Calculus Examples

$\frac{x + 9}{x^{2} - 81}$

Step 1

Write $\frac{x + 9}{x^{2} - 81}$ as a function.

$f (x) = \frac{x + 9}{x^{2} - 81}$

Step 2

Find the second 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 $f (x) = x + 9$ and $g (x) = x^{2} - 81$ .

$\frac{(x^{2} - 81) \frac{d}{d x} [x + 9] - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

$\frac{(x^{2} - 81) (\frac{d}{d x} [x] + \frac{d}{d x} [9]) - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

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 $n = 1$ .

$\frac{(x^{2} - 81) (1 + \frac{d}{d x} [9]) - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.3

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

$\frac{(x^{2} - 81) (1 + 0) - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.4

Simplify the expression.

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Step 2.1.2.4.1

Add $1$ and $0$ .

$\frac{(x^{2} - 81) \cdot 1 - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.4.2

Multiply $x^{2} - 81$ by $1$ .

$\frac{x^{2} - 81 - (x + 9) \frac{d}{d x} [x^{2} - 81]}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.5

By the Sum Rule, the derivative of $x^{2} - 81$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 81]$ .

$\frac{x^{2} - 81 - (x + 9) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 81])}{{(x^{2} - 81)}^{2}}$

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 $n = 2$ .

$\frac{x^{2} - 81 - (x + 9) (2 x + \frac{d}{d x} [- 81])}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.7

Since $- 81$ is constant with respect to $x$ , the derivative of $- 81$ with respect to $x$ is $0$ .

$\frac{x^{2} - 81 - (x + 9) (2 x + 0)}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.8

Simplify the expression.

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Step 2.1.2.8.1

Add $2 x$ and $0$ .

$\frac{x^{2} - 81 - (x + 9) (2 x)}{{(x^{2} - 81)}^{2}}$

Step 2.1.2.8.2

Multiply $2$ by $- 1$ .

$\frac{x^{2} - 81 - 2 (x + 9) x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3

Simplify.

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Step 2.1.3.1

Apply the distributive property.

$\frac{x^{2} - 81 + (- 2 x - 2 \cdot 9) x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.2

Apply the distributive property.

$\frac{x^{2} - 81 - 2 x \cdot x - 2 \cdot 9 x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.3

Simplify the numerator.

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Step 2.1.3.3.1

Simplify each term.

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Step 2.1.3.3.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.3.3.1.1.1

Move $x$ .

$\frac{x^{2} - 81 - 2 (x \cdot x) - 2 \cdot 9 x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.3.1.1.2

Multiply $x$ by $x$ .

$\frac{x^{2} - 81 - 2 x^{2} - 2 \cdot 9 x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.3.1.2

Multiply $- 2$ by $9$ .

$\frac{x^{2} - 81 - 2 x^{2} - 18 x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.3.2

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

$\frac{- x^{2} - 81 - 18 x}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.4

Reorder terms.

$\frac{- x^{2} - 18 x - 81}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5

Factor by grouping.

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Step 2.1.3.5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 81 = 81$ and whose sum is $b = - 18$ .

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Step 2.1.3.5.1.1

Factor $- 18$ out of $- 18 x$ .

$\frac{- x^{2} - 18 (x) - 81}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5.1.2

Rewrite $- 18$ as $- 9$ plus $- 9$

$\frac{- x^{2} + (- 9 - 9) x - 81}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5.1.3

Apply the distributive property.

$\frac{- x^{2} - 9 x - 9 x - 81}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5.2

Factor out the greatest common factor from each group.

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Step 2.1.3.5.2.1

Group the first two terms and the last two terms.

$\frac{(- x^{2} - 9 x) - 9 x - 81}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{x (- x - 9) + 9 (- x - 9)}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.5.3

Factor the polynomial by factoring out the greatest common factor, $- x - 9$ .

$\frac{(- x - 9) (x + 9)}{{(x^{2} - 81)}^{2}}$

Step 2.1.3.6

Simplify the denominator.

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Step 2.1.3.6.1

Rewrite $81$ as $9^{2}$ .

$\frac{(- x - 9) (x + 9)}{{(x^{2} - 9^{2})}^{2}}$

Step 2.1.3.6.2

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 $b = 9$ .

$\frac{(- x - 9) (x + 9)}{{((x + 9) (x - 9))}^{2}}$

Step 2.1.3.6.3

Apply the product rule to $(x + 9) (x - 9)$ .

$\frac{(- x - 9) (x + 9)}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7

Simplify the numerator.

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Step 2.1.3.7.1

Factor $- 1$ out of $- x$ .

$\frac{(- (x) - 9) (x + 9)}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.2

Rewrite $- 9$ as $- 1 (9)$ .

$\frac{(- (x) - 1 (9)) (x + 9)}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.3

Factor $- 1$ out of $- (x) - 1 (9)$ .

$\frac{- (x + 9) (x + 9)}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.4

Rewrite $- (x + 9)$ as $- 1 (x + 9)$ .

$\frac{- 1 (x + 9) (x + 9)}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.5

Raise $x + 9$ to the power of $1$ .

$\frac{- 1 ({(x + 9)}^{1} (x + 9))}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.6

Raise $x + 9$ to the power of $1$ .

$\frac{- 1 ({(x + 9)}^{1} {(x + 9)}^{1})}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.7

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

$\frac{- 1 {(x + 9)}^{1 + 1}}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.7.8

Add $1$ and $1$ .

$\frac{- 1 {(x + 9)}^{2}}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.8

Cancel the common factor of ${(x + 9)}^{2}$ .

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Step 2.1.3.8.1

Cancel the common factor.

$\frac{- 1 {(x + 9)}^{2}}{{(x + 9)}^{2} {(x - 9)}^{2}}$

Step 2.1.3.8.2

Rewrite the expression.

$\frac{- 1}{{(x - 9)}^{2}}$

Step 2.1.3.9

Move the negative in front of the fraction.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{{(x - 9)}^{2}}$ .

$- \frac{d}{d x} [\frac{1}{{(x - 9)}^{2}}] + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

Rewrite $\frac{1}{{(x - 9)}^{2}}$ as ${({(x - 9)}^{2})}^{- 1}$ .

$- \frac{d}{d x} [{({(x - 9)}^{2})}^{- 1}] + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.2.2

Multiply the exponents in ${({(x - 9)}^{2})}^{- 1}$ .

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Step 2.2.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{d}{d x} [{(x - 9)}^{2 \cdot - 1}] + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

$- \frac{d}{d x} [{(x - 9)}^{- 2}] + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x - 9$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u$ as $x - 9$ .

$- (\frac{d}{d u} [u^{- 2}] \frac{d}{d x} [x - 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.3.2

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

$- (- 2 u^{- 3} \frac{d}{d x} [x - 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.3.3

Replace all occurrences of $u$ with $x - 9$ .

$- (- 2 {(x - 9)}^{- 3} \frac{d}{d x} [x - 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4

Differentiate.

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Step 2.2.4.1

Multiply $- 2$ by $- 1$ .

$2 ({(x - 9)}^{- 3} \frac{d}{d x} [x - 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.2

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

$2 {(x - 9)}^{- 3} (\frac{d}{d x} [x] + \frac{d}{d x} [- 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.3

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

$2 {(x - 9)}^{- 3} (1 + \frac{d}{d x} [- 9]) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.4

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

$2 {(x - 9)}^{- 3} (1 + 0) + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.5

Simplify the expression.

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Step 2.2.4.5.1

Add $1$ and $0$ .

$2 {(x - 9)}^{- 3} \cdot 1 + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.5.2

Multiply $2$ by $1$ .

$2 {(x - 9)}^{- 3} + \frac{1}{{(x - 9)}^{2}} \frac{d}{d x} [- 1]$

Step 2.2.4.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 {(x - 9)}^{- 3} + \frac{1}{{(x - 9)}^{2}} \cdot 0$

Step 2.2.4.7

Simplify the expression.

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Step 2.2.4.7.1

Multiply $\frac{1}{{(x - 9)}^{2}}$ by $0$ .

$2 {(x - 9)}^{- 3} + 0$

Step 2.2.4.7.2

Add $2 {(x - 9)}^{- 3}$ and $0$ .

$2 {(x - 9)}^{- 3}$

Step 2.2.5

Simplify.

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Step 2.2.5.1

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

$2 \frac{1}{{(x - 9)}^{3}}$

Step 2.2.5.2

Combine $2$ and $\frac{1}{{(x - 9)}^{3}}$ .

$f'' (x) = \frac{2}{{(x - 9)}^{3}}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2}{{(x - 9)}^{3}}$ .

$\frac{2}{{(x - 9)}^{3}}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{2}{{(x - 9)}^{3}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$\frac{2}{{(x - 9)}^{3}} = 0$

Step 3.2

Set the numerator equal to zero.

$2 = 0$

Step 3.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 4

No values found that can make the second derivative equal to $0$ .

No Inflection Points