{1}{2}$ by $1$ .

$- \sin (x) + \frac{1}{2}$

Step 2.4

Reorder terms.

$\frac{1}{2} - \sin (x)$

Step 3

Find the second derivative of the function.

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Step 3.1

Differentiate.

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Step 3.1.1

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

$f'' (x) = \frac{d}{d x} (\frac{1}{2}) + \frac{d}{d x} (- \sin (x))$

Step 3.1.2

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2}$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- \sin (x))$

Step 3.2

Evaluate $\frac{d}{d x} [- \sin (x)]$ .

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Step 3.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

$f'' (x) = 0 - \frac{d}{d x} \sin (x)$

Step 3.2.2

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

$f'' (x) = 0 - \cos (x)$

Step 3.3

Subtract $\cos (x)$ from $0$ .

$f'' (x) = - \cos (x)$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{1}{2} - \sin (x) = 0$

Step 5

Subtract $\frac{1}{2}$ from both sides of the equation.

$- \sin (x) = - \frac{1}{2}$

Step 6

Divide each term in $- \sin (x) = - \frac{1}{2}$ by $- 1$ and simplify.

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Step 6.1

Divide each term in $- \sin (x) = - \frac{1}{2}$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{- \frac{1}{2}}{- 1}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{- \frac{1}{2}}{- 1}$

Step 6.2.2

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

$\sin (x) = \frac{- \frac{1}{2}}{- 1}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Dividing two negative values results in a positive value.

$\sin (x) = \frac{\frac{1}{2}}{1}$

Step 6.3.2

Divide $\frac{1}{2}$ by $1$ .

$\sin (x) = \frac{1}{2}$

Step 7

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

$x = \arcsin (\frac{1}{2})$

Step 8

Simplify the right side.

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Step 8.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 9

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.

$x = π - \frac{π}{6}$

Step 10

Simplify $π - \frac{π}{6}$ .

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Step 10.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 10.2

Combine fractions.

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Step 10.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 10.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 10.3

Simplify the numerator.

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Step 10.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 10.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 11

The solution to the equation $x = \frac{π}{6}$ .

$x = \frac{π}{6}, \frac{5 π}{6}$

Step 12

Evaluate the second derivative at $x = \frac{π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \cos (\frac{π}{6})$

Step 13

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- \frac{\sqrt{3}}{2}$

Step 14

$x = \frac{π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{6}$ is a local maximum

Step 15

Find the y-value when $x = \frac{π}{6}$ .

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Step 15.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = \cos (\frac{π}{6}) + \frac{\frac{π}{6}}{2}$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{\frac{π}{6}}{2}$

Step 15.2.1.2

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{π}{6} \cdot \frac{1}{2}$

Step 15.2.1.3

Multiply $\frac{π}{6} \cdot \frac{1}{2}$ .

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Step 15.2.1.3.1

Multiply $\frac{π}{6}$ by $\frac{1}{2}$ .

$f (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{π}{6 \cdot 2}$

Step 15.2.1.3.2

Multiply $6$ by $2$ .

$f (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{π}{12}$

Step 15.2.2

The final answer is $\frac{\sqrt{3}}{2} + \frac{π}{12}$ .

$y = \frac{\sqrt{3}}{2} + \frac{π}{12}$

Step 16

Evaluate the second derivative at $x = \frac{5 π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \cos (\frac{5 π}{6})$

Step 17

Evaluate the second derivative.

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Step 17.1

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.

$- - \cos (\frac{π}{6})$

Step 17.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- - \frac{\sqrt{3}}{2}$

Step 17.3

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 17.3.1

Multiply $- 1$ by $- 1$ .

$1 \frac{\sqrt{3}}{2}$

Step 17.3.2

Multiply $\frac{\sqrt{3}}{2}$ by $1$ .

$\frac{\sqrt{3}}{2}$

Step 18

$x = \frac{5 π}{6}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{6}$ is a local minimum

Step 19

Find the y-value when $x = \frac{5 π}{6}$ .

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Step 19.1

Replace the variable $x$ with $\frac{5 π}{6}$ in the expression.

$f (\frac{5 π}{6}) = \cos (\frac{5 π}{6}) + \frac{\frac{5 π}{6}}{2}$

Step 19.2

Simplify the result.

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Step 19.2.1

Simplify each term.

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Step 19.2.1.1

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.

$f (\frac{5 π}{6}) = - \cos (\frac{π}{6}) + \frac{\frac{5 π}{6}}{2}$

Step 19.2.1.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} + \frac{\frac{5 π}{6}}{2}$

Step 19.2.1.3

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} + \frac{5 π}{6} \cdot \frac{1}{2}$

Step 19.2.1.4

Multiply $\frac{5 π}{6} \cdot \frac{1}{2}$ .

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Step 19.2.1.4.1

Multiply $\frac{5 π}{6}$ by $\frac{1}{2}$ .

$f (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} + \frac{5 π}{6 \cdot 2}$

Step 19.2.1.4.2

Multiply $6$ by $2$ .

$f (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} + \frac{5 π}{12}$

Step 19.2.2

The final answer is $- \frac{\sqrt{3}}{2} + \frac{5 π}{12}$ .

$y = - \frac{\sqrt{3}}{2} + \frac{5 π}{12}$

Step 20

These are the local extrema for $f (x) = \cos (x) + \frac{x}{2}$ .

$(\frac{π}{6}, \frac{\sqrt{3}}{2} + \frac{π}{12})$ is a local maxima

$(\frac{5 π}{6}, - \frac{\sqrt{3}}{2} + \frac{5 π}{12})$ is a local minima

Step 21