Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = - \frac{x}{2}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $- \frac{x}{2}$ .

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

Step 1.2.2

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

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

Step 1.2.3

Replace all occurrences of $u$ with $- \frac{x}{2}$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $- 2$ .

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

Step 1.3.2

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

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

Step 1.3.3

Simplify terms.

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Step 1.3.3.1

Multiply $- 1$ by $2$ .

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

Step 1.3.3.2

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 1.3.3.3

Combine $\frac{- 2}{2}$ and $\sin (- \frac{x}{2})$ .

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

Step 1.3.3.4

Cancel the common factor of $- 2$ and $2$ .

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Step 1.3.3.4.1

Factor $2$ out of $- 2 \sin (- \frac{x}{2})$ .

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

Step 1.3.3.4.2

Cancel the common factors.

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Step 1.3.3.4.2.1

Factor $2$ out of $2$ .

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

Step 1.3.3.4.2.2

Cancel the common factor.

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

Step 1.3.3.4.2.3

Rewrite the expression.

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

Step 1.3.3.4.2.4

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

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

Step 1.3.4

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

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

Step 1.3.5

Multiply $- 1$ by $1$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Since $\sin (- \frac{x}{2})$ is an odd function, rewrite $\sin (- \frac{x}{2})$ as $- \sin (\frac{x}{2})$ .

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

Step 1.4.2

Multiply $- - \sin (\frac{x}{2})$ .

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Step 1.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 1.4.2.2

Multiply $\sin (\frac{x}{2})$ by $1$ .

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

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = \frac{x}{2}$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $\frac{x}{2}$ .

$f'' (x) = \frac{d}{d u} (\sin (u)) \frac{d}{d x} (\frac{x}{2})$

Step 2.1.2

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

$f'' (x) = \cos (u) \frac{d}{d x} (\frac{x}{2})$

Step 2.1.3

Replace all occurrences of $u$ with $\frac{x}{2}$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Combine $\frac{1}{2}$ and $\cos (\frac{x}{2})$ .

$f'' (x) = \frac{\cos (\frac{x}{2})}{2} \cdot \frac{d}{d x} (x)$

Step 2.2.3

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

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

Step 2.2.4

Multiply $\frac{\cos (\frac{x}{2})}{2}$ by $1$ .

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

Step 3

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

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

Step 4

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

$\frac{x}{2} = \arcsin (0)$

Step 5

Simplify the right side.

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Step 5.1

The exact value of $\arcsin (0)$ is $0$ .

$\frac{x}{2} = 0$

Step 6

Set the numerator equal to zero.

$x = 0$

Step 7

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.

$\frac{x}{2} = π - 0$

Step 8

Solve for $x$ .

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Step 8.1

Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 (π - 0)$

Step 8.2

Simplify both sides of the equation.

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Step 8.2.1

Simplify the left side.

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Step 8.2.1.1

Cancel the common factor of $2$ .

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Step 8.2.1.1.1

Cancel the common factor.

$2 \frac{x}{2} = 2 (π - 0)$

Step 8.2.1.1.2

Rewrite the expression.

$x = 2 (π - 0)$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

Subtract $0$ from $π$ .

$x = 2 π$

Step 9

The solution to the equation $\frac{x}{2} = 0$ .

$x = 0, 2 π$

Step 10

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{\cos (\frac{0}{2})}{2}$

Step 11

Evaluate the second derivative.

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Step 11.1

Cancel the common factor of $0$ and $2$ .

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Step 11.1.1

Factor $2$ out of $0$ .

$\frac{\cos (\frac{2 (0)}{2})}{2}$

Step 11.1.2

Cancel the common factors.

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Step 11.1.2.1

Factor $2$ out of $2$ .

$\frac{\cos (\frac{2 \cdot 0}{2 \cdot 1})}{2}$

Step 11.1.2.2

Cancel the common factor.

$\frac{\cos (\frac{2 \cdot 0}{2 \cdot 1})}{2}$

Step 11.1.2.3

Rewrite the expression.

$\frac{\cos (\frac{0}{1})}{2}$

Step 11.1.2.4

Divide $0$ by $1$ .

$\frac{\cos (0)}{2}$

Step 11.2

The exact value of $\cos (0)$ is $1$ .

$\frac{1}{2}$

Step 12

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 13

Find the y-value when $x = 0$ .

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Step 13.1

Replace the variable $x$ with $0$ in the expression.

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

Step 13.2

Simplify the result.

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Step 13.2.1

Divide $0$ by $2$ .

$f (0) = - 2 \cos (- 0)$

Step 13.2.2

Multiply $- 1$ by $0$ .

$f (0) = - 2 \cos (0)$

Step 13.2.3

The exact value of $\cos (0)$ is $1$ .

$f (0) = - 2 \cdot 1$

Step 13.2.4

Multiply $- 2$ by $1$ .

$f (0) = - 2$

Step 13.2.5

The final answer is $- 2$ .

$y = - 2$

Step 14

Evaluate the second derivative at $x = 2 π$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{\cos (\frac{2 π}{2})}{2}$

Step 15

Evaluate the second derivative.

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Step 15.1

Cancel the common factor of $2$ .

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Step 15.1.1

Cancel the common factor.

$\frac{\cos (\frac{2 π}{2})}{2}$

Step 15.1.2

Divide $π$ by $1$ .

$\frac{\cos (π)}{2}$

Step 15.2

Simplify the numerator.

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Step 15.2.1

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.

$\frac{- \cos (0)}{2}$

Step 15.2.2

The exact value of $\cos (0)$ is $1$ .

$\frac{- 1 \cdot 1}{2}$

Step 15.2.3

Multiply $- 1$ by $1$ .

$\frac{- 1}{2}$

Step 15.3

Move the negative in front of the fraction.

$- \frac{1}{2}$

Step 16

$x = 2 π$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 2 π$ is a local maximum

Step 17

Find the y-value when $x = 2 π$ .

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Step 17.1

Replace the variable $x$ with $2 π$ in the expression.

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

Step 17.2

Simplify the result.

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Step 17.2.1

Cancel the common factor of $2$ .

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Step 17.2.1.1

Cancel the common factor.

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

Step 17.2.1.2

Divide $π$ by $1$ .

$f (2 π) = - 2 \cos (- π)$

Step 17.2.2

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.

$f (2 π) = - 2 (- \cos (0))$

Step 17.2.3

The exact value of $\cos (0)$ is $1$ .

$f (2 π) = - 2 (- 1 \cdot 1)$

Step 17.2.4

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 17.2.4.1

Multiply $- 1$ by $1$ .

$f (2 π) = - 2 \cdot - 1$

Step 17.2.4.2

Multiply $- 2$ by $- 1$ .

$f (2 π) = 2$

Step 17.2.5

The final answer is $2$ .

$y = 2$

Step 18

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

$(0, - 2)$ is a local minima

$(2 π, 2)$ is a local maxima

Step 19