Calculus Examples

$y = x e^{\frac{1}{x}}$

Step 1

Write $y = x e^{\frac{1}{x}}$ as a function.

$f (x) = x e^{\frac{1}{x}}$

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 $f (x) = x$ and $g (x) = e^{\frac{1}{x}}$ .

$x \frac{d}{d x} [e^{\frac{1}{x}}] + e^{\frac{1}{x}} \frac{d}{d x} [x]$

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 $g (x) = \frac{1}{x}$ .

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Step 2.2.1

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

$x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [\frac{1}{x}]) + e^{\frac{1}{x}} \frac{d}{d x} [x]$

Step 2.2.2

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

$x (e^{u} \frac{d}{d x} [\frac{1}{x}]) + e^{\frac{1}{x}} \frac{d}{d x} [x]$

Step 2.2.3

Replace all occurrences of $u$ with $\frac{1}{x}$ .

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 2.3.2

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $- 2$ and $1$ .

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

Step 2.6.2

Move $- 1$ to the left of $e^{\frac{1}{x}}$ .

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

Step 2.6.3

Rewrite $- 1 e^{\frac{1}{x}}$ as $- e^{\frac{1}{x}}$ .

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

Simplify.

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Step 2.9.1

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

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

Step 2.9.2

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

$- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}$ with respect to $x$ is $\frac{d}{d x} [- \frac{e^{\frac{1}{x}}}{x}] + \frac{d}{d x} [e^{\frac{1}{x}}]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [- \frac{e^{\frac{1}{x}}}{x}]$ .

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Step 3.2.1

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 $g (x) = \frac{e^{\frac{1}{x}}}{x}$ .

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

Step 3.2.2

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 $f (x) = e^{\frac{1}{x}}$ and $g (x) = x$ .

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

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 $g (x) = \frac{1}{x}$ .

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Step 3.2.3.1

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

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

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 $a$ = $e$ .

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

Step 3.2.3.3

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

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

Step 3.2.4

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

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

$f'' (x) = - \frac{x (e^{\frac{1}{x}} (- x^{- 2})) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.8

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

$f'' (x) = - \frac{x^{- 2} x (e^{\frac{1}{x}} \cdot (- 1)) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.9

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

$f'' (x) = - \frac{x^{- 2 + 1} (e^{\frac{1}{x}} \cdot (- 1)) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.10

Add $- 2$ and $1$ .

$f'' (x) = - \frac{x^{- 1} (e^{\frac{1}{x}} \cdot (- 1)) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.11

Move $- 1$ to the left of $e^{\frac{1}{x}}$ .

$f'' (x) = - \frac{x^{- 1} (- 1 \cdot e^{\frac{1}{x}}) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.12

Rewrite $- 1 e^{\frac{1}{x}}$ as $- e^{\frac{1}{x}}$ .

$f'' (x) = - \frac{x^{- 1} (- e^{\frac{1}{x}}) - e^{\frac{1}{x}} \cdot 1}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.13

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{x^{- 1} (- e^{\frac{1}{x}}) - e^{\frac{1}{x}}}{x^{2}} + \frac{e^{\frac{1}{x}}}{x} \cdot 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.14

Multiply $\frac{e^{\frac{1}{x}}}{x}$ by $0$ .

$f'' (x) = - \frac{x^{- 1} (- e^{\frac{1}{x}}) - e^{\frac{1}{x}}}{x^{2}} + 0 + \frac{d}{d x} (e^{\frac{1}{x}})$

Step 3.2.15

Add $- \frac{x^{- 1} (- e^{\frac{1}{x}}) - e^{\frac{1}{x}}}{x^{2}}$ and $0$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [e^{\frac{1}{x}}]$ .

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Step 3.3.1

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 $g (x) = \frac{1}{x}$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u_{2}$ as $\frac{1}{x}$ .

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

Step 3.3.1.2

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

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

Step 3.3.1.3

Replace all occurrences of $u_{2}$ with $\frac{1}{x}$ .

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

Step 3.3.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 3.3.3

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

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

Step 3.4

Simplify.

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Combine terms.

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Step 3.4.3.1

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

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

Step 3.4.3.2

Combine $e^{\frac{1}{x}}$ and $\frac{1}{x^{2}}$ .

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

Step 3.4.3.3

Combine the numerators over the common denominator.

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

Step 3.4.4

Simplify the numerator.

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Step 3.4.4.1

Apply the distributive property.

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

Step 3.4.4.2

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

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Step 3.4.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.2.2

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

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

Step 3.4.4.3

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

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Step 3.4.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.3.2

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

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

Step 3.4.4.4

Subtract $e^{\frac{1}{x}}$ from $e^{\frac{1}{x}}$ .

$f'' (x) = \frac{\frac{e^{\frac{1}{x}}}{x} + 0}{x^{2}}$

Step 3.4.4.5

Add $\frac{e^{\frac{1}{x}}}{x}$ and $0$ .

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

Step 3.4.5

Multiply the numerator by the reciprocal of the denominator.

$f'' (x) = \frac{e^{\frac{1}{x}}}{x} \cdot \frac{1}{x^{2}}$

Step 3.4.6

Combine.

$f'' (x) = \frac{e^{\frac{1}{x}} \cdot 1}{x \cdot x^{2}}$

Step 3.4.7

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

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Step 3.4.7.1

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

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Step 3.4.7.1.1

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

$f'' (x) = \frac{e^{\frac{1}{x}} \cdot 1}{x \cdot x^{2}}$

Step 3.4.7.1.2

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

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

Step 3.4.7.2

Add $1$ and $2$ .

$f'' (x) = \frac{e^{\frac{1}{x}} \cdot 1}{x^{3}}$

Step 3.4.8

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

$f'' (x) = \frac{e^{\frac{1}{x}}}{x^{3}}$

Step 4

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

$- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}} = 0$

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

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 $g (x) = e^{\frac{1}{x}}$ .

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

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 $g (x) = \frac{1}{x}$ .

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Step 5.1.2.1

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

$f' (x) = x (\frac{d}{d u} (e^{u}) \frac{d}{d x} (\frac{1}{x})) + e^{\frac{1}{x}} \frac{d}{d x} (x)$

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 $a$ = $e$ .

$f' (x) = x (e^{u} \frac{d}{d x} (\frac{1}{x})) + e^{\frac{1}{x}} \frac{d}{d x} (x)$

Step 5.1.2.3

Replace all occurrences of $u$ with $\frac{1}{x}$ .

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

Step 5.1.3

Differentiate using the Power Rule.

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Step 5.1.3.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 5.1.3.2

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

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.1.6

Simplify the expression.

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Step 5.1.6.1

Add $- 2$ and $1$ .

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

Step 5.1.6.2

Move $- 1$ to the left of $e^{\frac{1}{x}}$ .

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

Step 5.1.6.3

Rewrite $- 1 e^{\frac{1}{x}}$ as $- e^{\frac{1}{x}}$ .

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

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

Simplify.

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Step 5.1.9.1

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

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

Step 5.1.9.2

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

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

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}$ .

$- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}$

Step 6

Set the first derivative equal to $0$ then solve the equation $- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}} = 0$

Step 6.2

Factor $e^{\frac{1}{x}}$ out of $- \frac{e^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}$ .

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Step 6.2.1

Factor $e^{\frac{1}{x}}$ out of $- \frac{e^{\frac{1}{x}}}{x}$ .

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

Step 6.2.2

Multiply by $1$ .

$e^{\frac{1}{x}} (- \frac{1}{x}) + e^{\frac{1}{x}} \cdot 1 = 0$

Step 6.2.3

Factor $e^{\frac{1}{x}}$ out of $e^{\frac{1}{x}} (- \frac{1}{x}) + e^{\frac{1}{x}} \cdot 1$ .

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

Step 6.3

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

$e^{\frac{1}{x}} = 0$

$- \frac{1}{x} + 1 = 0$

Step 6.4

Set $e^{\frac{1}{x}}$ equal to $0$ and solve for $x$ .

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Step 6.4.1

Set $e^{\frac{1}{x}}$ equal to $0$ .

$e^{\frac{1}{x}} = 0$

Step 6.4.2

Solve $e^{\frac{1}{x}} = 0$ for $x$ .

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Step 6.4.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{\frac{1}{x}}) = \ln (0)$

Step 6.4.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 6.4.2.3

There is no solution for $e^{\frac{1}{x}} = 0$

No solution

Step 6.5

Set $- \frac{1}{x} + 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $- \frac{1}{x} + 1$ equal to $0$ .

$- \frac{1}{x} + 1 = 0$

Step 6.5.2

Solve $- \frac{1}{x} + 1 = 0$ for $x$ .

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Step 6.5.2.1

Subtract $1$ from both sides of the equation.

$- \frac{1}{x} = - 1$

Step 6.5.2.2

Find the LCD of the terms in the equation.

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Step 6.5.2.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x, 1$

Step 6.5.2.2.2

The LCM of one and any expression is the expression.

$x$

Step 6.5.2.3

Multiply each term in $- \frac{1}{x} = - 1$ by $x$ to eliminate the fractions.

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Step 6.5.2.3.1

Multiply each term in $- \frac{1}{x} = - 1$ by $x$ .

$- \frac{1}{x} x = - x$

Step 6.5.2.3.2

Simplify the left side.

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Step 6.5.2.3.2.1

Cancel the common factor of $x$ .

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Step 6.5.2.3.2.1.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

$\frac{- 1}{x} x = - x$

Step 6.5.2.3.2.1.2

Cancel the common factor.

$\frac{- 1}{x} x = - x$

Step 6.5.2.3.2.1.3

Rewrite the expression.

$- 1 = - x$

Step 6.5.2.4

Solve the equation.

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Step 6.5.2.4.1

Rewrite the equation as $- x = - 1$ .

$- x = - 1$

Step 6.5.2.4.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 6.5.2.4.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 6.5.2.4.2.2

Simplify the left side.

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Step 6.5.2.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 6.5.2.4.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 6.5.2.4.2.3

Simplify the right side.

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Step 6.5.2.4.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 6.6

The final solution is all the values that make $e^{\frac{1}{x}} (- \frac{1}{x} + 1) = 0$ true.

$x = 1$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the denominator in $\frac{1}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 8

Critical points to evaluate.

$x = 1$

Step 9

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{e^{\frac{1}{1}}}{{(1)}^{3}}$

Step 10

Evaluate the second derivative.

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Step 10.1

Cancel the common factor of $1$ .

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Step 10.1.1

Cancel the common factor.

$\frac{e^{\frac{1}{1}}}{{(1)}^{3}}$

Step 10.1.2

Rewrite the expression.

$\frac{e^{1}}{{(1)}^{3}}$

Step 10.2

Simplify.

$\frac{e}{1^{3}}$

Step 10.3

One to any power is one.

$\frac{e}{1}$

Step 10.4

Divide $e$ by $1$ .

$e$

Step 11

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 12

Find the y-value when $x = 1$ .

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Step 12.1

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

$f (1) = (1) \cdot e^{\frac{1}{1}}$

Step 12.2

Simplify the result.

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Step 12.2.1

Multiply $e^{\frac{1}{1}}$ by $1$ .

$f (1) = e^{\frac{1}{1}}$

Step 12.2.2

Divide $1$ by $1$ .

$f (1) = e$

Step 12.2.3

The final answer is $e$ .

$y = e$

Step 13

These are the local extrema for $f (x) = x e^{\frac{1}{x}}$ .

$(1, e)$ is a local minima

Step 14