Calculus Examples

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

Step 1

Write $\sin (x) - \cos (x)$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

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 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $- 1$ .

$\cos (x) + 1 \sin (x)$

Step 2.3.4

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

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

Step 3

Find the second derivative of the function.

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Step 3.1

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 3.2

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

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

Step 3.3

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

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

Step 4

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 5

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

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

Step 6

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

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Step 6.1

Cancel the common factor.

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

Step 6.2

Rewrite the expression.

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

Step 7

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

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

Step 8

Separate fractions.

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

Step 9

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

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

Step 10

Divide $0$ by $1$ .

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

Step 11

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

$1 + \tan (x) = 0$

Step 12

Subtract $1$ from both sides of the equation.

$\tan (x) = - 1$

Step 13

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

$x = \arctan (- 1)$

Step 14

Simplify the right side.

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Step 14.1

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

$x = - \frac{π}{4}$

Step 15

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

$x = - \frac{π}{4} - π$

Step 16

Simplify the expression to find the second solution.

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Step 16.1

Add $2 π$ to $- \frac{π}{4} - π$ .

$x = - \frac{π}{4} - π + 2 π$

Step 16.2

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$x = \frac{3 π}{4}$

Step 17

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

$x = - \frac{π}{4}, \frac{3 π}{4}$

Step 18

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 19

Evaluate the second derivative.

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Step 19.1

Simplify each term.

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Step 19.1.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \sin (\frac{7 π}{4}) + \cos (- \frac{π}{4})$

Step 19.1.2

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.

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

Step 19.1.3

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

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

Step 19.1.4

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

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Step 19.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 19.1.4.2

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

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

Step 19.1.5

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{\sqrt{2}}{2} + \cos (\frac{7 π}{4})$

Step 19.1.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 19.1.7

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

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

Step 19.2

Simplify terms.

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Step 19.2.1

Combine the numerators over the common denominator.

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

Step 19.2.2

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

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

Step 19.2.3

Cancel the common factor of $2$ .

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Step 19.2.3.1

Cancel the common factor.

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

Step 19.2.3.2

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

$\sqrt{2}$

Step 20

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

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

Step 21

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

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Step 21.1

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

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

Step 21.2

Simplify the result.

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Step 21.2.1

Simplify each term.

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Step 21.2.1.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (- \frac{π}{4}) = \sin (\frac{7 π}{4}) - \cos (- \frac{π}{4})$

Step 21.2.1.2

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.

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

Step 21.2.1.3

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 21.2.1.4

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (- \frac{π}{4}) = - \frac{\sqrt{2}}{2} - \cos (\frac{7 π}{4})$

Step 21.2.1.5

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 21.2.1.6

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 21.2.2

Simplify terms.

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Step 21.2.2.1

Combine the numerators over the common denominator.

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

Step 21.2.2.2

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

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

Step 21.2.2.3

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

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Step 21.2.2.3.1

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

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

Step 21.2.2.3.2

Cancel the common factors.

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Step 21.2.2.3.2.1

Factor $2$ out of $2$ .

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

Step 21.2.2.3.2.2

Cancel the common factor.

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

Step 21.2.2.3.2.3

Rewrite the expression.

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

Step 21.2.2.3.2.4

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

$f (- \frac{π}{4}) = - \sqrt{2}$

Step 21.2.3

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

$y = - \sqrt{2}$

Step 22

Evaluate the second derivative at $x = \frac{3 π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{3 π}{4}) + \cos (\frac{3 π}{4})$

Step 23

Evaluate the second derivative.

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Step 23.1

Simplify each term.

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Step 23.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- \sin (\frac{π}{4}) + \cos (\frac{3 π}{4})$

Step 23.1.2

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

$- \frac{\sqrt{2}}{2} + \cos (\frac{3 π}{4})$

Step 23.1.3

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.

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

Step 23.1.4

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

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

Step 23.2

Simplify terms.

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Step 23.2.1

Combine the numerators over the common denominator.

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

Step 23.2.2

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

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

Step 23.2.3

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

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Step 23.2.3.1

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

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

Step 23.2.3.2

Cancel the common factors.

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Step 23.2.3.2.1

Factor $2$ out of $2$ .

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

Step 23.2.3.2.2

Cancel the common factor.

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

Step 23.2.3.2.3

Rewrite the expression.

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

Step 23.2.3.2.4

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

$- \sqrt{2}$

Step 24

$x = \frac{3 π}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{3 π}{4}$ is a local maximum

Step 25

Find the y-value when $x = \frac{3 π}{4}$ .

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Step 25.1

Replace the variable $x$ with $\frac{3 π}{4}$ in the expression.

$f (\frac{3 π}{4}) = \sin (\frac{3 π}{4}) - \cos (\frac{3 π}{4})$

Step 25.2

Simplify the result.

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Step 25.2.1

Simplify each term.

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Step 25.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{3 π}{4}) = \sin (\frac{π}{4}) - \cos (\frac{3 π}{4})$

Step 25.2.1.2

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

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} - \cos (\frac{3 π}{4})$

Step 25.2.1.3

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.

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} + \cos (\frac{π}{4})$

Step 25.2.1.4

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

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 25.2.1.5

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

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Step 25.2.1.5.1

Multiply $- 1$ by $- 1$ .

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} + 1 (\frac{\sqrt{2}}{2})$

Step 25.2.1.5.2

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

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 25.2.2

Simplify terms.

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Step 25.2.2.1

Combine the numerators over the common denominator.

$f (\frac{3 π}{4}) = \frac{\sqrt{2} + \sqrt{2}}{2}$

Step 25.2.2.2

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

$f (\frac{3 π}{4}) = \frac{2 \sqrt{2}}{2}$

Step 25.2.2.3

Cancel the common factor of $2$ .

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Step 25.2.2.3.1

Cancel the common factor.

$f (\frac{3 π}{4}) = \frac{2 \sqrt{2}}{2}$

Step 25.2.2.3.2

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

$f (\frac{3 π}{4}) = \sqrt{2}$

Step 25.2.3

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

$y = \sqrt{2}$

Step 26

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

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

$(\frac{3 π}{4}, \sqrt{2})$ is a local maxima

Step 27