Calculus Examples

$f (x) = \frac{x^{2} - 9}{x^{2} - 4}$

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 $f (x) = x^{2} - 9$ and $g (x) = x^{2} - 4$ .

$f' (x) = \frac{(x^{2} - 4) \frac{d}{d x} (x^{2} - 9) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9]$ .

$f' (x) = \frac{(x^{2} - 4) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9)) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.2

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

$f' (x) = \frac{(x^{2} - 4) (2 x + \frac{d}{d x} (- 9)) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.3

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$f' (x) = \frac{(x^{2} - 4) (2 x + 0) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.4

Simplify the expression.

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Step 1.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{(x^{2} - 4) (2 x) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.4.2

Move $2$ to the left of $x^{2} - 4$ .

$f' (x) = \frac{2 \cdot ((x^{2} - 4) x) - (x^{2} - 9) \frac{d}{d x} (x^{2} - 4)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.5

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

$f' (x) = \frac{2 (x^{2} - 4) x - (x^{2} - 9) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 4))}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.6

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

$f' (x) = \frac{2 (x^{2} - 4) x - (x^{2} - 9) (2 x + \frac{d}{d x} (- 4))}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.7

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

$f' (x) = \frac{2 (x^{2} - 4) x - (x^{2} - 9) (2 x + 0)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.8

Simplify the expression.

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Step 1.1.2.8.1

Add $2 x$ and $0$ .

$f' (x) = \frac{2 (x^{2} - 4) x - (x^{2} - 9) (2 x)}{{(x^{2} - 4)}^{2}}$

Step 1.1.2.8.2

Multiply $2$ by $- 1$ .

$f' (x) = \frac{2 (x^{2} - 4) x - 2 (x^{2} - 9) x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

$f' (x) = \frac{(2 x^{2} + 2 \cdot - 4) x - 2 (x^{2} - 9) x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.2

Apply the distributive property.

$f' (x) = \frac{2 x^{2} x + 2 \cdot (- 4 x) - 2 (x^{2} - 9) x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.3

Apply the distributive property.

$f' (x) = \frac{2 x^{2} x + 2 \cdot (- 4 x) + (- 2 x^{2} - 2 \cdot - 9) x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.4

Apply the distributive property.

$f' (x) = \frac{2 x^{2} x + 2 \cdot (- 4 x) - 2 x^{2} x - 2 \cdot (- 9 x)}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.5

Simplify the numerator.

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Step 1.1.3.5.1

Combine the opposite terms in $2 x^{2} x + 2 \cdot - 4 x - 2 x^{2} x - 2 \cdot - 9 x$ .

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Step 1.1.3.5.1.1

Subtract $2 x^{2} x$ from $2 x^{2} x$ .

$f' (x) = \frac{2 \cdot (- 4 x) + 0 - 2 \cdot (- 9 x)}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.5.1.2

Add $2 \cdot - 4 x$ and $0$ .

$f' (x) = \frac{2 \cdot (- 4 x) - 2 \cdot (- 9 x)}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.5.2

Simplify each term.

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Step 1.1.3.5.2.1

Multiply $2$ by $- 4$ .

$f' (x) = \frac{- 8 x - 2 \cdot (- 9 x)}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.5.2.2

Multiply $- 2$ by $- 9$ .

$f' (x) = \frac{- 8 x + 18 x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.5.3

Add $- 8 x$ and $18 x$ .

$f' (x) = \frac{10 x}{{(x^{2} - 4)}^{2}}$

Step 1.1.3.6

Simplify the denominator.

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Step 1.1.3.6.1

Rewrite $4$ as $2^{2}$ .

$f' (x) = \frac{10 x}{{(x^{2} - 2^{2})}^{2}}$

Step 1.1.3.6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$f' (x) = \frac{10 x}{{((x + 2) (x - 2))}^{2}}$

Step 1.1.3.6.3

Apply the product rule to $(x + 2) (x - 2)$ .

$f' (x) = \frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}}$ .

$\frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$10 x = 0$

Step 2.3

Divide each term in $10 x = 0$ by $10$ and simplify.

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Step 2.3.1

Divide each term in $10 x = 0$ by $10$ .

$\frac{10 x}{10} = \frac{0}{10}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $10$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{10 x}{10} = \frac{0}{10}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{10}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $10$ .

$x = 0$

Step 3

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

$0$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{10 x}{{(x + 2)}^{2} {(x - 2)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 4.2

Solve for $x$ .

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Step 4.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

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

${(x - 2)}^{2} = 0$

Step 4.2.2

Set ${(x + 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.2.1

Set ${(x + 2)}^{2}$ equal to $0$ .

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

Step 4.2.2.2

Solve ${(x + 2)}^{2} = 0$ for $x$ .

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Step 4.2.2.2.1

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.2.2.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.2.3

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.3.1

Set ${(x - 2)}^{2}$ equal to $0$ .

${(x - 2)}^{2} = 0$

Step 4.2.3.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 4.2.3.2.1

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.2.3.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.2.4

The final solution is all the values that make ${(x + 2)}^{2} {(x - 2)}^{2} = 0$ true.

$x = - 2, 2$

Step 4.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 2, x = 2$

Step 5

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 0) \cup (0, 2) \cup (2, \infty)$

Step 6

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

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Step 6.1

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

$f' (- 3) = \frac{10 (- 3)}{{((- 3) + 2)}^{2} {((- 3) - 2)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $10$ by $- 3$ .

$f' (- 3) = \frac{- 30}{{(- 3 + 2)}^{2} {(- 3 - 2)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Add $- 3$ and $2$ .

$f' (- 3) = \frac{- 30}{{(- 1)}^{2} {(- 3 - 2)}^{2}}$

Step 6.2.2.2

Subtract $2$ from $- 3$ .

$f' (- 3) = \frac{- 30}{{(- 1)}^{2} {(- 5)}^{2}}$

Step 6.2.2.3

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

$f' (- 3) = \frac{- 30}{1 {(- 5)}^{2}}$

Step 6.2.2.4

Raise $- 5$ to the power of $2$ .

$f' (- 3) = \frac{- 30}{1 \cdot 25}$

Step 6.2.2.5

Multiply $25$ by $1$ .

$f' (- 3) = \frac{- 30}{25}$

Step 6.2.3

Reduce the expression by cancelling the common factors.

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Step 6.2.3.1

Cancel the common factor of $- 30$ and $25$ .

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Step 6.2.3.1.1

Factor $5$ out of $- 30$ .

$f' (- 3) = \frac{5 (- 6)}{25}$

Step 6.2.3.1.2

Cancel the common factors.

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Step 6.2.3.1.2.1

Factor $5$ out of $25$ .

$f' (- 3) = \frac{5 \cdot - 6}{5 \cdot 5}$

Step 6.2.3.1.2.2

Cancel the common factor.

$f' (- 3) = \frac{5 \cdot - 6}{5 \cdot 5}$

Step 6.2.3.1.2.3

Rewrite the expression.

$f' (- 3) = \frac{- 6}{5}$

Step 6.2.3.2

Move the negative in front of the fraction.

$f' (- 3) = - \frac{6}{5}$

Step 6.2.4

The final answer is $- \frac{6}{5}$ .

$- \frac{6}{5}$

Step 6.3

At $x = - 3$ the derivative is $- \frac{6}{5}$ . Since this is negative, the function is decreasing on $(- \infty, - 2)$ .

Decreasing on $(- \infty, - 2)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(- 2, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

$f' (- 1) = \frac{10 (- 1)}{{((- 1) + 2)}^{2} {((- 1) - 2)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $10$ by $- 1$ .

$f' (- 1) = \frac{- 10}{{(- 1 + 2)}^{2} {(- 1 - 2)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Add $- 1$ and $2$ .

$f' (- 1) = \frac{- 10}{1^{2} {(- 1 - 2)}^{2}}$

Step 7.2.2.2

Subtract $2$ from $- 1$ .

$f' (- 1) = \frac{- 10}{1^{2} {(- 3)}^{2}}$

Step 7.2.2.3

One to any power is one.

$f' (- 1) = \frac{- 10}{1 {(- 3)}^{2}}$

Step 7.2.2.4

Raise $- 3$ to the power of $2$ .

$f' (- 1) = \frac{- 10}{1 \cdot 9}$

Step 7.2.2.5

Multiply $9$ by $1$ .

$f' (- 1) = \frac{- 10}{9}$

Step 7.2.3

Move the negative in front of the fraction.

$f' (- 1) = - \frac{10}{9}$

Step 7.2.4

The final answer is $- \frac{10}{9}$ .

$- \frac{10}{9}$

Step 7.3

At $x = - 1$ the derivative is $- \frac{10}{9}$ . Since this is negative, the function is decreasing on $(- 2, 0)$ .

Decreasing on $(- 2, 0)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(0, 2)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = \frac{10 (1)}{{((1) + 2)}^{2} {((1) - 2)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Multiply $10$ by $1$ .

$f' (1) = \frac{10}{{(1 + 2)}^{2} {(1 - 2)}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Add $1$ and $2$ .

$f' (1) = \frac{10}{3^{2} {(1 - 2)}^{2}}$

Step 8.2.2.2

Subtract $2$ from $1$ .

$f' (1) = \frac{10}{3^{2} {(- 1)}^{2}}$

Step 8.2.2.3

Raise $3$ to the power of $2$ .

$f' (1) = \frac{10}{9 {(- 1)}^{2}}$

Step 8.2.2.4

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

$f' (1) = \frac{10}{9 \cdot 1}$

Step 8.2.3

Multiply $9$ by $1$ .

$f' (1) = \frac{10}{9}$

Step 8.2.4

The final answer is $\frac{10}{9}$ .

$\frac{10}{9}$

Step 8.3

At $x = 1$ the derivative is $\frac{10}{9}$ . Since this is positive, the function is increasing on $(0, 2)$ .

Increasing on $(0, 2)$ since $f' (x) > 0$

Step 9

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

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Step 9.1

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

$f' (3) = \frac{10 (3)}{{((3) + 2)}^{2} {((3) - 2)}^{2}}$

Step 9.2

Simplify the result.

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Step 9.2.1

Multiply $10$ by $3$ .

$f' (3) = \frac{30}{{(3 + 2)}^{2} {(3 - 2)}^{2}}$

Step 9.2.2

Simplify the denominator.

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Step 9.2.2.1

Add $3$ and $2$ .

$f' (3) = \frac{30}{5^{2} {(3 - 2)}^{2}}$

Step 9.2.2.2

Subtract $2$ from $3$ .

$f' (3) = \frac{30}{5^{2} \cdot 1^{2}}$

Step 9.2.2.3

Raise $5$ to the power of $2$ .

$f' (3) = \frac{30}{25 \cdot 1^{2}}$

Step 9.2.2.4

One to any power is one.

$f' (3) = \frac{30}{25 \cdot 1}$

Step 9.2.3

Reduce the expression by cancelling the common factors.

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Step 9.2.3.1

Multiply $25$ by $1$ .

$f' (3) = \frac{30}{25}$

Step 9.2.3.2

Cancel the common factor of $30$ and $25$ .

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Step 9.2.3.2.1

Factor $5$ out of $30$ .

$f' (3) = \frac{5 (6)}{25}$

Step 9.2.3.2.2

Cancel the common factors.

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Step 9.2.3.2.2.1

Factor $5$ out of $25$ .

$f' (3) = \frac{5 \cdot 6}{5 \cdot 5}$

Step 9.2.3.2.2.2

Cancel the common factor.

$f' (3) = \frac{5 \cdot 6}{5 \cdot 5}$

Step 9.2.3.2.2.3

Rewrite the expression.

$f' (3) = \frac{6}{5}$

Step 9.2.4

The final answer is $\frac{6}{5}$ .

$\frac{6}{5}$

Step 9.3

At $x = 3$ the derivative is $\frac{6}{5}$ . Since this is positive, the function is increasing on $(2, \infty)$ .

Increasing on $(2, \infty)$ since $f' (x) > 0$

Step 10

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

Increasing on: $(0, 2), (2, \infty)$

Decreasing on: $(- \infty, - 2), (- 2, 0)$

Step 11