Calculus Examples

$y = \cos (x)$

Step 1

Write $y = \cos (x)$ as a function.

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

Step 2

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

$- \sin (x)$

Step 3

Find the second derivative of the function.

Tap for more steps...

Step 3.1

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

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

Step 3.2

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

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

Step 4

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

$- \sin (x) = 0$

Step 5

Divide each term in $- \sin (x) = 0$ by $- 1$ and simplify.

Tap for more steps...

Step 5.1

Divide each term in $- \sin (x) = 0$ by $- 1$ .

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

Step 5.2

Simplify the left side.

Tap for more steps...

Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

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

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

Step 5.3

Simplify the right side.

Tap for more steps...

Step 5.3.1

Divide $0$ by $- 1$ .

$\sin (x) = 0$

Step 6

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

$x = \arcsin (0)$

Step 7

Simplify the right side.

Tap for more steps...

Step 7.1

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

$x = 0$

Step 8

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.

$x = π - 0$

Step 9

Subtract $0$ from $π$ .

$x = π$

Step 10

The solution to the equation $x = 0$ .

$x = 0, π$

Step 11

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.

$- \cos (0)$

Step 12

Evaluate the second derivative.

Tap for more steps...

Step 12.1

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

$- 1 \cdot 1$

Step 12.2

Multiply $- 1$ by $1$ .

$- 1$

Step 13

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

$x = 0$ is a local maximum

Step 14

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

Tap for more steps...

Step 14.1

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

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

Step 14.2

Simplify the result.

Tap for more steps...

Step 14.2.1

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

$f (0) = 1$

Step 14.2.2

The final answer is $1$ .

$y = 1$

Step 15

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

$- \cos (π)$

Step 16

Evaluate the second derivative.

Tap for more steps...

Step 16.1

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.

$- - \cos (0)$

Step 16.2

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

$- (- 1 \cdot 1)$

Step 16.3

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

Tap for more steps...

Step 16.3.1

Multiply $- 1$ by $1$ .

$- - 1$

Step 16.3.2

Multiply $- 1$ by $- 1$ .

$1$

Step 17

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

$x = π$ is a local minimum

Step 18

Find the y-value when $x = π$ .

Tap for more steps...

Step 18.1

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

$f (π) = \cos (π)$

Step 18.2

Simplify the result.

Tap for more steps...

Step 18.2.1

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.

$f (π) = - \cos (0)$

Step 18.2.2

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

$f (π) = - 1 \cdot 1$

Step 18.2.3

Multiply $- 1$ by $1$ .

$f (π) = - 1$

Step 18.2.4

The final answer is $- 1$ .

$y = - 1$

Step 19

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

$(0, 1)$ is a local maxima

$(π, - 1)$ is a local minima

Step 20