Calculus Examples

$f (x) = 12 x - 20 \ln (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $12 x - 20 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [12 x] + \frac{d}{d x} [- 20 \ln (x)]$ .

$\frac{d}{d x} [12 x] + \frac{d}{d x} [- 20 \ln (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [12 x]$ .

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Step 1.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

$12 \frac{d}{d x} [x] + \frac{d}{d x} [- 20 \ln (x)]$

Step 1.2.2

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

$12 \cdot 1 + \frac{d}{d x} [- 20 \ln (x)]$

Step 1.2.3

Multiply $12$ by $1$ .

$12 + \frac{d}{d x} [- 20 \ln (x)]$

Step 1.3

Evaluate $\frac{d}{d x} [- 20 \ln (x)]$ .

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Step 1.3.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 \ln (x)$ with respect to $x$ is $- 20 \frac{d}{d x} [\ln (x)]$ .

$12 - 20 \frac{d}{d x} [\ln (x)]$

Step 1.3.2

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

$12 - 20 \frac{1}{x}$

Step 1.3.3

Combine $- 20$ and $\frac{1}{x}$ .

$12 + \frac{- 20}{x}$

Step 1.3.4

Move the negative in front of the fraction.

$12 - \frac{20}{x}$

Step 1.4

Reorder terms.

$- \frac{20}{x} + 12$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- \frac{20}{x} + 12$ with respect to $x$ is $\frac{d}{d x} [- \frac{20}{x}] + \frac{d}{d x} [12]$ .

$f'' (x) = \frac{d}{d x} (- \frac{20}{x}) + \frac{d}{d x} (12)$

Step 2.2

Evaluate $\frac{d}{d x} [- \frac{20}{x}]$ .

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Step 2.2.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- \frac{20}{x}$ with respect to $x$ is $- 20 \frac{d}{d x} [\frac{1}{x}]$ .

$f'' (x) = - 20 \frac{d}{d x} \frac{1}{x} + \frac{d}{d x} (12)$

Step 2.2.2

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

$f'' (x) = - 20 \frac{d}{d x} x^{- 1} + \frac{d}{d x} (12)$

Step 2.2.3

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

$f'' (x) = - 20 (- x^{- 2}) + \frac{d}{d x} (12)$

Step 2.2.4

Multiply $- 1$ by $- 20$ .

$f'' (x) = 20 x^{- 2} + \frac{d}{d x} (12)$

Step 2.3

Since $12$ is constant with respect to $x$ , the derivative of $12$ with respect to $x$ is $0$ .

$f'' (x) = 20 x^{- 2} + 0$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = 20 (\frac{1}{x^{2}}) + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $20$ and $\frac{1}{x^{2}}$ .

$f'' (x) = \frac{20}{x^{2}} + 0$

Step 2.4.2.2

Add $\frac{20}{x^{2}}$ and $0$ .

$f'' (x) = \frac{20}{x^{2}}$

Step 3

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

$- \frac{20}{x} + 12 = 0$

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

By the Sum Rule, the derivative of $12 x - 20 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [12 x] + \frac{d}{d x} [- 20 \ln (x)]$ .

$\frac{d}{d x} [12 x] + \frac{d}{d x} [- 20 \ln (x)]$

Step 4.1.2

Evaluate $\frac{d}{d x} [12 x]$ .

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Step 4.1.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

$12 \frac{d}{d x} [x] + \frac{d}{d x} [- 20 \ln (x)]$

Step 4.1.2.2

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

$12 \cdot 1 + \frac{d}{d x} [- 20 \ln (x)]$

Step 4.1.2.3

Multiply $12$ by $1$ .

$12 + \frac{d}{d x} [- 20 \ln (x)]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- 20 \ln (x)]$ .

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Step 4.1.3.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 \ln (x)$ with respect to $x$ is $- 20 \frac{d}{d x} [\ln (x)]$ .

$12 - 20 \frac{d}{d x} [\ln (x)]$

Step 4.1.3.2

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

$12 - 20 \frac{1}{x}$

Step 4.1.3.3

Combine $- 20$ and $\frac{1}{x}$ .

$12 + \frac{- 20}{x}$

Step 4.1.3.4

Move the negative in front of the fraction.

$12 - \frac{20}{x}$

Step 4.1.4

Reorder terms.

$f' (x) = - \frac{20}{x} + 12$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{20}{x} + 12$ .

$- \frac{20}{x} + 12$

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{20}{x} + 12 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{20}{x} + 12 = 0$

Step 5.2

Subtract $12$ from both sides of the equation.

$- \frac{20}{x} = - 12$

Step 5.3

Find the LCD of the terms in the equation.

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Step 5.3.1

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

$x, 1$

Step 5.3.2

The LCM of one and any expression is the expression.

$x$

Step 5.4

Multiply each term in $- \frac{20}{x} = - 12$ by $x$ to eliminate the fractions.

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Step 5.4.1

Multiply each term in $- \frac{20}{x} = - 12$ by $x$ .

$- \frac{20}{x} x = - 12 x$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $x$ .

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Step 5.4.2.1.1

Move the leading negative in $- \frac{20}{x}$ into the numerator.

$\frac{- 20}{x} x = - 12 x$

Step 5.4.2.1.2

Cancel the common factor.

$\frac{- 20}{x} x = - 12 x$

Step 5.4.2.1.3

Rewrite the expression.

$- 20 = - 12 x$

Step 5.5

Solve the equation.

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Step 5.5.1

Rewrite the equation as $- 12 x = - 20$ .

$- 12 x = - 20$

Step 5.5.2

Divide each term in $- 12 x = - 20$ by $- 12$ and simplify.

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Step 5.5.2.1

Divide each term in $- 12 x = - 20$ by $- 12$ .

$\frac{- 12 x}{- 12} = \frac{- 20}{- 12}$

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Cancel the common factor of $- 12$ .

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Step 5.5.2.2.1.1

Cancel the common factor.

$\frac{- 12 x}{- 12} = \frac{- 20}{- 12}$

Step 5.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 20}{- 12}$

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

Cancel the common factor of $- 20$ and $- 12$ .

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Step 5.5.2.3.1.1

Factor $- 4$ out of $- 20$ .

$x = \frac{- 4 (5)}{- 12}$

Step 5.5.2.3.1.2

Cancel the common factors.

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Step 5.5.2.3.1.2.1

Factor $- 4$ out of $- 12$ .

$x = \frac{- 4 \cdot 5}{- 4 \cdot 3}$

Step 5.5.2.3.1.2.2

Cancel the common factor.

$x = \frac{- 4 \cdot 5}{- 4 \cdot 3}$

Step 5.5.2.3.1.2.3

Rewrite the expression.

$x = \frac{5}{3}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{20}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 7

Critical points to evaluate.

$x = \frac{5}{3}$

Step 8

Evaluate the second derivative at $x = \frac{5}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{20}{{(\frac{5}{3})}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Apply the product rule to $\frac{5}{3}$ .

$\frac{20}{\frac{5^{2}}{3^{2}}}$

Step 9.1.2

Raise $5$ to the power of $2$ .

$\frac{20}{\frac{25}{3^{2}}}$

Step 9.1.3

Raise $3$ to the power of $2$ .

$\frac{20}{\frac{25}{9}}$

Step 9.2

Multiply the numerator by the reciprocal of the denominator.

$20 (\frac{9}{25})$

Step 9.3

Cancel the common factor of $5$ .

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Step 9.3.1

Factor $5$ out of $20$ .

$5 (4) \frac{9}{25}$

Step 9.3.2

Factor $5$ out of $25$ .

$5 \cdot 4 \frac{9}{5 \cdot 5}$

Step 9.3.3

Cancel the common factor.

$5 \cdot 4 \frac{9}{5 \cdot 5}$

Step 9.3.4

Rewrite the expression.

$4 (\frac{9}{5})$

Step 9.4

Combine $4$ and $\frac{9}{5}$ .

$\frac{4 \cdot 9}{5}$

Step 9.5

Multiply $4$ by $9$ .

$\frac{36}{5}$

Step 10

$x = \frac{5}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5}{3}$ is a local minimum

Step 11

Find the y-value when $x = \frac{5}{3}$ .

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Step 11.1

Replace the variable $x$ with $\frac{5}{3}$ in the expression.

$f (\frac{5}{3}) = 12 (\frac{5}{3}) - 20 \ln (\frac{5}{3})$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Cancel the common factor of $3$ .

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Step 11.2.1.1.1

Factor $3$ out of $12$ .

$f (\frac{5}{3}) = 3 (4) (\frac{5}{3}) - 20 \ln (\frac{5}{3})$

Step 11.2.1.1.2

Cancel the common factor.

$f (\frac{5}{3}) = 3 \cdot (4 (\frac{5}{3})) - 20 \ln (\frac{5}{3})$

Step 11.2.1.1.3

Rewrite the expression.

$f (\frac{5}{3}) = 4 \cdot 5 - 20 \ln (\frac{5}{3})$

Step 11.2.1.2

Multiply $4$ by $5$ .

$f (\frac{5}{3}) = 20 - 20 \ln (\frac{5}{3})$

Step 11.2.1.3

Simplify $- 20 \ln (\frac{5}{3})$ by moving $20$ inside the logarithm.

$f (\frac{5}{3}) = 20 - \ln ({(\frac{5}{3})}^{20})$

Step 11.2.1.4

Apply the product rule to $\frac{5}{3}$ .

$f (\frac{5}{3}) = 20 - \ln (\frac{5^{20}}{3^{20}})$

Step 11.2.1.5

Raise $5$ to the power of $20$ .

$f (\frac{5}{3}) = 20 - \ln (\frac{95367431640625}{3^{20}})$

Step 11.2.1.6

Raise $3$ to the power of $20$ .

$f (\frac{5}{3}) = 20 - \ln (\frac{95367431640625}{3486784401})$

Step 11.2.2

The final answer is $20 - \ln (\frac{95367431640625}{3486784401})$ .

$y = 20 - \ln (\frac{95367431640625}{3486784401})$

Step 12

These are the local extrema for $f (x) = 12 x - 20 \ln (x)$ .

$(\frac{5}{3}, 20 - \ln (\frac{95367431640625}{3486784401}))$ is a local minima

Step 13