Calculus Examples

$f (x) = \frac{2 x^{2}}{x^{2} - 16}$

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

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x^{2}}{x^{2} - 16}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x^{2}}{x^{2} - 16}]$ .

$f' (x) = 2 \frac{d}{d x} (\frac{x^{2}}{x^{2} - 16})$

Step 1.1.2

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 $f (x) = x^{2}$ and $g (x) = x^{2} - 16$ .

$f' (x) = 2 (\frac{(x^{2} - 16) \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}})$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

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

$f' (x) = 2 (\frac{(x^{2} - 16) (2 x) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.2

Move $2$ to the left of $x^{2} - 16$ .

$f' (x) = 2 (\frac{2 \cdot ((x^{2} - 16) x) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.3

By the Sum Rule, the derivative of $x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - x^{2} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 16))}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.4

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

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - x^{2} (2 x + \frac{d}{d x} (- 16))}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.5

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - x^{2} (2 x + 0)}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.6

Simplify the expression.

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Step 1.1.3.6.1

Add $2 x$ and $0$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - x^{2} (2 x)}{{(x^{2} - 16)}^{2}})$

Step 1.1.3.6.2

Multiply $2$ by $- 1$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - 2 x^{2} x}{{(x^{2} - 16)}^{2}})$

Step 1.1.4

Raise $x$ to the power of $1$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - 2 (x \cdot x^{2})}{{(x^{2} - 16)}^{2}})$

Step 1.1.5

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

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - 2 x^{1 + 2}}{{(x^{2} - 16)}^{2}})$

Step 1.1.6

Add $1$ and $2$ .

$f' (x) = 2 (\frac{2 (x^{2} - 16) x - 2 x^{3}}{{(x^{2} - 16)}^{2}})$

Step 1.1.7

Combine $2$ and $\frac{2 (x^{2} - 16) x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$ .

$f' (x) = \frac{2 (2 (x^{2} - 16) x - 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8

Simplify.

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Step 1.1.8.1

Apply the distributive property.

$f' (x) = \frac{2 ((2 x^{2} + 2 \cdot - 16) x - 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.2

Apply the distributive property.

$f' (x) = \frac{2 (2 x^{2} x + 2 \cdot (- 16 x) - 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.3

Apply the distributive property.

$f' (x) = \frac{2 (2 x^{2} x) + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4

Simplify the numerator.

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Step 1.1.8.4.1

Simplify each term.

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Step 1.1.8.4.1.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.8.4.1.1.1

Move $x$ .

$f' (x) = \frac{2 (2 (x \cdot x^{2})) + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.1.2

Multiply $x$ by $x^{2}$ .

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Step 1.1.8.4.1.1.2.1

Raise $x$ to the power of $1$ .

$f' (x) = \frac{2 (2 (x \cdot x^{2})) + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.1.2.2

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

$f' (x) = \frac{2 (2 x^{1 + 2}) + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.1.3

Add $1$ and $2$ .

$f' (x) = \frac{2 (2 x^{3}) + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.2

Multiply $2$ by $2$ .

$f' (x) = \frac{4 x^{3} + 2 (2 \cdot (- 16 x)) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.3

Multiply $2$ by $- 16$ .

$f' (x) = \frac{4 x^{3} + 2 (- 32 x) + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.4

Multiply $- 32$ by $2$ .

$f' (x) = \frac{4 x^{3} - 64 x + 2 (- 2 x^{3})}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.1.5

Multiply $- 2$ by $2$ .

$f' (x) = \frac{4 x^{3} - 64 x - 4 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.2

Combine the opposite terms in $4 x^{3} - 64 x - 4 x^{3}$ .

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Step 1.1.8.4.2.1

Subtract $4 x^{3}$ from $4 x^{3}$ .

$f' (x) = \frac{- 64 x + 0}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.4.2.2

Add $- 64 x$ and $0$ .

$f' (x) = \frac{- 64 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.5

Move the negative in front of the fraction.

$f' (x) = - \frac{64 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.8.6

Simplify the denominator.

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Step 1.1.8.6.1

Rewrite $16$ as $4^{2}$ .

$f' (x) = - \frac{64 x}{{(x^{2} - 4^{2})}^{2}}$

Step 1.1.8.6.2

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 $b = 4$ .

$f' (x) = - \frac{64 x}{{((x + 4) (x - 4))}^{2}}$

Step 1.1.8.6.3

Apply the product rule to $(x + 4) (x - 4)$ .

$f' (x) = - \frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$ .

$- \frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- \frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$64 x = 0$

Step 2.3

Divide each term in $64 x = 0$ by $64$ and simplify.

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Step 2.3.1

Divide each term in $64 x = 0$ by $64$ .

$\frac{64 x}{64} = \frac{0}{64}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $64$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{64 x}{64} = \frac{0}{64}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{64}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $64$ .

$x = 0$

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 $\frac{64 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

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 $0$ .

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

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

Step 4.2.2

Set ${(x + 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.2.1

Set ${(x + 4)}^{2}$ equal to $0$ .

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

Step 4.2.2.2

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

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Step 4.2.2.2.1

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

$x + 4 = 0$

Step 4.2.2.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4.2.3

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.3.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 4.2.3.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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Step 4.2.3.2.1

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

$x - 4 = 0$

Step 4.2.3.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 4.2.4

The final solution is all the values that make ${(x + 4)}^{2} {(x - 4)}^{2} = 0$ true.

$x = - 4, 4$

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 $0$ .

$x = - 4, x = 4$

Step 5

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

$(- \infty, - 4) \cup (- 4, 0) \cup (0, 4) \cup (4, \infty)$

Step 6

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 6.1

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

$f' (- 5) = - \frac{64 (- 5)}{{((- 5) + 4)}^{2} {((- 5) - 4)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $64$ by $- 5$ .

$f' (- 5) = - \frac{- 320}{{(- 5 + 4)}^{2} {(- 5 - 4)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Add $- 5$ and $4$ .

$f' (- 5) = - \frac{- 320}{{(- 1)}^{2} {(- 5 - 4)}^{2}}$

Step 6.2.2.2

Subtract $4$ from $- 5$ .

$f' (- 5) = - \frac{- 320}{{(- 1)}^{2} {(- 9)}^{2}}$

Step 6.2.2.3

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

$f' (- 5) = - \frac{- 320}{1 {(- 9)}^{2}}$

Step 6.2.2.4

Raise $- 9$ to the power of $2$ .

$f' (- 5) = - \frac{- 320}{1 \cdot 81}$

Step 6.2.2.5

Multiply $81$ by $1$ .

$f' (- 5) = - \frac{- 320}{81}$

Step 6.2.3

Move the negative in front of the fraction.

$f' (- 5) = \frac{320}{81}$

Step 6.2.4

The final answer is $\frac{320}{81}$ .

$\frac{320}{81}$

Step 6.3

At $x = - 5$ the derivative is $\frac{320}{81}$ . Since this is positive, the function is increasing on $(- \infty, - 4)$ .

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

Step 7

Substitute a value from the interval $(- 4, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = - \frac{64 (- 2)}{{((- 2) + 4)}^{2} {((- 2) - 4)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $64$ by $- 2$ .

$f' (- 2) = - \frac{- 128}{{(- 2 + 4)}^{2} {(- 2 - 4)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Add $- 2$ and $4$ .

$f' (- 2) = - \frac{- 128}{2^{2} {(- 2 - 4)}^{2}}$

Step 7.2.2.2

Subtract $4$ from $- 2$ .

$f' (- 2) = - \frac{- 128}{2^{2} {(- 6)}^{2}}$

Step 7.2.2.3

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

$f' (- 2) = - \frac{- 128}{4 {(- 6)}^{2}}$

Step 7.2.2.4

Raise $- 6$ to the power of $2$ .

$f' (- 2) = - \frac{- 128}{4 \cdot 36}$

Step 7.2.3

Reduce the expression by cancelling the common factors.

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Step 7.2.3.1

Multiply $4$ by $36$ .

$f' (- 2) = - \frac{- 128}{144}$

Step 7.2.3.2

Cancel the common factor of $- 128$ and $144$ .

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Step 7.2.3.2.1

Factor $16$ out of $- 128$ .

$f' (- 2) = - \frac{16 (- 8)}{144}$

Step 7.2.3.2.2

Cancel the common factors.

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Step 7.2.3.2.2.1

Factor $16$ out of $144$ .

$f' (- 2) = - \frac{16 \cdot - 8}{16 \cdot 9}$

Step 7.2.3.2.2.2

Cancel the common factor.

$f' (- 2) = - \frac{16 \cdot - 8}{16 \cdot 9}$

Step 7.2.3.2.2.3

Rewrite the expression.

$f' (- 2) = - \frac{- 8}{9}$

Step 7.2.3.3

Move the negative in front of the fraction.

$f' (- 2) = \frac{8}{9}$

Step 7.2.4

The final answer is $\frac{8}{9}$ .

$\frac{8}{9}$

Step 7.3

At $x = - 2$ the derivative is $\frac{8}{9}$ . Since this is positive, the function is increasing on $(- 4, 0)$ .

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

Step 8

Substitute a value from the interval $(0, 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = - \frac{64 (2)}{{((2) + 4)}^{2} {((2) - 4)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Multiply $64$ by $2$ .

$f' (2) = - \frac{128}{{(2 + 4)}^{2} {(2 - 4)}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Add $2$ and $4$ .

$f' (2) = - \frac{128}{6^{2} {(2 - 4)}^{2}}$

Step 8.2.2.2

Subtract $4$ from $2$ .

$f' (2) = - \frac{128}{6^{2} {(- 2)}^{2}}$

Step 8.2.2.3

Raise $6$ to the power of $2$ .

$f' (2) = - \frac{128}{36 {(- 2)}^{2}}$

Step 8.2.2.4

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

$f' (2) = - \frac{128}{36 \cdot 4}$

Step 8.2.3

Reduce the expression by cancelling the common factors.

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Step 8.2.3.1

Multiply $36$ by $4$ .

$f' (2) = - \frac{128}{144}$

Step 8.2.3.2

Cancel the common factor of $128$ and $144$ .

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Step 8.2.3.2.1

Factor $16$ out of $128$ .

$f' (2) = - \frac{16 (8)}{144}$

Step 8.2.3.2.2

Cancel the common factors.

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Step 8.2.3.2.2.1

Factor $16$ out of $144$ .

$f' (2) = - \frac{16 \cdot 8}{16 \cdot 9}$

Step 8.2.3.2.2.2

Cancel the common factor.

$f' (2) = - \frac{16 \cdot 8}{16 \cdot 9}$

Step 8.2.3.2.2.3

Rewrite the expression.

$f' (2) = - \frac{8}{9}$

Step 8.2.4

The final answer is $- \frac{8}{9}$ .

$- \frac{8}{9}$

Step 8.3

At $x = 2$ the derivative is $- \frac{8}{9}$ . Since this is negative, the function is decreasing on $(0, 4)$ .

Decreasing on $(0, 4)$ since $f' (x) < 0$

Step 9

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 9.1

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

$f' (5) = - \frac{64 (5)}{{((5) + 4)}^{2} {((5) - 4)}^{2}}$

Step 9.2

Simplify the result.

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Step 9.2.1

Multiply $64$ by $5$ .

$f' (5) = - \frac{320}{{(5 + 4)}^{2} {(5 - 4)}^{2}}$

Step 9.2.2

Simplify the denominator.

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Step 9.2.2.1

Add $5$ and $4$ .

$f' (5) = - \frac{320}{9^{2} {(5 - 4)}^{2}}$

Step 9.2.2.2

Subtract $4$ from $5$ .

$f' (5) = - \frac{320}{9^{2} \cdot 1^{2}}$

Step 9.2.2.3

Raise $9$ to the power of $2$ .

$f' (5) = - \frac{320}{81 \cdot 1^{2}}$

Step 9.2.2.4

One to any power is one.

$f' (5) = - \frac{320}{81 \cdot 1}$

Step 9.2.3

Multiply $81$ by $1$ .

$f' (5) = - \frac{320}{81}$

Step 9.2.4

The final answer is $- \frac{320}{81}$ .

$- \frac{320}{81}$

Step 9.3

At $x = 5$ the derivative is $- \frac{320}{81}$ . Since this is negative, the function is decreasing on $(4, \infty)$ .

Decreasing on $(4, \infty)$ since $f' (x) < 0$

Step 10

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

Increasing on: $(- \infty, - 4), (- 4, 0)$

Decreasing on: $(0, 4), (4, \infty)$

Step 11