Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \sin (x)$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.3

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

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

Step 1.4

Combine terms.

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Step 1.4.1

Subtract $\sin (x)$ from $\sin (x)$ .

$x \cos (x) + 0$

Step 1.4.2

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

$x \cos (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \cos (x)$ .

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

Step 2.2

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

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

$f'' (x) = x (- \sin (x)) + \cos (x) \cdot 1$

Step 2.3.2

Simplify the expression.

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Step 2.3.2.1

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

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

Step 2.3.2.2

Reorder terms.

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

Step 3

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

$x \cos (x) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$\cos (x) = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 6.2

Solve $\cos (x) = 0$ for $x$ .

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Step 6.2.1

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

$x = \arccos (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 6.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 6.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 6.2.4.1

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

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

Step 6.2.4.2

Combine fractions.

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Step 6.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 6.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 6.2.4.3

Simplify the numerator.

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Step 6.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 6.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 6.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 7

The final solution is all the values that make $x \cos (x) = 0$ true.

$x = 0, \frac{π}{2}, \frac{3 π}{2}$

Step 8

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.

$0 \sin (0) + \cos (0)$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

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

$0 \cdot 0 + \cos (0)$

Step 9.1.2

Multiply $0$ by $0$ .

$0 + \cos (0)$

Step 9.1.3

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

$0 + 1$

Step 9.2

Add $0$ and $1$ .

$1$

Step 10

$x = 0$ 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 11

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

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Step 11.1

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

$f (0) = (0) \sin (0) + \cos (0)$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

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

$f (0) = 0 \cdot 0 + \cos (0)$

Step 11.2.1.2

Multiply $0$ by $0$ .

$f (0) = 0 + \cos (0)$

Step 11.2.1.3

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

$f (0) = 0 + 1$

Step 11.2.2

Add $0$ and $1$ .

$f (0) = 1$

Step 11.2.3

The final answer is $1$ .

$y = 1$

Step 12

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

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

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

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

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

Step 13.1.2

Multiply $- 1$ by $1$ .

$- \frac{π}{2} + \cos (\frac{π}{2})$

Step 13.1.3

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$- \frac{π}{2} + 0$

Step 13.2

Add $- \frac{π}{2}$ and $0$ .

$- \frac{π}{2}$

Step 14

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

$x = \frac{π}{2}$ is a local maximum

Step 15

Find the y-value when $x = \frac{π}{2}$ .

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Step 15.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

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

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 $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = \frac{π}{2} \cdot 1 + \cos (\frac{π}{2})$

Step 15.2.1.2

Multiply $\frac{π}{2}$ by $1$ .

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

Step 15.2.1.3

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{π}{2}) = \frac{π}{2} + 0$

Step 15.2.2

Add $\frac{π}{2}$ and $0$ .

$f (\frac{π}{2}) = \frac{π}{2}$

Step 15.2.3

The final answer is $\frac{π}{2}$ .

$y = \frac{π}{2}$

Step 16

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 17

Evaluate the second derivative.

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Step 17.1

Simplify each term.

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Step 17.1.1

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 fourth quadrant.

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

Step 17.1.2

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

$- \frac{3 π}{2} (- 1 \cdot 1) + \cos (\frac{3 π}{2})$

Step 17.1.3

Multiply $- 1$ by $1$ .

$- \frac{3 π}{2} \cdot - 1 + \cos (\frac{3 π}{2})$

Step 17.1.4

Multiply $- \frac{3 π}{2} \cdot - 1$ .

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Step 17.1.4.1

Multiply $- 1$ by $- 1$ .

$1 \frac{3 π}{2} + \cos (\frac{3 π}{2})$

Step 17.1.4.2

Multiply $\frac{3 π}{2}$ by $1$ .

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

Step 17.1.5

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

$\frac{3 π}{2} + \cos (\frac{π}{2})$

Step 17.1.6

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$\frac{3 π}{2} + 0$

Step 17.2

Add $\frac{3 π}{2}$ and $0$ .

$\frac{3 π}{2}$

Step 18

$x = \frac{3 π}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3 π}{2}$ is a local minimum

Step 19

Find the y-value when $x = \frac{3 π}{2}$ .

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Step 19.1

Replace the variable $x$ with $\frac{3 π}{2}$ in the expression.

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

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 sine is negative in the fourth quadrant.

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

Step 19.2.1.2

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

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

Step 19.2.1.3

Multiply $- 1$ by $1$ .

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

Step 19.2.1.4

Multiply $\frac{3 π}{2} \cdot - 1$ .

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Step 19.2.1.4.1

Combine $\frac{3 π}{2}$ and $- 1$ .

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

Step 19.2.1.4.2

Multiply $- 1$ by $3$ .

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

Step 19.2.1.5

Move the negative in front of the fraction.

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

Step 19.2.1.6

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

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

Step 19.2.1.7

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{3 π}{2}) = - \frac{3 π}{2} + 0$

Step 19.2.2

Add $- \frac{3 π}{2}$ and $0$ .

$f (\frac{3 π}{2}) = - \frac{3 π}{2}$

Step 19.2.3

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

$y = - \frac{3 π}{2}$

Step 20

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

$(0, 1)$ is a local minima

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

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

Step 21