Calculus Examples

$f (x) = \frac{- x - 2}{{(x - 5)}^{2}} + 9$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

By the Sum Rule, the derivative of $\frac{- x - 2}{{(x - 5)}^{2}} + 9$ with respect to $x$ is $\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}] + \frac{d}{d x} [9]$ .

$\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}] + \frac{d}{d x} [9]$

Step 1.2

Evaluate $\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}]$ .

Tap for more steps...

Step 1.2.1

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 $f (x) = - x - 2$ and $g (x) = {(x - 5)}^{2}$ .

$\frac{{(x - 5)}^{2} \frac{d}{d x} [- x - 2] - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.2

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

$\frac{{(x - 5)}^{2} (\frac{d}{d x} [- x] + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{{(x - 5)}^{2} (- \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.4

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.6

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 $g (x) = x - 5$ .

Tap for more steps...

Step 1.2.6.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (\frac{d}{d u} [u^{2}] \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.6.2

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 u \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.6.3

Replace all occurrences of $u$ with $x - 5$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.7

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 5]))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.8

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (1 + \frac{d}{d x} [- 5]))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.9

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.10

Multiply $- 1$ by $1$ .

$\frac{{(x - 5)}^{2} (- 1 + 0) - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.11

Add $- 1$ and $0$ .

$\frac{{(x - 5)}^{2} \cdot - 1 - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.12

Move $- 1$ to the left of ${(x - 5)}^{2}$ .

$\frac{- 1 \cdot {(x - 5)}^{2} - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.13

Rewrite $- 1 {(x - 5)}^{2}$ as $- {(x - 5)}^{2}$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.14

Add $1$ and $0$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5) \cdot 1)}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.15

Multiply $2$ by $1$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.16

Multiply $2$ by $- 1$ .

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 1.2.17

Multiply the exponents in ${({(x - 5)}^{2})}^{2}$ .

Tap for more steps...

Step 1.2.17.1

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

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{(x - 5)}^{2 \cdot 2}} + \frac{d}{d x} [9]$

Step 1.2.17.2

Multiply $2$ by $2$ .

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 1.2.18

Factor $x - 5$ out of $- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)$ .

Tap for more steps...

Step 1.2.18.1

Factor $x - 5$ out of $- {(x - 5)}^{2}$ .

$\frac{(x - 5) (- (x - 5)) - 2 (- x - 2) (x - 5)}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 1.2.18.2

Factor $x - 5$ out of $- 2 (- x - 2) (x - 5)$ .

$\frac{(x - 5) (- (x - 5)) + (x - 5) (- 2 (- x - 2))}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 1.2.18.3

Factor $x - 5$ out of $(x - 5) (- (x - 5)) + (x - 5) (- 2 (- x - 2))$ .

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 1.2.19

Cancel the common factors.

Tap for more steps...

Step 1.2.19.1

Factor $x - 5$ out of ${(x - 5)}^{4}$ .

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{(x - 5) {(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 1.2.19.2

Cancel the common factor.

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{(x - 5) {(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 1.2.19.3

Rewrite the expression.

$\frac{- (x - 5) - 2 (- x - 2)}{{(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 1.3

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

$\frac{- (x - 5) - 2 (- x - 2)}{{(x - 5)}^{3}} + 0$

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Apply the distributive property.

$\frac{- x - - 5 - 2 (- x - 2)}{{(x - 5)}^{3}} + 0$

Step 1.4.2

Apply the distributive property.

$\frac{- x - - 5 - 2 (- x) - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 1.4.3

Combine terms.

Tap for more steps...

Step 1.4.3.1

Multiply $- 1$ by $- 5$ .

$\frac{- x + 5 - 2 (- x) - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 1.4.3.2

Multiply $- 1$ by $- 2$ .

$\frac{- x + 5 + 2 x - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 1.4.3.3

Multiply $- 2$ by $- 2$ .

$\frac{- x + 5 + 2 x + 4}{{(x - 5)}^{3}} + 0$

Step 1.4.3.4

Add $- x$ and $2 x$ .

$\frac{x + 5 + 4}{{(x - 5)}^{3}} + 0$

Step 1.4.3.5

Add $5$ and $4$ .

$\frac{x + 9}{{(x - 5)}^{3}} + 0$

Step 1.4.3.6

Add $\frac{x + 9}{{(x - 5)}^{3}}$ and $0$ .

$\frac{x + 9}{{(x - 5)}^{3}}$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

$f'' (x) = \frac{{(x - 5)}^{3} \frac{d}{d x} (x + 9) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{({(x - 5)}^{3})}^{2}}$

Step 2.2

Differentiate.

Tap for more steps...

Step 2.2.1

Multiply the exponents in ${({(x - 5)}^{3})}^{2}$ .

Tap for more steps...

Step 2.2.1.1

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

$f'' (x) = \frac{{(x - 5)}^{3} \frac{d}{d x} (x + 9) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{3 \cdot 2}}$

Step 2.2.1.2

Multiply $3$ by $2$ .

$f'' (x) = \frac{{(x - 5)}^{3} \frac{d}{d x} (x + 9) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.2.2

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

$f'' (x) = \frac{{(x - 5)}^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (9)) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.2.3

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

$f'' (x) = \frac{{(x - 5)}^{3} (1 + \frac{d}{d x} (9)) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.2.4

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

$f'' (x) = \frac{{(x - 5)}^{3} (1 + 0) - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.2.5

Simplify the expression.

Tap for more steps...

Step 2.2.5.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 5)}^{3} \cdot 1 - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.2.5.2

Multiply ${(x - 5)}^{3}$ by $1$ .

$f'' (x) = \frac{{(x - 5)}^{3} - (x + 9) \frac{d}{d x} {(x - 5)}^{3}}{{(x - 5)}^{6}}$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x - 5$ .

Tap for more steps...

Step 2.3.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$f'' (x) = \frac{{(x - 5)}^{3} - (x + 9) (\frac{d}{d u} (u^{3}) \frac{d}{d x} (x - 5))}{{(x - 5)}^{6}}$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$f'' (x) = \frac{{(x - 5)}^{3} - (x + 9) (3 u^{2} \frac{d}{d x} (x - 5))}{{(x - 5)}^{6}}$

Step 2.3.3

Replace all occurrences of $u$ with $x - 5$ .

$f'' (x) = \frac{{(x - 5)}^{3} - (x + 9) (3 {(x - 5)}^{2} \frac{d}{d x} (x - 5))}{{(x - 5)}^{6}}$

Step 2.4

Simplify with factoring out.

Tap for more steps...

Step 2.4.1

Multiply $3$ by $- 1$ .

$f'' (x) = \frac{{(x - 5)}^{3} - 3 (x + 9) ({(x - 5)}^{2} \frac{d}{d x} (x - 5))}{{(x - 5)}^{6}}$

Step 2.4.2

Factor ${(x - 5)}^{2}$ out of ${(x - 5)}^{3} - 3 (x + 9) ({(x - 5)}^{2} \frac{d}{d x} [x - 5])$ .

Tap for more steps...

Step 2.4.2.1

Factor ${(x - 5)}^{2}$ out of ${(x - 5)}^{3}$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x - 5) - 3 (x + 9) ({(x - 5)}^{2} \frac{d}{d x} (x - 5))}{{(x - 5)}^{6}}$

Step 2.4.2.2

Factor ${(x - 5)}^{2}$ out of $- 3 (x + 9) ({(x - 5)}^{2} \frac{d}{d x} [x - 5])$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x - 5) + {(x - 5)}^{2} (- 3 (x + 9) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{6}}$

Step 2.4.2.3

Factor ${(x - 5)}^{2}$ out of ${(x - 5)}^{2} (x - 5) + {(x - 5)}^{2} (- 3 (x + 9) (\frac{d}{d x} [x - 5]))$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x - 5 - 3 (x + 9) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{6}}$

Step 2.5

Cancel the common factors.

Tap for more steps...

Step 2.5.1

Factor ${(x - 5)}^{2}$ out of ${(x - 5)}^{6}$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x - 5 - 3 (x + 9) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{2} {(x - 5)}^{4}}$

Step 2.5.2

Cancel the common factor.

$f'' (x) = \frac{{(x - 5)}^{2} (x - 5 - 3 (x + 9) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{2} {(x - 5)}^{4}}$

Step 2.5.3

Rewrite the expression.

$f'' (x) = \frac{x - 5 - 3 (x + 9) (\frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.6

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$f'' (x) = \frac{x - 5 - 3 (x + 9) (\frac{d}{d x} (x) + \frac{d}{d x} (- 5))}{{(x - 5)}^{4}}$

Step 2.7

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

$f'' (x) = \frac{x - 5 - 3 (x + 9) (1 + \frac{d}{d x} (- 5))}{{(x - 5)}^{4}}$

Step 2.8

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f'' (x) = \frac{x - 5 - 3 (x + 9) (1 + 0)}{{(x - 5)}^{4}}$

Step 2.9

Simplify the expression.

Tap for more steps...

Step 2.9.1

Add $1$ and $0$ .

$f'' (x) = \frac{x - 5 - 3 (x + 9) \cdot 1}{{(x - 5)}^{4}}$

Step 2.9.2

Multiply $- 3$ by $1$ .

$f'' (x) = \frac{x - 5 - 3 (x + 9)}{{(x - 5)}^{4}}$

Step 2.10

Simplify.

Tap for more steps...

Step 2.10.1

Apply the distributive property.

$f'' (x) = \frac{x - 5 - 3 x - 3 \cdot 9}{{(x - 5)}^{4}}$

Step 2.10.2

Simplify the numerator.

Tap for more steps...

Step 2.10.2.1

Multiply $- 3$ by $9$ .

$f'' (x) = \frac{x - 5 - 3 x - 27}{{(x - 5)}^{4}}$

Step 2.10.2.2

Subtract $3 x$ from $x$ .

$f'' (x) = \frac{- 2 x - 5 - 27}{{(x - 5)}^{4}}$

Step 2.10.2.3

Subtract $27$ from $- 5$ .

$f'' (x) = \frac{- 2 x - 32}{{(x - 5)}^{4}}$

Step 2.10.3

Factor $2$ out of $- 2 x - 32$ .

Tap for more steps...

Step 2.10.3.1

Factor $2$ out of $- 2 x$ .

$f'' (x) = \frac{2 (- x) - 32}{{(x - 5)}^{4}}$

Step 2.10.3.2

Factor $2$ out of $- 32$ .

$f'' (x) = \frac{2 (- x) + 2 (- 16)}{{(x - 5)}^{4}}$

Step 2.10.3.3

Factor $2$ out of $2 (- x) + 2 (- 16)$ .

$f'' (x) = \frac{2 (- x - 16)}{{(x - 5)}^{4}}$

Step 2.10.4

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

$f'' (x) = \frac{2 (- (x) - 16)}{{(x - 5)}^{4}}$

Step 2.10.5

Rewrite $- 16$ as $- 1 (16)$ .

$f'' (x) = \frac{2 (- (x) - 1 \cdot 16)}{{(x - 5)}^{4}}$

Step 2.10.6

Factor $- 1$ out of $- (x) - 1 (16)$ .

$f'' (x) = \frac{2 (- (x + 16))}{{(x - 5)}^{4}}$

Step 2.10.7

Rewrite $- (x + 16)$ as $- 1 (x + 16)$ .

$f'' (x) = \frac{2 (- 1 (x + 16))}{{(x - 5)}^{4}}$

Step 2.10.8

Move the negative in front of the fraction.

$f'' (x) = - \frac{2 (x + 16)}{{(x - 5)}^{4}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{x + 9}{{(x - 5)}^{3}} = 0$

Step 4

Find the first derivative.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

By the Sum Rule, the derivative of $\frac{- x - 2}{{(x - 5)}^{2}} + 9$ with respect to $x$ is $\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}] + \frac{d}{d x} [9]$ .

$\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}] + \frac{d}{d x} [9]$

Step 4.1.2

Evaluate $\frac{d}{d x} [\frac{- x - 2}{{(x - 5)}^{2}}]$ .

Tap for more steps...

Step 4.1.2.1

$\frac{{(x - 5)}^{2} \frac{d}{d x} [- x - 2] - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.2

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

$\frac{{(x - 5)}^{2} (\frac{d}{d x} [- x] + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{{(x - 5)}^{2} (- \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.4

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + \frac{d}{d x} [- 2]) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) \frac{d}{d x} [{(x - 5)}^{2}]}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.6

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 $g (x) = x - 5$ .

Tap for more steps...

Step 4.1.2.6.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (\frac{d}{d u} [u^{2}] \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.6.2

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 u \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.6.3

Replace all occurrences of $u$ with $x - 5$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) \frac{d}{d x} [x - 5])}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.7

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 5]))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.8

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

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (1 + \frac{d}{d x} [- 5]))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.9

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{{(x - 5)}^{2} (- 1 \cdot 1 + 0) - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.10

Multiply $- 1$ by $1$ .

$\frac{{(x - 5)}^{2} (- 1 + 0) - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.11

Add $- 1$ and $0$ .

$\frac{{(x - 5)}^{2} \cdot - 1 - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.12

Move $- 1$ to the left of ${(x - 5)}^{2}$ .

$\frac{- 1 \cdot {(x - 5)}^{2} - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.13

Rewrite $- 1 {(x - 5)}^{2}$ as $- {(x - 5)}^{2}$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5) (1 + 0))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.14

Add $1$ and $0$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5) \cdot 1)}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.15

Multiply $2$ by $1$ .

$\frac{- {(x - 5)}^{2} - (- x - 2) (2 (x - 5))}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.16

Multiply $2$ by $- 1$ .

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{({(x - 5)}^{2})}^{2}} + \frac{d}{d x} [9]$

Step 4.1.2.17

Multiply the exponents in ${({(x - 5)}^{2})}^{2}$ .

Tap for more steps...

Step 4.1.2.17.1

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

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{(x - 5)}^{2 \cdot 2}} + \frac{d}{d x} [9]$

Step 4.1.2.17.2

Multiply $2$ by $2$ .

$\frac{- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 4.1.2.18

Factor $x - 5$ out of $- {(x - 5)}^{2} - 2 (- x - 2) (x - 5)$ .

Tap for more steps...

Step 4.1.2.18.1

Factor $x - 5$ out of $- {(x - 5)}^{2}$ .

$\frac{(x - 5) (- (x - 5)) - 2 (- x - 2) (x - 5)}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 4.1.2.18.2

Factor $x - 5$ out of $- 2 (- x - 2) (x - 5)$ .

$\frac{(x - 5) (- (x - 5)) + (x - 5) (- 2 (- x - 2))}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 4.1.2.18.3

Factor $x - 5$ out of $(x - 5) (- (x - 5)) + (x - 5) (- 2 (- x - 2))$ .

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{{(x - 5)}^{4}} + \frac{d}{d x} [9]$

Step 4.1.2.19

Cancel the common factors.

Tap for more steps...

Step 4.1.2.19.1

Factor $x - 5$ out of ${(x - 5)}^{4}$ .

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{(x - 5) {(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 4.1.2.19.2

Cancel the common factor.

$\frac{(x - 5) (- (x - 5) - 2 (- x - 2))}{(x - 5) {(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 4.1.2.19.3

Rewrite the expression.

$\frac{- (x - 5) - 2 (- x - 2)}{{(x - 5)}^{3}} + \frac{d}{d x} [9]$

Step 4.1.3

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

$\frac{- (x - 5) - 2 (- x - 2)}{{(x - 5)}^{3}} + 0$

Step 4.1.4

Simplify.

Tap for more steps...

Step 4.1.4.1

Apply the distributive property.

$\frac{- x - - 5 - 2 (- x - 2)}{{(x - 5)}^{3}} + 0$

Step 4.1.4.2

Apply the distributive property.

$\frac{- x - - 5 - 2 (- x) - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3

Combine terms.

Tap for more steps...

Step 4.1.4.3.1

Multiply $- 1$ by $- 5$ .

$\frac{- x + 5 - 2 (- x) - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3.2

Multiply $- 1$ by $- 2$ .

$\frac{- x + 5 + 2 x - 2 \cdot - 2}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3.3

Multiply $- 2$ by $- 2$ .

$\frac{- x + 5 + 2 x + 4}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3.4

Add $- x$ and $2 x$ .

$\frac{x + 5 + 4}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3.5

Add $5$ and $4$ .

$\frac{x + 9}{{(x - 5)}^{3}} + 0$

Step 4.1.4.3.6

Add $\frac{x + 9}{{(x - 5)}^{3}}$ and $0$ .

$f' (x) = \frac{x + 9}{{(x - 5)}^{3}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x + 9}{{(x - 5)}^{3}}$ .

$\frac{x + 9}{{(x - 5)}^{3}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{x + 9}{{(x - 5)}^{3}} = 0$ .

Tap for more steps...

Step 5.1

Set the first derivative equal to $0$ .

$\frac{x + 9}{{(x - 5)}^{3}} = 0$

Step 5.2

Set the numerator equal to zero.

$x + 9 = 0$

Step 5.3

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

Step 6.1

Set the denominator in $\frac{x + 9}{{(x - 5)}^{3}}$ equal to $0$ to find where the expression is undefined.

${(x - 5)}^{3} = 0$

Step 6.2

Solve for $x$ .

Tap for more steps...

Step 6.2.1

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Critical points to evaluate.

$x = - 9$

Step 8

Evaluate the second derivative at $x = - 9$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2 ((- 9) + 16)}{{((- 9) - 5)}^{4}}$

Step 9

Evaluate the second derivative.

Tap for more steps...

Step 9.1

Add $- 9$ and $16$ .

$- \frac{2 \cdot 7}{{(- 9 - 5)}^{4}}$

Step 9.2

Simplify the denominator.

Tap for more steps...

Step 9.2.1

Subtract $5$ from $- 9$ .

$- \frac{2 \cdot 7}{{(- 14)}^{4}}$

Step 9.2.2

Raise $- 14$ to the power of $4$ .

$- \frac{2 \cdot 7}{38416}$

Step 9.3

Reduce the expression by cancelling the common factors.

Tap for more steps...

Step 9.3.1

Multiply $2$ by $7$ .

$- \frac{14}{38416}$

Step 9.3.2

Cancel the common factor of $14$ and $38416$ .

Tap for more steps...

Step 9.3.2.1

Factor $14$ out of $14$ .

$- \frac{14 (1)}{38416}$

Step 9.3.2.2

Cancel the common factors.

Tap for more steps...

Step 9.3.2.2.1

Factor $14$ out of $38416$ .

$- \frac{14 \cdot 1}{14 \cdot 2744}$

Step 9.3.2.2.2

Cancel the common factor.

$- \frac{14 \cdot 1}{14 \cdot 2744}$

Step 9.3.2.2.3

Rewrite the expression.

$- \frac{1}{2744}$

Step 10

$x = - 9$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 9$ is a local maximum

Step 11

Find the y-value when $x = - 9$ .

Tap for more steps...

Step 11.1

Replace the variable $x$ with $- 9$ in the expression.

$f (- 9) = \frac{- (- 9) - 2}{{((- 9) - 5)}^{2}} + 9$

Step 11.2

Simplify the result.

Tap for more steps...

Step 11.2.1

Simplify the expression.

Tap for more steps...

Step 11.2.1.1

Multiply $- 1$ by $- 9$ .

$f (- 9) = \frac{9 - 2}{{((- 9) - 5)}^{2}} + 9$

Step 11.2.1.2

Subtract $2$ from $9$ .

$f (- 9) = \frac{7}{{((- 9) - 5)}^{2}} + 9$

Step 11.2.2

Simplify each term.

Tap for more steps...

Step 11.2.2.1

Simplify the denominator.

Tap for more steps...

Step 11.2.2.1.1

Subtract $5$ from $- 9$ .

$f (- 9) = \frac{7}{{(- 14)}^{2}} + 9$

Step 11.2.2.1.2

Raise $- 14$ to the power of $2$ .

$f (- 9) = \frac{7}{196} + 9$

Step 11.2.2.2

Cancel the common factor of $7$ and $196$ .

Tap for more steps...

Step 11.2.2.2.1

Factor $7$ out of $7$ .

$f (- 9) = \frac{7 (1)}{196} + 9$

Step 11.2.2.2.2

Cancel the common factors.

Tap for more steps...

Step 11.2.2.2.2.1

Factor $7$ out of $196$ .

$f (- 9) = \frac{7 \cdot 1}{7 \cdot 28} + 9$

Step 11.2.2.2.2.2

Cancel the common factor.

$f (- 9) = \frac{7 \cdot 1}{7 \cdot 28} + 9$

Step 11.2.2.2.2.3

Rewrite the expression.

$f (- 9) = \frac{1}{28} + 9$

Step 11.2.3

To write $9$ as a fraction with a common denominator, multiply by $\frac{28}{28}$ .

$f (- 9) = \frac{1}{28} + 9 \cdot \frac{28}{28}$

Step 11.2.4

Combine $9$ and $\frac{28}{28}$ .

$f (- 9) = \frac{1}{28} + \frac{9 \cdot 28}{28}$

Step 11.2.5

Combine the numerators over the common denominator.

$f (- 9) = \frac{1 + 9 \cdot 28}{28}$

Step 11.2.6

Simplify the numerator.

Tap for more steps...

Step 11.2.6.1

Multiply $9$ by $28$ .

$f (- 9) = \frac{1 + 252}{28}$

Step 11.2.6.2

Add $1$ and $252$ .

$f (- 9) = \frac{253}{28}$

Step 11.2.7

The final answer is $\frac{253}{28}$ .

$y = \frac{253}{28}$

Step 12

These are the local extrema for $f (x) = \frac{- x - 2}{{(x - 5)}^{2}} + 9$ .

$(- 9, \frac{253}{28})$ is a local maxima

Step 13