Calculus Examples

$f (x) = \frac{5}{x + 5}$

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

Differentiate using the Constant Multiple Rule.

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Step 1.1.1.1

Since $5$ is constant with respect to $x$ , the derivative of $\frac{5}{x + 5}$ with respect to $x$ is $5 \frac{d}{d x} [\frac{1}{x + 5}]$ .

$5 \frac{d}{d x} [\frac{1}{x + 5}]$

Step 1.1.1.2

Rewrite $\frac{1}{x + 5}$ as ${(x + 5)}^{- 1}$ .

$5 \frac{d}{d x} [{(x + 5)}^{- 1}]$

Step 1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x + 5$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u$ as $x + 5$ .

$5 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x + 5])$

Step 1.1.2.2

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

$5 (- u^{- 2} \frac{d}{d x} [x + 5])$

Step 1.1.2.3

Replace all occurrences of $u$ with $x + 5$ .

$5 (- {(x + 5)}^{- 2} \frac{d}{d x} [x + 5])$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

Multiply $- 1$ by $5$ .

$- 5 ({(x + 5)}^{- 2} \frac{d}{d x} [x + 5])$

Step 1.1.3.2

By the Sum Rule, the derivative of $x + 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [5]$ .

$- 5 {(x + 5)}^{- 2} (\frac{d}{d x} [x] + \frac{d}{d x} [5])$

Step 1.1.3.3

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

$- 5 {(x + 5)}^{- 2} (1 + \frac{d}{d x} [5])$

Step 1.1.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$- 5 {(x + 5)}^{- 2} (1 + 0)$

Step 1.1.3.5

Simplify the expression.

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Step 1.1.3.5.1

Add $1$ and $0$ .

$- 5 {(x + 5)}^{- 2} \cdot 1$

Step 1.1.3.5.2

Multiply $- 5$ by $1$ .

$- 5 {(x + 5)}^{- 2}$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 5 \frac{1}{{(x + 5)}^{2}}$

Step 1.1.4.2

Combine terms.

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Step 1.1.4.2.1

Combine $- 5$ and $\frac{1}{{(x + 5)}^{2}}$ .

$\frac{- 5}{{(x + 5)}^{2}}$

Step 1.1.4.2.2

Move the negative in front of the fraction.

$f' (x) = - \frac{5}{{(x + 5)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{5}{{(x + 5)}^{2}}$ .

$- \frac{5}{{(x + 5)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{5}{{(x + 5)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- \frac{5}{{(x + 5)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$5 = 0$

Step 2.3

Since $5 \neq 0$ , there are no solutions.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{5}{{(x + 5)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 5)}^{2} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

Set the $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 5

After finding the point that makes the derivative $f' (x) = - \frac{5}{{(x + 5)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{5}{x + 5}$ is increasing and where it is decreasing is $(- \infty, - 5) \cup (- 5, \infty)$ .

$(- \infty, - 5) \cup (- 5, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 6$ in the expression.

$f' (- 6) = - \frac{5}{{((- 6) + 5)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the denominator.

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Step 6.2.1.1

Add $- 6$ and $5$ .

$f' (- 6) = - \frac{5}{{(- 1)}^{2}}$

Step 6.2.1.2

Raise $- 1$ to the power of $2$ .

$f' (- 6) = - \frac{5}{1}$

Step 6.2.2

Simplify the expression.

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Step 6.2.2.1

Divide $5$ by $1$ .

$f' (- 6) = - 1 \cdot 5$

Step 6.2.2.2

Multiply $- 1$ by $5$ .

$f' (- 6) = - 5$

Step 6.2.3

The final answer is $- 5$ .

$- 5$

Step 6.3

At $x = - 6$ the derivative is $- 5$ . Since this is negative, the function is decreasing on $(- \infty, - 5)$ .

Decreasing on $(- \infty, - 5)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(- 5, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = - \frac{5}{{((- 4) + 5)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Add $- 4$ and $5$ .

$f' (- 4) = - \frac{5}{1^{2}}$

Step 7.2.1.2

One to any power is one.

$f' (- 4) = - \frac{5}{1}$

Step 7.2.2

Simplify the expression.

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Step 7.2.2.1

Divide $5$ by $1$ .

$f' (- 4) = - 1 \cdot 5$

Step 7.2.2.2

Multiply $- 1$ by $5$ .

$f' (- 4) = - 5$

Step 7.2.3

The final answer is $- 5$ .

$- 5$

Step 7.3

At $x = - 4$ the derivative is $- 5$ . Since this is negative, the function is decreasing on $(- 5, \infty)$ .

Decreasing on $(- 5, \infty)$ since $f' (x) < 0$

Step 8

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

Decreasing on: $(- \infty, - 5), (- 5, \infty)$

Step 9