Calculus Examples

$- \cos^{2} (x)$

Step 1

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

$f (x) = - \cos^{2} (x)$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos^{2} (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos^{2} (x)]$ .

$f' (x) = - \frac{d}{d x} \cos^{2} (x)$

Step 2.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \cos (x)$ .

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Step 2.1.2.1

To apply the Chain Rule, set $u$ as $\cos (x)$ .

$f' (x) = - (\frac{d}{d u} (u^{2}) \frac{d}{d x} (\cos (x)))$

Step 2.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$f' (x) = - (2 u \frac{d}{d x} (\cos (x)))$

Step 2.1.2.3

Replace all occurrences of $u$ with $\cos (x)$ .

$f' (x) = - (2 \cos (x) \frac{d}{d x} (\cos (x)))$

Step 2.1.3

Multiply $2$ by $- 1$ .

$f' (x) = - 2 (\cos (x) \frac{d}{d x} (\cos (x)))$

Step 2.1.4

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

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

Step 2.1.5

Multiply $- 1$ by $- 2$ .

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

Step 2.1.6

Simplify.

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Step 2.1.6.1

Reorder $2 \cos (x)$ and $\sin (x)$ .

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

Step 2.1.6.2

Reorder $\sin (x)$ and $2$ .

$f' (x) = 2 \cdot (\sin (x) \cos (x))$

Step 2.1.6.3

Apply the sine double-angle identity.

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

Step 2.2

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

$\sin (2 x)$

Step 3

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

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Step 3.1

Set the first derivative equal to $0$ .

$\sin (2 x) = 0$

Step 3.2

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

$2 x = \arcsin (0)$

Step 3.3

Simplify the right side.

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Step 3.3.1

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

$2 x = 0$

Step 3.4

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 3.4.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Cancel the common factor of $2$ .

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Step 3.4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 3.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Divide $0$ by $2$ .

$x = 0$

Step 3.5

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.

$2 x = π - 0$

Step 3.6

Solve for $x$ .

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Step 3.6.1

Simplify.

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Step 3.6.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 3.6.1.2

Add $π$ and $0$ .

$2 x = π$

Step 3.6.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 3.6.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 3.6.2.2

Simplify the left side.

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Step 3.6.2.2.1

Cancel the common factor of $2$ .

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Step 3.6.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 3.6.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 3.7

Find the period of $\sin (2 x)$ .

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Step 3.7.1

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

$\frac{2 π}{| b |}$

Step 3.7.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 3.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 3.7.4

Cancel the common factor of $2$ .

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Step 3.7.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 3.7.4.2

Divide $π$ by $1$ .

$π$

Step 3.8

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

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

Step 3.9

Consolidate the answers.

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

Step 4

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

$\frac{π n}{2}$

Step 5

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Apply the distributive property.

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

Step 6.2.2

Cancel the common factor of $2$ .

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Step 6.2.2.1

Cancel the common factor.

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

Step 6.2.2.2

Rewrite the expression.

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

Step 6.2.3

Multiply $2$ by $- 1$ .

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

Step 6.2.4

The final answer is $\sin (π n - 2)$ .

$\sin (π n - 2)$

Step 6.3

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Apply the distributive property.

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

Step 7.2.2

Cancel the common factor of $2$ .

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Step 7.2.2.1

Cancel the common factor.

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

Step 7.2.2.2

Rewrite the expression.

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

Step 7.2.3

Multiply $2$ by $1$ .

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

Step 7.2.4

The final answer is $\sin (π n + 2)$ .

$\sin (π n + 2)$

Step 7.3

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

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

Step 8

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

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

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

Step 9