Calculus Examples

$f (x) = 3 \cos (x) - \cos^{3} (x)$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $3 \cos (x) - \cos^{3} (x)$ with respect to $x$ is $\frac{d}{d x} [3 \cos (x)] + \frac{d}{d x} [- \cos^{3} (x)]$ .

$f' (x) = \frac{d}{d x} (3 \cos (x)) + \frac{d}{d x} (- \cos^{3} (x))$

Step 1.1.2

Evaluate $\frac{d}{d x} [3 \cos (x)]$ .

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Step 1.1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos (x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos (x)]$ .

$f' (x) = 3 \frac{d}{d x} (\cos (x)) + \frac{d}{d x} (- \cos^{3} (x))$

Step 1.1.2.2

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

$f' (x) = 3 (- \sin (x)) + \frac{d}{d x} (- \cos^{3} (x))$

Step 1.1.2.3

Multiply $- 1$ by $3$ .

$f' (x) = - 3 \sin (x) + \frac{d}{d x} (- \cos^{3} (x))$

Step 1.1.3

Evaluate $\frac{d}{d x} [- \cos^{3} (x)]$ .

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Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos^{3} (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos^{3} (x)]$ .

$f' (x) = - 3 \sin (x) - \frac{d}{d x} \cos^{3} (x)$

Step 1.1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = \cos (x)$ .

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Step 1.1.3.2.1

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

$f' (x) = - 3 \sin (x) - (\frac{d}{d u} (u^{3}) \frac{d}{d x} (\cos (x)))$

Step 1.1.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$f' (x) = - 3 \sin (x) - (3 u^{2} \frac{d}{d x} (\cos (x)))$

Step 1.1.3.2.3

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

$f' (x) = - 3 \sin (x) - (3 \cos^{2} (x) \frac{d}{d x} (\cos (x)))$

Step 1.1.3.3

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

$f' (x) = - 3 \sin (x) - (3 \cos^{2} (x) (- \sin (x)))$

Step 1.1.3.4

Multiply $- 1$ by $3$ .

$f' (x) = - 3 \sin (x) - (- 3 \cos^{2} (x) \sin (x))$

Step 1.1.3.5

Multiply $- 3$ by $- 1$ .

$f' (x) = - 3 \sin (x) + 3 \cos^{2} (x) \sin (x)$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder terms.

$f' (x) = 3 \cos^{2} (x) \sin (x) - 3 \sin (x)$

Step 1.1.4.2

Factor $3 \sin (x)$ out of $3 \cos^{2} (x) \sin (x) - 3 \sin (x)$ .

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Step 1.1.4.2.1

Factor $3 \sin (x)$ out of $3 \cos^{2} (x) \sin (x)$ .

$f' (x) = 3 \sin (x) (\cos^{2} (x)) - 3 \sin (x)$

Step 1.1.4.2.2

Factor $3 \sin (x)$ out of $- 3 \sin (x)$ .

$f' (x) = 3 \sin (x) (\cos^{2} (x)) + 3 \sin (x) (- 1)$

Step 1.1.4.2.3

Factor $3 \sin (x)$ out of $3 \sin (x) (\cos^{2} (x)) + 3 \sin (x) (- 1)$ .

$f' (x) = 3 \sin (x) (\cos^{2} (x) - 1)$

Step 1.1.4.3

Reorder $\cos^{2} (x)$ and $- 1$ .

$f' (x) = 3 \sin (x) (- 1 + \cos^{2} (x))$

Step 1.1.4.4

Rewrite $- 1$ as $- 1 (1)$ .

$f' (x) = 3 \sin (x) (- 1 \cdot 1 + \cos^{2} (x))$

Step 1.1.4.5

Factor $- 1$ out of $\cos^{2} (x)$ .

$f' (x) = 3 \sin (x) (- 1 \cdot 1 - 1 (- \cos^{2} (x)))$

Step 1.1.4.6

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (x))$ .

$f' (x) = 3 \sin (x) (- 1 (1 - \cos^{2} (x)))$

Step 1.1.4.7

Rewrite $- 1 (1 - \cos^{2} (x))$ as $- (1 - \cos^{2} (x))$ .

$f' (x) = 3 \sin (x) (- (1 - \cos^{2} (x)))$

Step 1.1.4.8

Apply pythagorean identity.

$f' (x) = 3 \sin (x) (- \sin^{2} (x))$

Step 1.1.4.9

Multiply $\sin (x)$ by $\sin^{2} (x)$ by adding the exponents.

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Step 1.1.4.9.1

Move $\sin^{2} (x)$ .

$f' (x) = 3 (\sin^{2} (x) \sin (x)) \cdot - 1$

Step 1.1.4.9.2

Multiply $\sin^{2} (x)$ by $\sin (x)$ .

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Step 1.1.4.9.2.1

Raise $\sin (x)$ to the power of $1$ .

$f' (x) = 3 (\sin^{2} (x) \sin (x)) \cdot - 1$

Step 1.1.4.9.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (x) = 3 {\sin (x)}^{2 + 1} \cdot - 1$

Step 1.1.4.9.3

Add $2$ and $1$ .

$f' (x) = 3 \sin^{3} (x) \cdot - 1$

Step 1.1.4.10

Multiply $- 1$ by $3$ .

$f' (x) = - 3 \sin^{3} (x)$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 3 \sin^{3} (x)$ .

$- 3 \sin^{3} (x)$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 3 \sin^{3} (x) = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 3 \sin^{3} (x) = 0$

Step 2.2

Divide each term in $- 3 \sin^{3} (x) = 0$ by $- 3$ and simplify.

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Step 2.2.1

Divide each term in $- 3 \sin^{3} (x) = 0$ by $- 3$ .

$\frac{- 3 \sin^{3} (x)}{- 3} = \frac{0}{- 3}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 3$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 3 \sin^{3} (x)}{- 3} = \frac{0}{- 3}$

Step 2.2.2.1.2

Divide $\sin^{3} (x)$ by $1$ .

$\sin^{3} (x) = \frac{0}{- 3}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $- 3$ .

$\sin^{3} (x) = 0$

Step 2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sin (x) = \sqrt[3]{0}$

Step 2.4

Simplify $\sqrt[3]{0}$ .

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Step 2.4.1

Rewrite $0$ as $0^{3}$ .

$\sin (x) = \sqrt[3]{0^{3}}$

Step 2.4.2

Pull terms out from under the radical, assuming real numbers.

$\sin (x) = 0$

Step 2.5

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

$x = \arcsin (0)$

Step 2.6

Simplify the right side.

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Step 2.6.1

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

$x = 0$

Step 2.7

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 2.8

Subtract $0$ from $π$ .

$x = π$

Step 2.9

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

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Step 2.9.1

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

$\frac{2 π}{| b |}$

Step 2.9.2

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

$\frac{2 π}{| 1 |}$

Step 2.9.3

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

$\frac{2 π}{1}$

Step 2.9.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.10

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

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 2.11

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 3

The values which make the derivative equal to $0$ are $π n$ .

$π n$

Step 4

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

$(- \infty, π n) \cup (π n, \infty)$

Step 5

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

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Step 5.1

Replace the variable $x$ with $π n - 1$ in the expression.

$f' (π n - 1) = - 3 \sin^{3} (π n - 1)$

Step 5.2

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

$- 3 \sin^{3} (π n - 1)$

Step 5.3

Simplify.

$- 3 \sin^{3} (π n - 1)$

Step 5.4

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

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $π n + 1$ in the expression.

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

Step 6.2

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

$- 3 \sin^{3} (π n + 1)$

Step 6.3

Simplify.

$- 3 \sin^{3} (π n + 1)$

Step 6.4

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

Decreasing on $(π n, \infty)$ since $f' (x) < 0$

Step 7

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

Decreasing on: $(- \infty, π n), (π n, \infty)$

Step 8