Calculus Examples

$2 x - 2 \cos (x)$

Step 1

Write $2 x - 2 \cos (x)$ as a function.

$f (x) = 2 x - 2 \cos (x)$

Step 2

Find the first derivative.

Tap for more steps...

Step 2.1

Find the first derivative.

Tap for more steps...

Step 2.1.1

By the Sum Rule, the derivative of $2 x - 2 \cos (x)$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 \cos (x)]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 \cos (x)]$

Step 2.1.2

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 2.1.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 \cos (x)]$

Step 2.1.2.2

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

$2 \cdot 1 + \frac{d}{d x} [- 2 \cos (x)]$

Step 2.1.2.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 2 \cos (x)]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 2 \cos (x)]$ .

Tap for more steps...

Step 2.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \cos (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [\cos (x)]$ .

$2 - 2 \frac{d}{d x} [\cos (x)]$

Step 2.1.3.2

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

$2 - 2 (- \sin (x))$

Step 2.1.3.3

Multiply $- 1$ by $- 2$ .

$f' (x) = 2 + 2 \sin (x)$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $2 + 2 \sin (x)$ .

$2 + 2 \sin (x)$

Step 3

Set the first derivative equal to $0$ then solve the equation $2 + 2 \sin (x) = 0$ .

Tap for more steps...

Step 3.1

Set the first derivative equal to $0$ .

$2 + 2 \sin (x) = 0$

Step 3.2

Subtract $2$ from both sides of the equation.

$2 \sin (x) = - 2$

Step 3.3

Divide each term in $2 \sin (x) = - 2$ by $2$ and simplify.

Tap for more steps...

Step 3.3.1

Divide each term in $2 \sin (x) = - 2$ by $2$ .

$\frac{2 \sin (x)}{2} = \frac{- 2}{2}$

Step 3.3.2

Simplify the left side.

Tap for more steps...

Step 3.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 \sin (x)}{2} = \frac{- 2}{2}$

Step 3.3.2.1.2

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

$\sin (x) = \frac{- 2}{2}$

Step 3.3.3

Simplify the right side.

Tap for more steps...

Step 3.3.3.1

Divide $- 2$ by $2$ .

$\sin (x) = - 1$

Step 3.4

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

$x = \arcsin (- 1)$

Step 3.5

Simplify the right side.

Tap for more steps...

Step 3.5.1

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

$x = - \frac{π}{2}$

Step 3.6

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 3.7

Simplify the expression to find the second solution.

Tap for more steps...

Step 3.7.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 3.7.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 3.8

Find the period of $\sin (x)$ .

Tap for more steps...

Step 3.8.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.8.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.8.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.8.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.9

Add $2 π$ to every negative angle to get positive angles.

Tap for more steps...

Step 3.9.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 3.9.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.9.3

Combine fractions.

Tap for more steps...

Step 3.9.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.9.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 3.9.4

Simplify the numerator.

Tap for more steps...

Step 3.9.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 3.9.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 3.9.5

List the new angles.

$x = \frac{3 π}{2}$

Step 3.10

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 3.11

Consolidate the answers.

$x = \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 4

The values which make the derivative equal to $0$ are $\frac{3 π}{2} + 2 π n$ .

$\frac{3 π}{2} + 2 π n$

Step 5

After finding the point that makes the derivative $f' (x) = 2 + 2 \sin (x)$ equal to $0$ or undefined, the interval to check where $f (x) = 2 x - 2 \cos (x)$ is increasing and where it is decreasing is $(- \infty, \frac{3 π}{2} + 2 π n) \cup (\frac{3 π}{2} + 2 π n, \infty)$ .

$(- \infty, \frac{3 π}{2} + 2 π n) \cup (\frac{3 π}{2} + 2 π n, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{3 π}{2} + 2 π n)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $\frac{3 π}{2} + 2 π n - 1$ in the expression.

$f' (\frac{3 π}{2} + 2 π n - 1) = 2 + 2 \sin (\frac{3 π}{2} + 2 π n - 1)$

Step 6.2

The final answer is $2 + 2 \sin (\frac{3 π}{2} + 2 π n - 1)$ .

$2 + 2 \sin (\frac{3 π}{2} + 2 π n - 1)$

Step 6.3

Simplify.

$2 + 2 \sin (\frac{3 π}{2} + 2 π n - 1)$

Step 6.4

At $x = \frac{3 π}{2} + 2 π n - 1$ the derivative is $2 + 2 \sin (\frac{3 π}{2} + 2 π n - 1)$ . Since this is negative, the function is decreasing on $(- \infty, \frac{3 π}{2} + 2 π n)$ .

Decreasing on $(- \infty, \frac{3 π}{2} + 2 π n)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{3 π}{2} + 2 π n, \infty)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

Step 7.1

Replace the variable $x$ with $\frac{3 π}{2} + 2 π n + 1$ in the expression.

$f' (\frac{3 π}{2} + 2 π n + 1) = 2 + 2 \sin (\frac{3 π}{2} + 2 π n + 1)$

Step 7.2

The final answer is $2 + 2 \sin (\frac{3 π}{2} + 2 π n + 1)$ .

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

Step 7.3

Simplify.

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

Step 7.4

At $x = \frac{3 π}{2} + 2 π n + 1$ the derivative is $2 + 2 \sin (\frac{3 π}{2} + 2 π n + 1)$ . Since this is positive, the function is increasing on $(\frac{3 π}{2} + 2 π n, \infty)$ .

Increasing on $(\frac{3 π}{2} + 2 π n, \infty)$ since $f' (x) > 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(\frac{3 π}{2} + 2 π n, \infty)$

Decreasing on: $(- \infty, \frac{3 π}{2} + 2 π n)$

Step 9