Calculus Examples

$y = (x + 9) e^{- \frac{x}{4}}$

Step 1

Write $y = (x + 9) e^{- \frac{x}{4}}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 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 $f (x) = x + 9$ and $g (x) = e^{- \frac{x}{4}}$ .

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - \frac{x}{4}$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- \frac{x}{4}$ .

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

Step 2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 2.2.3

Replace all occurrences of $u$ with $- \frac{x}{4}$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x]$ .

$(x + 9) e^{- \frac{x}{4}} (- \frac{1}{4} \frac{d}{d x} [x]) + e^{- \frac{x}{4}} \frac{d}{d x} [x + 9]$

Step 2.3.2

Combine $e^{- \frac{x}{4}}$ and $\frac{1}{4}$ .

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

Step 2.3.3

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4} \cdot 1) + e^{- \frac{x}{4}} \frac{d}{d x} [x + 9]$

Step 2.3.4

Multiply $- 1$ by $1$ .

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

Step 2.3.5

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

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

Step 2.3.6

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} (1 + \frac{d}{d x} [9])$

Step 2.3.7

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} (1 + 0)$

Step 2.3.8

Simplify the expression.

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Step 2.3.8.1

Add $1$ and $0$ .

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} \cdot 1$

Step 2.3.8.2

Multiply $e^{- \frac{x}{4}}$ by $1$ .

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$x (- \frac{e^{- \frac{x}{4}}}{4}) + 9 (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $x$ and $\frac{e^{- \frac{x}{4}}}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + 9 (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 2.4.2.2

Multiply $- 1$ by $9$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - 9 \frac{e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 2.4.2.3

Combine $- 9$ and $\frac{e^{- \frac{x}{4}}}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 2.4.2.4

Move the negative in front of the fraction.

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 2.4.2.5

To write $e^{- \frac{x}{4}}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}} \cdot \frac{4}{4}$

Step 2.4.2.6

Combine $e^{- \frac{x}{4}}$ and $\frac{4}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + \frac{e^{- \frac{x}{4}} \cdot 4}{4}$

Step 2.4.2.7

Combine the numerators over the common denominator.

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}} + e^{- \frac{x}{4}} \cdot 4}{4}$

Step 2.4.2.8

Move $4$ to the left of $e^{- \frac{x}{4}}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}} + 4 \cdot e^{- \frac{x}{4}}}{4}$

Step 2.4.2.9

Add $- 9 e^{- \frac{x}{4}}$ and $4 e^{- \frac{x}{4}}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 5 e^{- \frac{x}{4}}}{4}$

Step 2.4.2.10

Move the negative in front of the fraction.

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4}$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4}$ with respect to $x$ is $\frac{d}{d x} [- \frac{x e^{- \frac{x}{4}}}{4}] + \frac{d}{d x} [- \frac{5 e^{- \frac{x}{4}}}{4}]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [- \frac{x e^{- \frac{x}{4}}}{4}]$ .

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Step 3.2.1

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x e^{- \frac{x}{4}}}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x e^{- \frac{x}{4}}]$ .

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

Step 3.2.2

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 $f (x) = x$ and $g (x) = e^{- \frac{x}{4}}$ .

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

Step 3.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - \frac{x}{4}$ .

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Step 3.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- \frac{x}{4}$ .

$f'' (x) = - \frac{1}{4} \cdot (x (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- \frac{x}{4})) + e^{- \frac{x}{4}} \frac{d}{d x} (x)) + \frac{d}{d x} (- \frac{5 e^{- \frac{x}{4}}}{4})$

Step 3.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - \frac{1}{4} \cdot (x (e^{u_{1}} \frac{d}{d x} (- \frac{x}{4})) + e^{- \frac{x}{4}} \frac{d}{d x} (x)) + \frac{d}{d x} (- \frac{5 e^{- \frac{x}{4}}}{4})$

Step 3.2.3.3

Replace all occurrences of $u_{1}$ with $- \frac{x}{4}$ .

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

Step 3.2.4

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x]$ .

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

Multiply $- 1$ by $1$ .

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

Step 3.2.8

Combine $e^{- \frac{x}{4}}$ and $\frac{1}{4}$ .

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

Step 3.2.9

Combine $x$ and $\frac{e^{- \frac{x}{4}}}{4}$ .

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

Step 3.2.10

Multiply $e^{- \frac{x}{4}}$ by $1$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [- \frac{5 e^{- \frac{x}{4}}}{4}]$ .

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Step 3.3.1

Since $- \frac{5}{4}$ is constant with respect to $x$ , the derivative of $- \frac{5 e^{- \frac{x}{4}}}{4}$ with respect to $x$ is $- \frac{5}{4} \frac{d}{d x} [e^{- \frac{x}{4}}]$ .

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

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - \frac{x}{4}$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- \frac{x}{4}$ .

$f'' (x) = - \frac{1}{4} \cdot (- \frac{x e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}) - \frac{5}{4} \cdot (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- \frac{x}{4}))$

Step 3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - \frac{1}{4} \cdot (- \frac{x e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}) - \frac{5}{4} \cdot (e^{u_{2}} \frac{d}{d x} (- \frac{x}{4}))$

Step 3.3.2.3

Replace all occurrences of $u_{2}$ with $- \frac{x}{4}$ .

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

Step 3.3.3

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x]$ .

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

Step 3.3.4

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

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

Step 3.3.5

Multiply $- 1$ by $1$ .

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

Step 3.3.6

Combine $e^{- \frac{x}{4}}$ and $\frac{1}{4}$ .

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

Step 3.3.7

Multiply $- 1$ by $- 1$ .

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

Step 3.3.8

Multiply $\frac{5}{4}$ by $1$ .

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

Step 3.3.9

Multiply $\frac{5}{4}$ by $\frac{e^{- \frac{x}{4}}}{4}$ .

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

Step 3.3.10

Multiply $4$ by $4$ .

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

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

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

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.2.2

Multiply $\frac{1}{4}$ by $1$ .

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

Step 3.4.2.3

Multiply $\frac{1}{4}$ by $\frac{x e^{- \frac{x}{4}}}{4}$ .

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

Step 3.4.2.4

Multiply $4$ by $4$ .

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

Step 3.4.2.5

Combine $e^{- \frac{x}{4}}$ and $\frac{1}{4}$ .

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

Step 3.4.2.6

To write $- \frac{e^{- \frac{x}{4}}}{4}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 3.4.2.7

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.4.2.7.1

Multiply $\frac{e^{- \frac{x}{4}}}{4}$ by $\frac{4}{4}$ .

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

Step 3.4.2.7.2

Multiply $4$ by $4$ .

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

Step 3.4.2.8

Combine the numerators over the common denominator.

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

Step 3.4.2.9

Multiply $4$ by $- 1$ .

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

Step 3.4.2.10

Add $- 4 e^{- \frac{x}{4}}$ and $5 e^{- \frac{x}{4}}$ .

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

Step 4

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

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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

Step 5.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - \frac{x}{4}$ .

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Step 5.1.2.1

To apply the Chain Rule, set $u$ as $- \frac{x}{4}$ .

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

Step 5.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 5.1.2.3

Replace all occurrences of $u$ with $- \frac{x}{4}$ .

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

Step 5.1.3

Differentiate.

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Step 5.1.3.1

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x]$ .

$(x + 9) e^{- \frac{x}{4}} (- \frac{1}{4} \frac{d}{d x} [x]) + e^{- \frac{x}{4}} \frac{d}{d x} [x + 9]$

Step 5.1.3.2

Combine $e^{- \frac{x}{4}}$ and $\frac{1}{4}$ .

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

Step 5.1.3.3

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4} \cdot 1) + e^{- \frac{x}{4}} \frac{d}{d x} [x + 9]$

Step 5.1.3.4

Multiply $- 1$ by $1$ .

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

Step 5.1.3.5

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

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

Step 5.1.3.6

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} (1 + \frac{d}{d x} [9])$

Step 5.1.3.7

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

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} (1 + 0)$

Step 5.1.3.8

Simplify the expression.

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Step 5.1.3.8.1

Add $1$ and $0$ .

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}} \cdot 1$

Step 5.1.3.8.2

Multiply $e^{- \frac{x}{4}}$ by $1$ .

$(x + 9) (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 5.1.4

Simplify.

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Step 5.1.4.1

Apply the distributive property.

$x (- \frac{e^{- \frac{x}{4}}}{4}) + 9 (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 5.1.4.2

Combine terms.

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Step 5.1.4.2.1

Combine $x$ and $\frac{e^{- \frac{x}{4}}}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + 9 (- \frac{e^{- \frac{x}{4}}}{4}) + e^{- \frac{x}{4}}$

Step 5.1.4.2.2

Multiply $- 1$ by $9$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - 9 \frac{e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 5.1.4.2.3

Combine $- 9$ and $\frac{e^{- \frac{x}{4}}}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 5.1.4.2.4

Move the negative in front of the fraction.

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}}$

Step 5.1.4.2.5

To write $e^{- \frac{x}{4}}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + e^{- \frac{x}{4}} \cdot \frac{4}{4}$

Step 5.1.4.2.6

Combine $e^{- \frac{x}{4}}$ and $\frac{4}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{9 e^{- \frac{x}{4}}}{4} + \frac{e^{- \frac{x}{4}} \cdot 4}{4}$

Step 5.1.4.2.7

Combine the numerators over the common denominator.

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}} + e^{- \frac{x}{4}} \cdot 4}{4}$

Step 5.1.4.2.8

Move $4$ to the left of $e^{- \frac{x}{4}}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 9 e^{- \frac{x}{4}} + 4 \cdot e^{- \frac{x}{4}}}{4}$

Step 5.1.4.2.9

Add $- 9 e^{- \frac{x}{4}}$ and $4 e^{- \frac{x}{4}}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} + \frac{- 5 e^{- \frac{x}{4}}}{4}$

Step 5.1.4.2.10

Move the negative in front of the fraction.

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

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4}$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4}$

Step 6

Set the first derivative equal to $0$ then solve the equation $- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$- \frac{x e^{- \frac{x}{4}}}{4} - \frac{5 e^{- \frac{x}{4}}}{4} = 0$

Step 6.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = - 5$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 8

Critical points to evaluate.

$x = - 5$

Step 9

Evaluate the second derivative at $x = - 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 10

Evaluate the second derivative.

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Step 10.1

Combine the numerators over the common denominator.

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

Step 10.2

Simplify each term.

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Step 10.2.1

Move the negative in front of the fraction.

$\frac{- 5 e^{- - \frac{5}{4}} + e^{- \frac{- 5}{4}}}{16}$

Step 10.2.2

Multiply $- - \frac{5}{4}$ .

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Step 10.2.2.1

Multiply $- 1$ by $- 1$ .

$\frac{- 5 e^{1 (\frac{5}{4})} + e^{- \frac{- 5}{4}}}{16}$

Step 10.2.2.2

Multiply $\frac{5}{4}$ by $1$ .

$\frac{- 5 e^{\frac{5}{4}} + e^{- \frac{- 5}{4}}}{16}$

Step 10.2.3

Move the negative in front of the fraction.

$\frac{- 5 e^{\frac{5}{4}} + e^{- - \frac{5}{4}}}{16}$

Step 10.2.4

Multiply $- - \frac{5}{4}$ .

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Step 10.2.4.1

Multiply $- 1$ by $- 1$ .

$\frac{- 5 e^{\frac{5}{4}} + e^{1 (\frac{5}{4})}}{16}$

Step 10.2.4.2

Multiply $\frac{5}{4}$ by $1$ .

$\frac{- 5 e^{\frac{5}{4}} + e^{\frac{5}{4}}}{16}$

Step 10.3

Simplify terms.

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Step 10.3.1

Add $- 5 e^{\frac{5}{4}}$ and $e^{\frac{5}{4}}$ .

$\frac{- 4 e^{\frac{5}{4}}}{16}$

Step 10.3.2

Factor $4$ out of $- 4 e^{\frac{5}{4}}$ .

$\frac{4 (- e^{\frac{5}{4}})}{16}$

Step 10.4

Cancel the common factors.

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Step 10.4.1

Factor $4$ out of $16$ .

$\frac{4 (- e^{\frac{5}{4}})}{4 (4)}$

Step 10.4.2

Cancel the common factor.

$\frac{4 (- e^{\frac{5}{4}})}{4 \cdot 4}$

Step 10.4.3

Rewrite the expression.

$\frac{- e^{\frac{5}{4}}}{4}$

Step 10.5

Move the negative in front of the fraction.

$- \frac{e^{\frac{5}{4}}}{4}$

Step 11

$x = - 5$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 5$ is a local maximum

Step 12

Find the y-value when $x = - 5$ .

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Step 12.1

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

$f (- 5) = ((- 5) + 9) \cdot e^{- \frac{- 5}{4}}$

Step 12.2

Simplify the result.

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Step 12.2.1

Add $- 5$ and $9$ .

$f (- 5) = 4 \cdot e^{- \frac{- 5}{4}}$

Step 12.2.2

Move the negative in front of the fraction.

$f (- 5) = 4 e^{\frac{5}{4}}$

Step 12.2.3

The final answer is $4 e^{\frac{5}{4}}$ .

$y = 4 e^{\frac{5}{4}}$

Step 13

These are the local extrema for $f (x) = (x + 9) e^{- \frac{x}{4}}$ .

$(- 5, 4 e^{\frac{5}{4}})$ is a local maxima

Step 14