Calculus Examples

$y = 5 x^{2} \ln (\frac{x}{2})$

Step 1

Write $y = 5 x^{2} \ln (\frac{x}{2})$ as a function.

$f (x) = 5 x^{2} \ln (\frac{x}{2})$

Step 2

Find the first derivative of the function.

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Step 2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2} \ln (\frac{x}{2})$ with respect to $x$ is $5 \frac{d}{d x} [x^{2} \ln (\frac{x}{2})]$ .

$5 \frac{d}{d x} [x^{2} \ln (\frac{x}{2})]$

Step 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 $f (x) = x^{2}$ and $g (x) = \ln (\frac{x}{2})$ .

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

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 $f (x) = \ln (x)$ and $g (x) = \frac{x}{2}$ .

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Step 2.3.1

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

$5 (x^{2} (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{x}{2}]) + \ln (\frac{x}{2}) \frac{d}{d x} [x^{2}])$

Step 2.3.2

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

$5 (x^{2} (\frac{1}{u} \frac{d}{d x} [\frac{x}{2}]) + \ln (\frac{x}{2}) \frac{d}{d x} [x^{2}])$

Step 2.3.3

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

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

Step 2.4

Multiply by the reciprocal of the fraction to divide by $\frac{x}{2}$ .

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

Step 2.5

Simplify terms.

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Step 2.5.1

Multiply $\frac{2}{x}$ by $1$ .

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

Step 2.5.2

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

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

Step 2.5.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 2.5.3.1

Factor $x$ out of $2 x^{2}$ .

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

Step 2.5.3.2

Cancel the common factors.

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Step 2.5.3.2.1

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

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

Step 2.5.3.2.2

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

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

Step 2.5.3.2.3

Cancel the common factor.

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

Step 2.5.3.2.4

Rewrite the expression.

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

Step 2.5.3.2.5

Divide $2 x$ by $1$ .

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

Step 2.6

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

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

Step 2.7

Simplify terms.

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Step 2.7.1

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

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

Step 2.7.2

Combine $\frac{2}{2}$ and $x$ .

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

Step 2.7.3

Cancel the common factor of $2$ .

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Step 2.7.3.1

Cancel the common factor.

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

Step 2.7.3.2

Divide $x$ by $1$ .

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

Step 2.8

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

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

Step 2.9

Multiply $x$ by $1$ .

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

Step 2.10

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

$5 (x + \ln (\frac{x}{2}) (2 x))$

Step 2.11

Simplify.

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Step 2.11.1

Apply the distributive property.

$5 x + 5 (\ln (\frac{x}{2}) (2 x))$

Step 2.11.2

Multiply $2$ by $5$ .

$5 x + 10 \ln (\frac{x}{2}) x$

Step 2.11.3

Reorder terms.

$10 x \ln (\frac{x}{2}) + 5 x$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $10 x \ln (\frac{x}{2}) + 5 x$ with respect to $x$ is $\frac{d}{d x} [10 x \ln (\frac{x}{2})] + \frac{d}{d x} [5 x]$ .

$f'' (x) = \frac{d}{d x} (10 x \ln (\frac{x}{2})) + \frac{d}{d x} (5 x)$

Step 3.2

Evaluate $\frac{d}{d x} [10 x \ln (\frac{x}{2})]$ .

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Step 3.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x \ln (\frac{x}{2})$ with respect to $x$ is $10 \frac{d}{d x} [x \ln (\frac{x}{2})]$ .

$f'' (x) = 10 \frac{d}{d x} (x \ln (\frac{x}{2})) + \frac{d}{d x} (5 x)$

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 $g (x) = \ln (\frac{x}{2})$ .

$f'' (x) = 10 (x \frac{d}{d x} (\ln (\frac{x}{2})) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

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 $f (x) = \ln (x)$ and $g (x) = \frac{x}{2}$ .

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Step 3.2.3.1

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

$f'' (x) = 10 (x (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (\frac{x}{2})) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 3.2.3.2

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

$f'' (x) = 10 (x (\frac{1}{u} \cdot \frac{d}{d x} (\frac{x}{2})) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 3.2.3.3

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

$f'' (x) = 10 (x (\frac{1}{\frac{x}{2}} \cdot \frac{d}{d x} (\frac{x}{2})) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 3.2.4

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

$f'' (x) = 10 (x (\frac{1}{\frac{x}{2}} \cdot (\frac{1}{2} \cdot \frac{d}{d x} (x))) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 3.2.5

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

$f'' (x) = 10 (x (\frac{1}{\frac{x}{2}} \cdot (\frac{1}{2} \cdot 1)) + \ln (\frac{x}{2}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 3.2.6

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

$f'' (x) = 10 (x (\frac{1}{\frac{x}{2}} \cdot (\frac{1}{2} \cdot 1)) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.7

Multiply by the reciprocal of the fraction to divide by $\frac{x}{2}$ .

$f'' (x) = 10 (x (1 (\frac{2}{x}) \cdot (\frac{1}{2} \cdot 1)) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.8

Multiply $\frac{2}{x}$ by $1$ .

$f'' (x) = 10 (x (\frac{2}{x} \cdot (\frac{1}{2} \cdot 1)) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.9

Multiply $\frac{1}{2}$ by $1$ .

$f'' (x) = 10 (x (\frac{2}{x} \cdot \frac{1}{2}) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.10

Multiply $\frac{2}{x}$ by $\frac{1}{2}$ .

$f'' (x) = 10 (x (\frac{2}{x \cdot 2}) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.11

Move $2$ to the left of $x$ .

$f'' (x) = 10 (x (\frac{2}{2 \cdot x}) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.12

Cancel the common factor of $2$ .

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Step 3.2.12.1

Cancel the common factor.

$f'' (x) = 10 (x (\frac{2}{2 x}) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.12.2

Rewrite the expression.

$f'' (x) = 10 (x (\frac{1}{x}) + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.13

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

$f'' (x) = 10 (\frac{x}{x} + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.14

Cancel the common factor of $x$ .

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Step 3.2.14.1

Cancel the common factor.

$f'' (x) = 10 (\frac{x}{x} + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.14.2

Rewrite the expression.

$f'' (x) = 10 (1 + \ln (\frac{x}{2}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 3.2.15

Multiply $\ln (\frac{x}{2})$ by $1$ .

$f'' (x) = 10 (1 + \ln (\frac{x}{2})) + \frac{d}{d x} (5 x)$

Step 3.3

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

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Step 3.3.1

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

$f'' (x) = 10 (1 + \ln (\frac{x}{2})) + 5 \frac{d}{d x} (x)$

Step 3.3.2

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

$f'' (x) = 10 (1 + \ln (\frac{x}{2})) + 5 \cdot 1$

Step 3.3.3

Multiply $5$ by $1$ .

$f'' (x) = 10 (1 + \ln (\frac{x}{2})) + 5$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$f'' (x) = 10 \cdot 1 + 10 \ln (\frac{x}{2}) + 5$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Multiply $10$ by $1$ .

$f'' (x) = 10 + 10 \ln (\frac{x}{2}) + 5$

Step 3.4.2.2

Add $10$ and $5$ .

$f'' (x) = 10 \ln (\frac{x}{2}) + 15$

Step 4

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

$10 x \ln (\frac{x}{2}) + 5 x = 0$

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

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2} \ln (\frac{x}{2})$ with respect to $x$ is $5 \frac{d}{d x} [x^{2} \ln (\frac{x}{2})]$ .

$5 \frac{d}{d x} [x^{2} \ln (\frac{x}{2})]$

Step 5.1.2

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

Step 5.1.3

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

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Step 5.1.3.1

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

$5 (x^{2} (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{x}{2}]) + \ln (\frac{x}{2}) \frac{d}{d x} [x^{2}])$

Step 5.1.3.2

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

$5 (x^{2} (\frac{1}{u} \frac{d}{d x} [\frac{x}{2}]) + \ln (\frac{x}{2}) \frac{d}{d x} [x^{2}])$

Step 5.1.3.3

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

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

Step 5.1.4

Multiply by the reciprocal of the fraction to divide by $\frac{x}{2}$ .

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

Step 5.1.5

Simplify terms.

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Step 5.1.5.1

Multiply $\frac{2}{x}$ by $1$ .

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

Step 5.1.5.2

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

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

Step 5.1.5.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 5.1.5.3.1

Factor $x$ out of $2 x^{2}$ .

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

Step 5.1.5.3.2

Cancel the common factors.

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Step 5.1.5.3.2.1

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

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

Step 5.1.5.3.2.2

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

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

Step 5.1.5.3.2.3

Cancel the common factor.

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

Step 5.1.5.3.2.4

Rewrite the expression.

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

Step 5.1.5.3.2.5

Divide $2 x$ by $1$ .

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

Step 5.1.6

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

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

Step 5.1.7

Simplify terms.

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Step 5.1.7.1

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

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

Step 5.1.7.2

Combine $\frac{2}{2}$ and $x$ .

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

Step 5.1.7.3

Cancel the common factor of $2$ .

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Step 5.1.7.3.1

Cancel the common factor.

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

Step 5.1.7.3.2

Divide $x$ by $1$ .

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

Step 5.1.8

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

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

Step 5.1.9

Multiply $x$ by $1$ .

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

Step 5.1.10

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

$5 (x + \ln (\frac{x}{2}) (2 x))$

Step 5.1.11

Simplify.

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Step 5.1.11.1

Apply the distributive property.

$5 x + 5 (\ln (\frac{x}{2}) (2 x))$

Step 5.1.11.2

Multiply $2$ by $5$ .

$5 x + 10 \ln (\frac{x}{2}) x$

Step 5.1.11.3

Reorder terms.

$f' (x) = 10 x \ln (\frac{x}{2}) + 5 x$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $10 x \ln (\frac{x}{2}) + 5 x$ .

$10 x \ln (\frac{x}{2}) + 5 x$

Step 6

Set the first derivative equal to $0$ then solve the equation $10 x \ln (\frac{x}{2}) + 5 x = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$10 x \ln (\frac{x}{2}) + 5 x = 0$

Step 6.2

Subtract $5 x$ from both sides of the equation.

$10 x \ln (\frac{x}{2}) = - 5 x$

Step 6.3

Divide each term in $10 x \ln (\frac{x}{2}) = - 5 x$ by $10 x$ and simplify.

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Step 6.3.1

Divide each term in $10 x \ln (\frac{x}{2}) = - 5 x$ by $10 x$ .

$\frac{10 x \ln (\frac{x}{2})}{10 x} = \frac{- 5 x}{10 x}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Cancel the common factor of $10$ .

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Step 6.3.2.1.1

Cancel the common factor.

$\frac{10 x \ln (\frac{x}{2})}{10 x} = \frac{- 5 x}{10 x}$

Step 6.3.2.1.2

Rewrite the expression.

$\frac{x \ln (\frac{x}{2})}{x} = \frac{- 5 x}{10 x}$

Step 6.3.2.2

Cancel the common factor of $x$ .

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Step 6.3.2.2.1

Cancel the common factor.

$\frac{x \ln (\frac{x}{2})}{x} = \frac{- 5 x}{10 x}$

Step 6.3.2.2.2

Divide $\ln (\frac{x}{2})$ by $1$ .

$\ln (\frac{x}{2}) = \frac{- 5 x}{10 x}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Cancel the common factor of $- 5$ and $10$ .

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Step 6.3.3.1.1

Factor $5$ out of $- 5 x$ .

$\ln (\frac{x}{2}) = \frac{5 (- x)}{10 x}$

Step 6.3.3.1.2

Cancel the common factors.

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Step 6.3.3.1.2.1

Factor $5$ out of $10 x$ .

$\ln (\frac{x}{2}) = \frac{5 (- x)}{5 (2 x)}$

Step 6.3.3.1.2.2

Cancel the common factor.

$\ln (\frac{x}{2}) = \frac{5 (- x)}{5 (2 x)}$

Step 6.3.3.1.2.3

Rewrite the expression.

$\ln (\frac{x}{2}) = \frac{- x}{2 x}$

Step 6.3.3.2

Cancel the common factor of $x$ .

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Step 6.3.3.2.1

Cancel the common factor.

$\ln (\frac{x}{2}) = \frac{- x}{2 x}$

Step 6.3.3.2.2

Rewrite the expression.

$\ln (\frac{x}{2}) = \frac{- 1}{2}$

Step 6.3.3.3

Move the negative in front of the fraction.

$\ln (\frac{x}{2}) = - \frac{1}{2}$

Step 6.4

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

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

Step 6.5

Rewrite $\ln (\frac{x}{2}) = - \frac{1}{2}$ 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 $b^{y} = x$ .

$e^{- \frac{1}{2}} = \frac{x}{2}$

Step 6.6

Solve for $x$ .

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Step 6.6.1

Rewrite the equation as $\frac{x}{2} = e^{- \frac{1}{2}}$ .

$\frac{x}{2} = e^{- \frac{1}{2}}$

Step 6.6.2

Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 e^{- \frac{1}{2}}$

Step 6.6.3

Simplify both sides of the equation.

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Step 6.6.3.1

Simplify the left side.

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Step 6.6.3.1.1

Cancel the common factor of $2$ .

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Step 6.6.3.1.1.1

Cancel the common factor.

$2 \frac{x}{2} = 2 e^{- \frac{1}{2}}$

Step 6.6.3.1.1.2

Rewrite the expression.

$x = 2 e^{- \frac{1}{2}}$

Step 6.6.3.2

Simplify the right side.

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Step 6.6.3.2.1

Simplify $2 e^{- \frac{1}{2}}$ .

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Step 6.6.3.2.1.1

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

$x = 2 \frac{1}{e^{\frac{1}{2}}}$

Step 6.6.3.2.1.2

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

$x = \frac{2}{e^{\frac{1}{2}}}$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the argument in $\ln (\frac{x}{2})$ less than or equal to $0$ to find where the expression is undefined.

$\frac{x}{2} \leq 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

Multiply both sides by $2$ .

$\frac{x}{2} \cdot 2 \leq 0 \cdot 2$

Step 7.2.2

Simplify.

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Step 7.2.2.1

Simplify the left side.

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Step 7.2.2.1.1

Cancel the common factor of $2$ .

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Step 7.2.2.1.1.1

Cancel the common factor.

$\frac{x}{2} \cdot 2 \leq 0 \cdot 2$

Step 7.2.2.1.1.2

Rewrite the expression.

$x \leq 0 \cdot 2$

Step 7.2.2.2

Simplify the right side.

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Step 7.2.2.2.1

Multiply $0$ by $2$ .

$x \leq 0$

Step 7.3

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 8

Critical points to evaluate.

$x = \frac{2}{e^{\frac{1}{2}}}$

Step 9

Evaluate the second derivative at $x = \frac{2}{e^{\frac{1}{2}}}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$10 \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2}) + 15$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

Multiply the numerator by the reciprocal of the denominator.

$10 \ln (\frac{2}{e^{\frac{1}{2}}} \cdot \frac{1}{2}) + 15$

Step 10.1.2

Combine.

$10 \ln (\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2}) + 15$

Step 10.1.3

Reduce the expression $\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2}$ by cancelling the common factors.

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Step 10.1.3.1

Cancel the common factor.

$10 \ln (\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2}) + 15$

Step 10.1.3.2

Rewrite the expression.

$10 \ln (\frac{1}{e^{\frac{1}{2}}}) + 15$

Step 10.1.4

Move $e^{\frac{1}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$10 \ln (e^{- \frac{1}{2}}) + 15$

Step 10.1.5

Expand $\ln (e^{- \frac{1}{2}})$ by moving $- \frac{1}{2}$ outside the logarithm.

$10 (- \frac{1}{2} \ln (e)) + 15$

Step 10.1.6

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

$10 (- \frac{1}{2} \cdot 1) + 15$

Step 10.1.7

Multiply $- 1$ by $1$ .

$10 (- \frac{1}{2}) + 15$

Step 10.1.8

Cancel the common factor of $2$ .

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Step 10.1.8.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$10 (\frac{- 1}{2}) + 15$

Step 10.1.8.2

Factor $2$ out of $10$ .

$2 (5) \frac{- 1}{2} + 15$

Step 10.1.8.3

Cancel the common factor.

$2 \cdot 5 \frac{- 1}{2} + 15$

Step 10.1.8.4

Rewrite the expression.

$5 \cdot - 1 + 15$

Step 10.1.9

Multiply $5$ by $- 1$ .

$- 5 + 15$

Step 10.2

Add $- 5$ and $15$ .

$10$

Step 11

$x = \frac{2}{e^{\frac{1}{2}}}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{2}{e^{\frac{1}{2}}}$ is a local minimum

Step 12

Find the y-value when $x = \frac{2}{e^{\frac{1}{2}}}$ .

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Step 12.1

Replace the variable $x$ with $\frac{2}{e^{\frac{1}{2}}}$ in the expression.

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 {(\frac{2}{e^{\frac{1}{2}}})}^{2} \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify the expression.

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Step 12.2.1.1

Apply the product rule to $\frac{2}{e^{\frac{1}{2}}}$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{2^{2}}{{(e^{\frac{1}{2}})}^{2}}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.1.2

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

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{4}{{(e^{\frac{1}{2}})}^{2}}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.2

Simplify the denominator.

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Step 12.2.2.1

Multiply the exponents in ${(e^{\frac{1}{2}})}^{2}$ .

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Step 12.2.2.1.1

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

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{4}{e^{\frac{1}{2} \cdot 2}}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.2.1.2

Cancel the common factor of $2$ .

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Step 12.2.2.1.2.1

Cancel the common factor.

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{4}{e^{\frac{1}{2} \cdot 2}}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.2.1.2.2

Rewrite the expression.

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{4}{e}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.2.2

Simplify.

$f (\frac{2}{e^{\frac{1}{2}}}) = 5 (\frac{4}{e}) \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.3

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

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Step 12.2.3.1

Combine $5$ and $\frac{4}{e}$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{5 \cdot 4}{e} \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.3.2

Multiply $5$ by $4$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (\frac{\frac{2}{e^{\frac{1}{2}}}}{2})$

Step 12.2.4

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (\frac{2}{e^{\frac{1}{2}}} \cdot \frac{1}{2})$

Step 12.2.5

Combine.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2})$

Step 12.2.6

Reduce the expression $\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2}$ by cancelling the common factors.

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Step 12.2.6.1

Cancel the common factor.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (\frac{2 \cdot 1}{e^{\frac{1}{2}} \cdot 2})$

Step 12.2.6.2

Rewrite the expression.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (\frac{1}{e^{\frac{1}{2}}})$

Step 12.2.7

Move $e^{\frac{1}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \ln (e^{- \frac{1}{2}})$

Step 12.2.8

Expand $\ln (e^{- \frac{1}{2}})$ by moving $- \frac{1}{2}$ outside the logarithm.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot (- \frac{1}{2} \cdot \ln (e))$

Step 12.2.9

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

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot (- \frac{1}{2} \cdot 1)$

Step 12.2.10

Multiply $- 1$ by $1$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot (- \frac{1}{2})$

Step 12.2.11

Cancel the common factor of $2$ .

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Step 12.2.11.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{20}{e} \cdot \frac{- 1}{2}$

Step 12.2.11.2

Factor $2$ out of $20$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{2 (10)}{e} \cdot \frac{- 1}{2}$

Step 12.2.11.3

Cancel the common factor.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{2 \cdot 10}{e} \cdot \frac{- 1}{2}$

Step 12.2.11.4

Rewrite the expression.

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{10}{e} \cdot - 1$

Step 12.2.12

Combine $\frac{10}{e}$ and $- 1$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{10 \cdot - 1}{e}$

Step 12.2.13

Simplify the expression.

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Step 12.2.13.1

Multiply $10$ by $- 1$ .

$f (\frac{2}{e^{\frac{1}{2}}}) = \frac{- 10}{e}$

Step 12.2.13.2

Move the negative in front of the fraction.

$f (\frac{2}{e^{\frac{1}{2}}}) = - \frac{10}{e}$

Step 12.2.14

The final answer is $- \frac{10}{e}$ .

$y = - \frac{10}{e}$

Step 13

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

$(\frac{2}{e^{\frac{1}{2}}}, - \frac{10}{e})$ is a local minima

Step 14