Calculus Examples

$f (x) = \sin (x) + \cos (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

$\cos (x) - \sin (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 3

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

$\cos (x) - \sin (x) = 0$

Step 4

Divide each term in the equation by $\cos (x)$ .

$\frac{\cos (x)}{\cos (x)} + \frac{- \sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 5

Cancel the common factor of $\cos (x)$ .

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Step 5.1

Cancel the common factor.

$\frac{\cos (x)}{\cos (x)} + \frac{- \sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 5.2

Rewrite the expression.

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

Step 6

Separate fractions.

$1 + \frac{- 1}{1} \cdot \frac{\sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$1 + \frac{- 1}{1} \cdot \tan (x) = \frac{0}{\cos (x)}$

Step 8

Divide $- 1$ by $1$ .

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

Step 9

Separate fractions.

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

Step 10

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$1 - \tan (x) = \frac{0}{1} \cdot \sec (x)$

Step 11

Divide $0$ by $1$ .

$1 - \tan (x) = 0 \sec (x)$

Step 12

Multiply $0$ by $\sec (x)$ .

$1 - \tan (x) = 0$

Step 13

Subtract $1$ from both sides of the equation.

$- \tan (x) = - 1$

Step 14

Divide each term in $- \tan (x) = - 1$ by $- 1$ and simplify.

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Step 14.1

Divide each term in $- \tan (x) = - 1$ by $- 1$ .

$\frac{- \tan (x)}{- 1} = \frac{- 1}{- 1}$

Step 14.2

Simplify the left side.

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Step 14.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (x)}{1} = \frac{- 1}{- 1}$

Step 14.2.2

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

$\tan (x) = \frac{- 1}{- 1}$

Step 14.3

Simplify the right side.

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Step 14.3.1

Divide $- 1$ by $- 1$ .

$\tan (x) = 1$

Step 15

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

$x = \arctan (1)$

Step 16

Simplify the right side.

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Step 16.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 17

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{4}$

Step 18

Simplify $π + \frac{π}{4}$ .

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Step 18.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 18.2

Combine fractions.

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Step 18.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 18.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 18.3

Simplify the numerator.

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Step 18.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 18.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 19

The solution to the equation $x = \frac{π}{4}$ .

$x = \frac{π}{4}, \frac{5 π}{4}$

Step 20

Evaluate the second derivative at $x = \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{π}{4}) - \cos (\frac{π}{4})$

Step 21

Evaluate the second derivative.

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Step 21.1

Simplify each term.

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Step 21.1.1

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- \frac{\sqrt{2}}{2} - \cos (\frac{π}{4})$

Step 21.1.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

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

Step 21.2

Simplify terms.

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Step 21.2.1

Combine the numerators over the common denominator.

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

Step 21.2.2

Subtract $\sqrt{2}$ from $- \sqrt{2}$ .

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

Step 21.2.3

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

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Step 21.2.3.1

Factor $2$ out of $- 2 \sqrt{2}$ .

$\frac{2 (- \sqrt{2})}{2}$

Step 21.2.3.2

Cancel the common factors.

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Step 21.2.3.2.1

Factor $2$ out of $2$ .

$\frac{2 (- \sqrt{2})}{2 (1)}$

Step 21.2.3.2.2

Cancel the common factor.

$\frac{2 (- \sqrt{2})}{2 \cdot 1}$

Step 21.2.3.2.3

Rewrite the expression.

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

Step 21.2.3.2.4

Divide $- \sqrt{2}$ by $1$ .

$- \sqrt{2}$

Step 22

$x = \frac{π}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{4}$ is a local maximum

Step 23

Find the y-value when $x = \frac{π}{4}$ .

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Step 23.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = \sin (\frac{π}{4}) + \cos (\frac{π}{4})$

Step 23.2

Simplify the result.

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Step 23.2.1

Simplify each term.

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Step 23.2.1.1

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{4}) = \frac{\sqrt{2}}{2} + \cos (\frac{π}{4})$

Step 23.2.1.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 23.2.2

Simplify terms.

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Step 23.2.2.1

Combine the numerators over the common denominator.

$f (\frac{π}{4}) = \frac{\sqrt{2} + \sqrt{2}}{2}$

Step 23.2.2.2

Add $\sqrt{2}$ and $\sqrt{2}$ .

$f (\frac{π}{4}) = \frac{2 \sqrt{2}}{2}$

Step 23.2.2.3

Cancel the common factor of $2$ .

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Step 23.2.2.3.1

Cancel the common factor.

$f (\frac{π}{4}) = \frac{2 \sqrt{2}}{2}$

Step 23.2.2.3.2

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

$f (\frac{π}{4}) = \sqrt{2}$

Step 23.2.3

The final answer is $\sqrt{2}$ .

$y = \sqrt{2}$

Step 24

Evaluate the second derivative at $x = \frac{5 π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{5 π}{4}) - \cos (\frac{5 π}{4})$

Step 25

Evaluate the second derivative.

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Step 25.1

Simplify each term.

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Step 25.1.1

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 third quadrant.

$- - \sin (\frac{π}{4}) - \cos (\frac{5 π}{4})$

Step 25.1.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- - \frac{\sqrt{2}}{2} - \cos (\frac{5 π}{4})$

Step 25.1.3

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

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Step 25.1.3.1

Multiply $- 1$ by $- 1$ .

$1 \frac{\sqrt{2}}{2} - \cos (\frac{5 π}{4})$

Step 25.1.3.2

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

$\frac{\sqrt{2}}{2} - \cos (\frac{5 π}{4})$

Step 25.1.4

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 third quadrant.

$\frac{\sqrt{2}}{2} - - \cos (\frac{π}{4})$

Step 25.1.5

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

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

Step 25.1.6

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

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Step 25.1.6.1

Multiply $- 1$ by $- 1$ .

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

Step 25.1.6.2

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

$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 25.2

Simplify terms.

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Step 25.2.1

Combine the numerators over the common denominator.

$\frac{\sqrt{2} + \sqrt{2}}{2}$

Step 25.2.2

Add $\sqrt{2}$ and $\sqrt{2}$ .

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

Step 25.2.3

Cancel the common factor of $2$ .

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Step 25.2.3.1

Cancel the common factor.

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

Step 25.2.3.2

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

$\sqrt{2}$

Step 26

$x = \frac{5 π}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{4}$ is a local minimum

Step 27

Find the y-value when $x = \frac{5 π}{4}$ .

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Step 27.1

Replace the variable $x$ with $\frac{5 π}{4}$ in the expression.

$f (\frac{5 π}{4}) = \sin (\frac{5 π}{4}) + \cos (\frac{5 π}{4})$

Step 27.2

Simplify the result.

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Step 27.2.1

Simplify each term.

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Step 27.2.1.1

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 third quadrant.

$f (\frac{5 π}{4}) = - \sin (\frac{π}{4}) + \cos (\frac{5 π}{4})$

Step 27.2.1.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} + \cos (\frac{5 π}{4})$

Step 27.2.1.3

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 third quadrant.

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} - \cos (\frac{π}{4})$

Step 27.2.1.4

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}$

Step 27.2.2

Simplify terms.

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Step 27.2.2.1

Combine the numerators over the common denominator.

$f (\frac{5 π}{4}) = \frac{- \sqrt{2} - \sqrt{2}}{2}$

Step 27.2.2.2

Subtract $\sqrt{2}$ from $- \sqrt{2}$ .

$f (\frac{5 π}{4}) = \frac{- 2 \sqrt{2}}{2}$

Step 27.2.2.3

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

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Step 27.2.2.3.1

Factor $2$ out of $- 2 \sqrt{2}$ .

$f (\frac{5 π}{4}) = \frac{2 (- \sqrt{2})}{2}$

Step 27.2.2.3.2

Cancel the common factors.

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Step 27.2.2.3.2.1

Factor $2$ out of $2$ .

$f (\frac{5 π}{4}) = \frac{2 (- \sqrt{2})}{2 (1)}$

Step 27.2.2.3.2.2

Cancel the common factor.

$f (\frac{5 π}{4}) = \frac{2 (- \sqrt{2})}{2 \cdot 1}$

Step 27.2.2.3.2.3

Rewrite the expression.

$f (\frac{5 π}{4}) = \frac{- \sqrt{2}}{1}$

Step 27.2.2.3.2.4

Divide $- \sqrt{2}$ by $1$ .

$f (\frac{5 π}{4}) = - \sqrt{2}$

Step 27.2.3

The final answer is $- \sqrt{2}$ .

$y = - \sqrt{2}$

Step 28

These are the local extrema for $f (x) = \sin (x) + \cos (x)$ .

$(\frac{π}{4}, \sqrt{2})$ is a local maxima

$(\frac{5 π}{4}, - \sqrt{2})$ is a local minima

Step 29