Calculus Examples

$f (x) = \frac{7 x - 7}{x + 7}$

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 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 $f (x) = 7 x - 7$ and $g (x) = x + 7$ .

$\frac{(x + 7) \frac{d}{d x} [7 x - 7] - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $7 x - 7$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [- 7]$ .

$\frac{(x + 7) (\frac{d}{d x} [7 x] + \frac{d}{d x} [- 7]) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.2

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$\frac{(x + 7) (7 \frac{d}{d x} [x] + \frac{d}{d x} [- 7]) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.3

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

$\frac{(x + 7) (7 \cdot 1 + \frac{d}{d x} [- 7]) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.4

Multiply $7$ by $1$ .

$\frac{(x + 7) (7 + \frac{d}{d x} [- 7]) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.5

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7$ with respect to $x$ is $0$ .

$\frac{(x + 7) (7 + 0) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.6

Simplify the expression.

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Step 1.1.2.6.1

Add $7$ and $0$ .

$\frac{(x + 7) \cdot 7 - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.6.2

Move $7$ to the left of $x + 7$ .

$\frac{7 \cdot (x + 7) - (7 x - 7) \frac{d}{d x} [x + 7]}{{(x + 7)}^{2}}$

Step 1.1.2.7

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$\frac{7 (x + 7) - (7 x - 7) (\frac{d}{d x} [x] + \frac{d}{d x} [7])}{{(x + 7)}^{2}}$

Step 1.1.2.8

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

$\frac{7 (x + 7) - (7 x - 7) (1 + \frac{d}{d x} [7])}{{(x + 7)}^{2}}$

Step 1.1.2.9

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$\frac{7 (x + 7) - (7 x - 7) (1 + 0)}{{(x + 7)}^{2}}$

Step 1.1.2.10

Simplify the expression.

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Step 1.1.2.10.1

Add $1$ and $0$ .

$\frac{7 (x + 7) - (7 x - 7) \cdot 1}{{(x + 7)}^{2}}$

Step 1.1.2.10.2

Multiply $- 1$ by $1$ .

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

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

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

Step 1.1.3.2

Apply the distributive property.

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

Step 1.1.3.3

Simplify the numerator.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Multiply $7$ by $7$ .

$\frac{7 x + 49 - (7 x) - - 7}{{(x + 7)}^{2}}$

Step 1.1.3.3.1.2

Multiply $7$ by $- 1$ .

$\frac{7 x + 49 - 7 x - - 7}{{(x + 7)}^{2}}$

Step 1.1.3.3.1.3

Multiply $- 1$ by $- 7$ .

$\frac{7 x + 49 - 7 x + 7}{{(x + 7)}^{2}}$

Step 1.1.3.3.2

Combine the opposite terms in $7 x + 49 - 7 x + 7$ .

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Step 1.1.3.3.2.1

Subtract $7 x$ from $7 x$ .

$\frac{0 + 49 + 7}{{(x + 7)}^{2}}$

Step 1.1.3.3.2.2

Add $0$ and $49$ .

$\frac{49 + 7}{{(x + 7)}^{2}}$

Step 1.1.3.3.3

Add $49$ and $7$ .

$f' (x) = \frac{56}{{(x + 7)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{56}{{(x + 7)}^{2}}$ .

$\frac{56}{{(x + 7)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{56}{{(x + 7)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{56}{{(x + 7)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$56 = 0$

Step 2.3

Since $56 \neq 0$ , 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 $\frac{56}{{(x + 7)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 4.2

Solve for $x$ .

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Step 4.2.1

Set the $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 4.2.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 5

After finding the point that makes the derivative $f' (x) = \frac{56}{{(x + 7)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{7 x - 7}{x + 7}$ is increasing and where it is decreasing is $(- \infty, - 7) \cup (- 7, \infty)$ .

$(- \infty, - 7) \cup (- 7, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 7)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 8$ in the expression.

$f' (- 8) = \frac{56}{{((- 8) + 7)}^{2}}$

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 $- 8$ and $7$ .

$f' (- 8) = \frac{56}{{(- 1)}^{2}}$

Step 6.2.1.2

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

$f' (- 8) = \frac{56}{1}$

Step 6.2.2

Divide $56$ by $1$ .

$f' (- 8) = 56$

Step 6.2.3

The final answer is $56$ .

$56$

Step 6.3

At $x = - 8$ the derivative is $56$ . Since this is positive, the function is increasing on $(- \infty, - 7)$ .

Increasing on $(- \infty, - 7)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(- 7, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

$f' (- 6) = \frac{56}{{((- 6) + 7)}^{2}}$

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 $- 6$ and $7$ .

$f' (- 6) = \frac{56}{1^{2}}$

Step 7.2.1.2

One to any power is one.

$f' (- 6) = \frac{56}{1}$

Step 7.2.2

Divide $56$ by $1$ .

$f' (- 6) = 56$

Step 7.2.3

The final answer is $56$ .

$56$

Step 7.3

At $x = - 6$ the derivative is $56$ . Since this is positive, the function is increasing on $(- 7, \infty)$ .

Increasing on $(- 7, \infty)$ since $f' (x) > 0$

Step 8

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

Increasing on: $(- \infty, - 7), (- 7, \infty)$

Step 9