Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

Evaluate $\frac{d}{d x} [\frac{1}{2} x]$ .

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

$f'' (x) = 0 + \sin (x)$

Step 2.2.3

Multiply $- 1$ by $- 1$ .

$f'' (x) = 0 + 1 \sin (x)$

Step 2.2.4

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

$f'' (x) = 0 + \sin (x)$

Step 2.3

Add $0$ and $\sin (x)$ .

$f'' (x) = \sin (x)$

Step 3

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

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

Step 4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

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

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Dividing two negative values results in a positive value.

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

Step 5.3.2

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

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

Step 6

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

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

Step 7

Simplify the right side.

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Step 7.1

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

$x = \frac{π}{3}$

Step 8

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{3}$

Step 9

Simplify $2 π - \frac{π}{3}$ .

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Step 9.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 9.2

Combine fractions.

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Step 9.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 9.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 9.3

Simplify the numerator.

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Step 9.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 9.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 10

The solution to the equation $x = \frac{π}{3}$ .

$x = \frac{π}{3}, \frac{5 π}{3}$

Step 11

Evaluate the second derivative at $x = \frac{π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\sin (\frac{π}{3})$

Step 12

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

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

Step 13

$x = \frac{π}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{3}$ is a local minimum

Step 14

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

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Step 14.1

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

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

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

Multiply $\frac{1}{2} (\frac{π}{3})$ .

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Step 14.2.1.1.1

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

$f (\frac{π}{3}) = \frac{π}{2 \cdot 3} - \sin (\frac{π}{3})$

Step 14.2.1.1.2

Multiply $2$ by $3$ .

$f (\frac{π}{3}) = \frac{π}{6} - \sin (\frac{π}{3})$

Step 14.2.1.2

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

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

Step 14.2.2

The final answer is $\frac{π}{6} - \frac{\sqrt{3}}{2}$ .

$y = \frac{π}{6} - \frac{\sqrt{3}}{2}$

Step 15

Evaluate the second derivative at $x = \frac{5 π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\sin (\frac{5 π}{3})$

Step 16

Evaluate the second derivative.

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Step 16.1

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

$- \sin (\frac{π}{3})$

Step 16.2

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

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

Step 17

$x = \frac{5 π}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{5 π}{3}$ is a local maximum

Step 18

Find the y-value when $x = \frac{5 π}{3}$ .

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Step 18.1

Replace the variable $x$ with $\frac{5 π}{3}$ in the expression.

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

Step 18.2

Simplify the result.

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Step 18.2.1

Simplify each term.

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Step 18.2.1.1

Multiply $\frac{1}{2} (\frac{5 π}{3})$ .

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Step 18.2.1.1.1

Multiply $\frac{1}{2}$ by $\frac{5 π}{3}$ .

$f (\frac{5 π}{3}) = \frac{5 π}{2 \cdot 3} - \sin (\frac{5 π}{3})$

Step 18.2.1.1.2

Multiply $2$ by $3$ .

$f (\frac{5 π}{3}) = \frac{5 π}{6} - \sin (\frac{5 π}{3})$

Step 18.2.1.2

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

$f (\frac{5 π}{3}) = \frac{5 π}{6} + \sin (\frac{π}{3})$

Step 18.2.1.3

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

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

Step 18.2.1.4

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

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Step 18.2.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 18.2.1.4.2

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

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

Step 18.2.2

The final answer is $\frac{5 π}{6} + \frac{\sqrt{3}}{2}$ .

$y = \frac{5 π}{6} + \frac{\sqrt{3}}{2}$

Step 19

These are the local extrema for $f (x) = \frac{1}{2} x - \sin (x)$ .

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

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

Step 20