Calculus Examples

$\frac{x - 3}{x + 4}$

Step 1

Write $\frac{x - 3}{x + 4}$ as a function.

$f (x) = \frac{x - 3}{x + 4}$

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

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x - 3$ and $g (x) = x + 4$ .

$\frac{(x + 4) \frac{d}{d x} [x - 3] - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

$\frac{(x + 4) (\frac{d}{d x} [x] + \frac{d}{d x} [- 3]) - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

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 $n = 1$ .

$\frac{(x + 4) (1 + \frac{d}{d x} [- 3]) - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

Step 2.1.2.3

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$\frac{(x + 4) (1 + 0) - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

Step 2.1.2.4

Simplify the expression.

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Step 2.1.2.4.1

Add $1$ and $0$ .

$\frac{(x + 4) \cdot 1 - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

Step 2.1.2.4.2

Multiply $x + 4$ by $1$ .

$\frac{x + 4 - (x - 3) \frac{d}{d x} [x + 4]}{{(x + 4)}^{2}}$

Step 2.1.2.5

By the Sum Rule, the derivative of $x + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4]$ .

$\frac{x + 4 - (x - 3) (\frac{d}{d x} [x] + \frac{d}{d x} [4])}{{(x + 4)}^{2}}$

Step 2.1.2.6

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

$\frac{x + 4 - (x - 3) (1 + \frac{d}{d x} [4])}{{(x + 4)}^{2}}$

Step 2.1.2.7

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{x + 4 - (x - 3) (1 + 0)}{{(x + 4)}^{2}}$

Step 2.1.2.8

Simplify the expression.

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Step 2.1.2.8.1

Add $1$ and $0$ .

$\frac{x + 4 - (x - 3) \cdot 1}{{(x + 4)}^{2}}$

Step 2.1.2.8.2

Multiply $- 1$ by $1$ .

$\frac{x + 4 - (x - 3)}{{(x + 4)}^{2}}$

Step 2.1.3

Simplify.

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Step 2.1.3.1

Apply the distributive property.

$\frac{x + 4 - x - - 3}{{(x + 4)}^{2}}$

Step 2.1.3.2

Simplify the numerator.

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Step 2.1.3.2.1

Combine the opposite terms in $x + 4 - x - - 3$ .

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Step 2.1.3.2.1.1

Subtract $x$ from $x$ .

$\frac{0 + 4 - - 3}{{(x + 4)}^{2}}$

Step 2.1.3.2.1.2

Add $0$ and $4$ .

$\frac{4 - - 3}{{(x + 4)}^{2}}$

Step 2.1.3.2.2

Multiply $- 1$ by $- 3$ .

$\frac{4 + 3}{{(x + 4)}^{2}}$

Step 2.1.3.2.3

Add $4$ and $3$ .

$f' (x) = \frac{7}{{(x + 4)}^{2}}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{7}{{(x + 4)}^{2}}$ .

$\frac{7}{{(x + 4)}^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{7}{{(x + 4)}^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{7}{{(x + 4)}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$7 = 0$

Step 3.3

Since $7 \neq 0$ , there are no solutions.

No solution

Step 4

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 5

Find where the derivative is undefined.

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Step 5.1

Set the denominator in $\frac{7}{{(x + 4)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 4)}^{2} = 0$

Step 5.2

Solve for $x$ .

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Step 5.2.1

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 5.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

After finding the point that makes the derivative $f' (x) = \frac{7}{{(x + 4)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{x - 3}{x + 4}$ is increasing and where it is decreasing is $(- \infty, - 4) \cup (- 4, \infty)$ .

$(- \infty, - 4) \cup (- 4, \infty)$

Step 7

Substitute a value from the interval $(- \infty, - 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- 5$ in the expression.

$f' (- 5) = \frac{7}{{((- 5) + 4)}^{2}}$

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 $- 5$ and $4$ .

$f' (- 5) = \frac{7}{{(- 1)}^{2}}$

Step 7.2.1.2

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

$f' (- 5) = \frac{7}{1}$

Step 7.2.2

Divide $7$ by $1$ .

$f' (- 5) = 7$

Step 7.2.3

The final answer is $7$ .

$7$

Step 7.3

At $x = - 5$ the derivative is $7$ . Since this is positive, the function is increasing on $(- \infty, - 4)$ .

Increasing on $(- \infty, - 4)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(- 4, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $- 3$ in the expression.

$f' (- 3) = \frac{7}{{((- 3) + 4)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the denominator.

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Step 8.2.1.1

Add $- 3$ and $4$ .

$f' (- 3) = \frac{7}{1^{2}}$

Step 8.2.1.2

One to any power is one.

$f' (- 3) = \frac{7}{1}$

Step 8.2.2

Divide $7$ by $1$ .

$f' (- 3) = 7$

Step 8.2.3

The final answer is $7$ .

$7$

Step 8.3

At $x = - 3$ the derivative is $7$ . Since this is positive, the function is increasing on $(- 4, \infty)$ .

Increasing on $(- 4, \infty)$ since $f' (x) > 0$

Step 9

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

Increasing on: $(- \infty, - 4), (- 4, \infty)$

Step 10