Calculus Examples

$f (x) = x^{- 4} \ln (x)$

Step 1

Find the first derivative of the function.

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Step 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 $f (x) = x^{- 4}$ and $g (x) = \ln (x)$ .

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

Step 1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 1.3

Combine fractions.

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Step 1.3.1

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

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

Step 1.3.2

Move $x^{- 4}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.4

Multiply $x$ by $x^{4}$ by adding the exponents.

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Step 1.4.1

Multiply $x$ by $x^{4}$ .

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Step 1.4.1.1

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

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

Step 1.4.1.2

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

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

Step 1.4.2

Add $1$ and $4$ .

$\frac{1}{x^{5}} + \ln (x) \frac{d}{d x} [x^{- 4}]$

Step 1.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 4$ .

$\frac{1}{x^{5}} + \ln (x) (- 4 x^{- 5})$

Step 1.6

Simplify.

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Step 1.6.1

Reorder terms.

$- 4 x^{- 5} \ln (x) + \frac{1}{x^{5}}$

Step 1.6.2

Simplify each term.

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Step 1.6.2.1

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

$- 4 \frac{1}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 1.6.2.2

Combine $- 4$ and $\frac{1}{x^{5}}$ .

$\frac{- 4}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 1.6.2.3

Move the negative in front of the fraction.

$- \frac{4}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 1.6.2.4

Combine $\ln (x)$ and $\frac{4}{x^{5}}$ .

$- \frac{\ln (x) \cdot 4}{x^{5}} + \frac{1}{x^{5}}$

Step 1.6.2.5

Move $4$ to the left of $\ln (x)$ .

$- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}}$ with respect to $x$ is $\frac{d}{d x} [- \frac{4 \ln (x)}{x^{5}}] + \frac{d}{d x} [\frac{1}{x^{5}}]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [- \frac{4 \ln (x)}{x^{5}}]$ .

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Step 2.2.1

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

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

Step 2.2.2

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 $f (x) = \ln (x)$ and $g (x) = x^{5}$ .

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

Step 2.2.3

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 2.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

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

Step 2.2.5

Combine $x^{5}$ and $\frac{1}{x}$ .

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

Step 2.2.6

Cancel the common factor of $x^{5}$ and $x$ .

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Step 2.2.6.1

Factor $x$ out of $x^{5}$ .

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

Step 2.2.6.2

Cancel the common factors.

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Step 2.2.6.2.1

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

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

Step 2.2.6.2.2

Factor $x$ out of $x^{1}$ .

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

Step 2.2.6.2.3

Cancel the common factor.

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

Step 2.2.6.2.4

Rewrite the expression.

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

Step 2.2.6.2.5

Divide $x^{4}$ by $1$ .

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

Step 2.2.7

Multiply $5$ by $- 1$ .

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

Step 2.2.8

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

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Step 2.2.8.1

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

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

Step 2.2.8.2

Multiply $5$ by $2$ .

$f'' (x) = - 4 \frac{x^{4} - 5 \ln (x) x^{4}}{x^{10}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.9

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

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Step 2.2.9.1

Multiply by $1$ .

$f'' (x) = - 4 \frac{x^{4} \cdot 1 - 5 \ln (x) x^{4}}{x^{10}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.9.2

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

$f'' (x) = - 4 \frac{x^{4} \cdot 1 + x^{4} (- 5 \ln (x))}{x^{10}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.9.3

Factor $x^{4}$ out of $x^{4} \cdot 1 + x^{4} (- 5 \ln (x))$ .

$f'' (x) = - 4 \frac{x^{4} (1 - 5 \ln (x))}{x^{10}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.10

Cancel the common factors.

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Step 2.2.10.1

Factor $x^{4}$ out of $x^{10}$ .

$f'' (x) = - 4 \frac{x^{4} (1 - 5 \ln (x))}{x^{4} x^{6}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.10.2

Cancel the common factor.

$f'' (x) = - 4 \frac{x^{4} (1 - 5 \ln (x))}{x^{4} x^{6}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.10.3

Rewrite the expression.

$f'' (x) = - 4 \frac{1 - 5 \ln (x)}{x^{6}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.11

Combine $- 4$ and $\frac{1 - 5 \ln (x)}{x^{6}}$ .

$f'' (x) = \frac{- 4 (1 - 5 \ln (x))}{x^{6}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.2.12

Move the negative in front of the fraction.

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} + \frac{d}{d x} (\frac{1}{x^{5}})$

Step 2.3

Evaluate $\frac{d}{d x} [\frac{1}{x^{5}}]$ .

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Step 2.3.1

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

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} + \frac{d}{d x} ({(x^{5})}^{- 1})$

Step 2.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{5}$ .

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $x^{5}$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} + \frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{5})$

Step 2.3.2.2

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

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - u^{- 2} \frac{d}{d x} x^{5}$

Step 2.3.2.3

Replace all occurrences of $u$ with $x^{5}$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - {(x^{5})}^{- 2} \frac{d}{d x} x^{5}$

Step 2.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - {(x^{5})}^{- 2} (5 x^{4})$

Step 2.3.4

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

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Step 2.3.4.1

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

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - x^{5 \cdot - 2} (5 x^{4})$

Step 2.3.4.2

Multiply $5$ by $- 2$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - x^{- 10} (5 x^{4})$

Step 2.3.5

Multiply $5$ by $- 1$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - 5 x^{- 10} x^{4}$

Step 2.3.6

Multiply $x^{- 10}$ by $x^{4}$ by adding the exponents.

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Step 2.3.6.1

Move $x^{4}$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - 5 (x^{4} x^{- 10})$

Step 2.3.6.2

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

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - 5 x^{4 - 10}$

Step 2.3.6.3

Subtract $10$ from $4$ .

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - 5 x^{- 6}$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = - \frac{4 (1 - 5 \ln (x))}{x^{6}} - 5 \frac{1}{x^{6}}$

Step 2.4.2

Apply the distributive property.

$f'' (x) = - \frac{4 \cdot 1 + 4 (- 5 \ln (x))}{x^{6}} - 5 \frac{1}{x^{6}}$

Step 2.4.3

Combine terms.

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Step 2.4.3.1

Multiply $4$ by $1$ .

$f'' (x) = - \frac{4 + 4 (- 5 \ln (x))}{x^{6}} - 5 \frac{1}{x^{6}}$

Step 2.4.3.2

Multiply $- 5$ by $4$ .

$f'' (x) = - \frac{4 - 20 \ln (x)}{x^{6}} - 5 \frac{1}{x^{6}}$

Step 2.4.3.3

Combine $- 5$ and $\frac{1}{x^{6}}$ .

$f'' (x) = - \frac{4 - 20 \ln (x)}{x^{6}} + \frac{- 5}{x^{6}}$

Step 2.4.3.4

Move the negative in front of the fraction.

$f'' (x) = - \frac{4 - 20 \ln (x)}{x^{6}} - \frac{5}{x^{6}}$

Step 2.4.3.5

Combine the numerators over the common denominator.

$f'' (x) = \frac{- (4 - 20 \ln (x)) - 5}{x^{6}}$

Step 2.4.4

Reorder terms.

$f'' (x) = \frac{- 5 - (4 - 20 \ln (x))}{x^{6}}$

Step 2.4.5

Simplify the numerator.

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Step 2.4.5.1

Factor $- 1$ out of $- 5 - (4 - 20 \ln (x))$ .

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Step 2.4.5.1.1

Rewrite $- 5$ as $- 1 (5)$ .

$f'' (x) = \frac{- 1 \cdot 5 - (4 - 20 \ln (x))}{x^{6}}$

Step 2.4.5.1.2

Factor $- 1$ out of $- 1 (5) - (4 - 20 \ln (x))$ .

$f'' (x) = \frac{- 1 (5 + 4 - 20 \ln (x))}{x^{6}}$

Step 2.4.5.1.3

Rewrite $- 1 (5 + 4 - 20 \ln (x))$ as $- (5 + 4 - 20 \ln (x))$ .

$f'' (x) = \frac{- (5 + 4 - 20 \ln (x))}{x^{6}}$

Step 2.4.5.2

Add $5$ and $4$ .

$f'' (x) = \frac{- (9 - 20 \ln (x))}{x^{6}}$

Step 2.4.6

Move the negative in front of the fraction.

$f'' (x) = - \frac{9 - 20 \ln (x)}{x^{6}}$

Step 3

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

$- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

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

Step 4.1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 4.1.3

Combine fractions.

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Step 4.1.3.1

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

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

Step 4.1.3.2

Move $x^{- 4}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.1.4

Multiply $x$ by $x^{4}$ by adding the exponents.

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Step 4.1.4.1

Multiply $x$ by $x^{4}$ .

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Step 4.1.4.1.1

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

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

Step 4.1.4.1.2

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

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

Step 4.1.4.2

Add $1$ and $4$ .

$\frac{1}{x^{5}} + \ln (x) \frac{d}{d x} [x^{- 4}]$

Step 4.1.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 4$ .

$\frac{1}{x^{5}} + \ln (x) (- 4 x^{- 5})$

Step 4.1.6

Simplify.

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Step 4.1.6.1

Reorder terms.

$- 4 x^{- 5} \ln (x) + \frac{1}{x^{5}}$

Step 4.1.6.2

Simplify each term.

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Step 4.1.6.2.1

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

$- 4 \frac{1}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 4.1.6.2.2

Combine $- 4$ and $\frac{1}{x^{5}}$ .

$\frac{- 4}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 4.1.6.2.3

Move the negative in front of the fraction.

$- \frac{4}{x^{5}} \ln (x) + \frac{1}{x^{5}}$

Step 4.1.6.2.4

Combine $\ln (x)$ and $\frac{4}{x^{5}}$ .

$- \frac{\ln (x) \cdot 4}{x^{5}} + \frac{1}{x^{5}}$

Step 4.1.6.2.5

Move $4$ to the left of $\ln (x)$ .

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}}$ .

$- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{4 \ln (x)}{x^{5}} + \frac{1}{x^{5}} = 0$

Step 5.2

Subtract $\frac{1}{x^{5}}$ from both sides of the equation.

$- \frac{4 \ln (x)}{x^{5}} = - \frac{1}{x^{5}}$

Step 5.3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- 4 \ln (x) = - 1$

Step 5.4

Divide each term in $- 4 \ln (x) = - 1$ by $- 4$ and simplify.

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Step 5.4.1

Divide each term in $- 4 \ln (x) = - 1$ by $- 4$ .

$\frac{- 4 \ln (x)}{- 4} = \frac{- 1}{- 4}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- 4$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{- 4 \ln (x)}{- 4} = \frac{- 1}{- 4}$

Step 5.4.2.1.2

Divide $\ln (x)$ by $1$ .

$\ln (x) = \frac{- 1}{- 4}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Dividing two negative values results in a positive value.

$\ln (x) = \frac{1}{4}$

Step 5.5

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x)} = e^{\frac{1}{4}}$

Step 5.6

Rewrite $\ln (x) = \frac{1}{4}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{1}{4}} = x$

Step 5.7

Rewrite the equation as $x = e^{\frac{1}{4}}$ .

$x = e^{\frac{1}{4}}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{4 \ln (x)}{x^{5}}$ equal to $0$ to find where the expression is undefined.

$x^{5} = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[5]{0}$

Step 6.2.2

Simplify $\sqrt[5]{0}$ .

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Step 6.2.2.1

Rewrite $0$ as $0^{5}$ .

$x = \sqrt[5]{0^{5}}$

Step 6.2.2.2

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 6.3

Set the argument in $\ln (x)$ less than or equal to $0$ to find where the expression is undefined.

$x \leq 0$

Step 6.4

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 $0$ .

$x \leq 0$

$(- \infty, 0]$

$x \leq 0$

$(- \infty, 0]$

Step 7

Critical points to evaluate.

$x = e^{\frac{1}{4}}$

Step 8

Evaluate the second derivative at $x = e^{\frac{1}{4}}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{9 - 20 \ln (e^{\frac{1}{4}})}{{(e^{\frac{1}{4}})}^{6}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Use logarithm rules to move $\frac{1}{4}$ out of the exponent.

$- \frac{9 - 20 (\frac{1}{4} \ln (e))}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.2

The natural logarithm of $e$ is $1$ .

$- \frac{9 - 20 (\frac{1}{4} \cdot 1)}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.3

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

$- \frac{9 - 20 (\frac{1}{4})}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.4

Cancel the common factor of $4$ .

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Step 9.1.4.1

Factor $4$ out of $- 20$ .

$- \frac{9 + 4 (- 5) \frac{1}{4}}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.4.2

Cancel the common factor.

$- \frac{9 + 4 \cdot - 5 \frac{1}{4}}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.4.3

Rewrite the expression.

$- \frac{9 - 5}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.1.5

Subtract $5$ from $9$ .

$- \frac{4}{{(e^{\frac{1}{4}})}^{6}}$

Step 9.2

Multiply the exponents in ${(e^{\frac{1}{4}})}^{6}$ .

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Step 9.2.1

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

$- \frac{4}{e^{\frac{1}{4} \cdot 6}}$

Step 9.2.2

Cancel the common factor of $2$ .

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Step 9.2.2.1

Factor $2$ out of $4$ .

$- \frac{4}{e^{\frac{1}{2 (2)} \cdot 6}}$

Step 9.2.2.2

Factor $2$ out of $6$ .

$- \frac{4}{e^{\frac{1}{2 \cdot 2} \cdot (2 \cdot 3)}}$

Step 9.2.2.3

Cancel the common factor.

$- \frac{4}{e^{\frac{1}{2 \cdot 2} \cdot (2 \cdot 3)}}$

Step 9.2.2.4

Rewrite the expression.

$- \frac{4}{e^{\frac{1}{2} \cdot 3}}$

Step 9.2.3

Combine $\frac{1}{2}$ and $3$ .

$- \frac{4}{e^{\frac{3}{2}}}$

Step 10

$x = e^{\frac{1}{4}}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = e^{\frac{1}{4}}$ is a local maximum

Step 11

Find the y-value when $x = e^{\frac{1}{4}}$ .

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Step 11.1

Replace the variable $x$ with $e^{\frac{1}{4}}$ in the expression.

$f (e^{\frac{1}{4}}) = {(e^{\frac{1}{4}})}^{- 4} \ln (e^{\frac{1}{4}})$

Step 11.2

Simplify the result.

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Step 11.2.1

Multiply the exponents in ${(e^{\frac{1}{4}})}^{- 4}$ .

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Step 11.2.1.1

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

$f (e^{\frac{1}{4}}) = e^{\frac{1}{4} \cdot - 4} \ln (e^{\frac{1}{4}})$

Step 11.2.1.2

Cancel the common factor of $4$ .

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Step 11.2.1.2.1

Factor $4$ out of $- 4$ .

$f (e^{\frac{1}{4}}) = e^{\frac{1}{4} \cdot (4 (- 1))} \ln (e^{\frac{1}{4}})$

Step 11.2.1.2.2

Cancel the common factor.

$f (e^{\frac{1}{4}}) = e^{\frac{1}{4} \cdot (4 \cdot - 1)} \ln (e^{\frac{1}{4}})$

Step 11.2.1.2.3

Rewrite the expression.

$f (e^{\frac{1}{4}}) = e^{- 1} \ln (e^{\frac{1}{4}})$

Step 11.2.2

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

$f (e^{\frac{1}{4}}) = \frac{1}{e} \cdot \ln (e^{\frac{1}{4}})$

Step 11.2.3

Use logarithm rules to move $\frac{1}{4}$ out of the exponent.

$f (e^{\frac{1}{4}}) = \frac{1}{e} \cdot (\frac{1}{4} \cdot \ln (e))$

Step 11.2.4

The natural logarithm of $e$ is $1$ .

$f (e^{\frac{1}{4}}) = \frac{1}{e} \cdot (\frac{1}{4} \cdot 1)$

Step 11.2.5

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

$f (e^{\frac{1}{4}}) = \frac{1}{e} \cdot \frac{1}{4}$

Step 11.2.6

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

$f (e^{\frac{1}{4}}) = \frac{1}{e \cdot 4}$

Step 11.2.7

Move $4$ to the left of $e$ .

$f (e^{\frac{1}{4}}) = \frac{1}{4 e}$

Step 11.2.8

The final answer is $\frac{1}{4 e}$ .

$y = \frac{1}{4 e}$

Step 12

These are the local extrema for $f (x) = x^{- 4} \ln (x)$ .

$(e^{\frac{1}{4}}, \frac{1}{4 e})$ is a local maxima

Step 13