Calculus Examples

$\sin (x)$

Step 1

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

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

Step 2

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

$\cos (x)$

Step 3

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

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

Step 4

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

$\cos (x) = 0$

Step 5

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

$x = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$x = \frac{π}{2}$

Step 7

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 8

Simplify $2 π - \frac{π}{2}$ .

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Step 8.1

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 8.2

Combine fractions.

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Step 8.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 8.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 8.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 9

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 10

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.

$- \sin (\frac{π}{2})$

Step 11

Evaluate the second derivative.

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Step 11.1

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

$- 1 \cdot 1$

Step 11.2

Multiply $- 1$ by $1$ .

$- 1$

Step 12

$x = \frac{π}{2}$ 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 13

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

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Step 13.1

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

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

Step 13.2

Simplify the result.

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Step 13.2.1

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

$f (\frac{π}{2}) = 1$

Step 13.2.2

The final answer is $1$ .

$y = 1$

Step 14

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.

$- \sin (\frac{3 π}{2})$

Step 15

Evaluate the second derivative.

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Step 15.1

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.

$- - \sin (\frac{π}{2})$

Step 15.2

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

$- (- 1 \cdot 1)$

Step 15.3

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

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Step 15.3.1

Multiply $- 1$ by $1$ .

$- - 1$

Step 15.3.2

Multiply $- 1$ by $- 1$ .

$1$

Step 16

$x = \frac{3 π}{2}$ 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 17

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

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Step 17.1

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

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

Step 17.2

Simplify the result.

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Step 17.2.1

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.

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

Step 17.2.2

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

$f (\frac{3 π}{2}) = - 1 \cdot 1$

Step 17.2.3

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = - 1$

Step 17.2.4

The final answer is $- 1$ .

$y = - 1$

Step 18

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

$(\frac{π}{2}, 1)$ is a local maxima

$(\frac{3 π}{2}, - 1)$ is a local minima

Step 19