Calculus Examples

$f (x) = - 3 x + 6 \ln (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $- 3 x + 6 \ln (2 x)$ with respect to $x$ is $\frac{d}{d x} [- 3 x] + \frac{d}{d x} [6 \ln (2 x)]$ .

$\frac{d}{d x} [- 3 x] + \frac{d}{d x} [6 \ln (2 x)]$

Step 1.2

Evaluate $\frac{d}{d x} [- 3 x]$ .

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Step 1.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$- 3 \frac{d}{d x} [x] + \frac{d}{d x} [6 \ln (2 x)]$

Step 1.2.2

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

$- 3 \cdot 1 + \frac{d}{d x} [6 \ln (2 x)]$

Step 1.2.3

Multiply $- 3$ by $1$ .

$- 3 + \frac{d}{d x} [6 \ln (2 x)]$

Step 1.3

Evaluate $\frac{d}{d x} [6 \ln (2 x)]$ .

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Step 1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \ln (2 x)$ with respect to $x$ is $6 \frac{d}{d x} [\ln (2 x)]$ .

$- 3 + 6 \frac{d}{d x} [\ln (2 x)]$

Step 1.3.2

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

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$- 3 + 6 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x])$

Step 1.3.2.2

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

$- 3 + 6 (\frac{1}{u} \frac{d}{d x} [2 x])$

Step 1.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$- 3 + 6 (\frac{1}{2 x} \frac{d}{d x} [2 x])$

Step 1.3.3

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

$- 3 + 6 (\frac{1}{2 x} (2 \frac{d}{d x} [x]))$

Step 1.3.4

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

$- 3 + 6 (\frac{1}{2 x} (2 \cdot 1))$

Step 1.3.5

Multiply $2$ by $1$ .

$- 3 + 6 (\frac{1}{2 x} \cdot 2)$

Step 1.3.6

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

$- 3 + 6 \frac{2}{2 x}$

Step 1.3.7

Cancel the common factor of $2$ .

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Step 1.3.7.1

Cancel the common factor.

$- 3 + 6 \frac{2}{2 x}$

Step 1.3.7.2

Rewrite the expression.

$- 3 + 6 \frac{1}{x}$

Step 1.3.8

Combine $6$ and $\frac{1}{x}$ .

$- 3 + \frac{6}{x}$

Step 1.4

Reorder terms.

$\frac{6}{x} - 3$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $\frac{6}{x} - 3$ with respect to $x$ is $\frac{d}{d x} [\frac{6}{x}] + \frac{d}{d x} [- 3]$ .

$f'' (x) = \frac{d}{d x} (\frac{6}{x}) + \frac{d}{d x} (- 3)$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{6}{x}]$ .

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Step 2.2.1

Since $6$ is constant with respect to $x$ , the derivative of $\frac{6}{x}$ with respect to $x$ is $6 \frac{d}{d x} [\frac{1}{x}]$ .

$f'' (x) = 6 \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (- 3)$

Step 2.2.2

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

$f'' (x) = 6 \frac{d}{d x} (x^{- 1}) + \frac{d}{d x} (- 3)$

Step 2.2.3

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

$f'' (x) = 6 (- x^{- 2}) + \frac{d}{d x} (- 3)$

Step 2.2.4

Multiply $- 1$ by $6$ .

$f'' (x) = - 6 x^{- 2} + \frac{d}{d x} (- 3)$

Step 2.3

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = - 6 x^{- 2} + 0$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = - 6 \frac{1}{x^{2}} + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 6$ and $\frac{1}{x^{2}}$ .

$f'' (x) = \frac{- 6}{x^{2}} + 0$

Step 2.4.2.2

Move the negative in front of the fraction.

$f'' (x) = - \frac{6}{x^{2}} + 0$

Step 2.4.2.3

Add $- \frac{6}{x^{2}}$ and $0$ .

$f'' (x) = - \frac{6}{x^{2}}$

Step 3

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

$\frac{6}{x} - 3 = 0$

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 $- 3 x + 6 \ln (2 x)$ with respect to $x$ is $\frac{d}{d x} [- 3 x] + \frac{d}{d x} [6 \ln (2 x)]$ .

$\frac{d}{d x} [- 3 x] + \frac{d}{d x} [6 \ln (2 x)]$

Step 4.1.2

Evaluate $\frac{d}{d x} [- 3 x]$ .

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Step 4.1.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$- 3 \frac{d}{d x} [x] + \frac{d}{d x} [6 \ln (2 x)]$

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 $n = 1$ .

$- 3 \cdot 1 + \frac{d}{d x} [6 \ln (2 x)]$

Step 4.1.2.3

Multiply $- 3$ by $1$ .

$- 3 + \frac{d}{d x} [6 \ln (2 x)]$

Step 4.1.3

Evaluate $\frac{d}{d x} [6 \ln (2 x)]$ .

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Step 4.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \ln (2 x)$ with respect to $x$ is $6 \frac{d}{d x} [\ln (2 x)]$ .

$- 3 + 6 \frac{d}{d x} [\ln (2 x)]$

Step 4.1.3.2

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

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Step 4.1.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$- 3 + 6 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x])$

Step 4.1.3.2.2

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

$- 3 + 6 (\frac{1}{u} \frac{d}{d x} [2 x])$

Step 4.1.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$- 3 + 6 (\frac{1}{2 x} \frac{d}{d x} [2 x])$

Step 4.1.3.3

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

$- 3 + 6 (\frac{1}{2 x} (2 \frac{d}{d x} [x]))$

Step 4.1.3.4

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

$- 3 + 6 (\frac{1}{2 x} (2 \cdot 1))$

Step 4.1.3.5

Multiply $2$ by $1$ .

$- 3 + 6 (\frac{1}{2 x} \cdot 2)$

Step 4.1.3.6

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

$- 3 + 6 \frac{2}{2 x}$

Step 4.1.3.7

Cancel the common factor of $2$ .

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Step 4.1.3.7.1

Cancel the common factor.

$- 3 + 6 \frac{2}{2 x}$

Step 4.1.3.7.2

Rewrite the expression.

$- 3 + 6 \frac{1}{x}$

Step 4.1.3.8

Combine $6$ and $\frac{1}{x}$ .

$- 3 + \frac{6}{x}$

Step 4.1.4

Reorder terms.

$f' (x) = \frac{6}{x} - 3$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{6}{x} - 3$ .

$\frac{6}{x} - 3$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{6}{x} - 3 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{6}{x} - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$\frac{6}{x} = 3$

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 $\frac{6}{x} = 3$ by $x$ to eliminate the fractions.

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Step 5.4.1

Multiply each term in $\frac{6}{x} = 3$ by $x$ .

$\frac{6}{x} x = 3 x$

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

Cancel the common factor.

$\frac{6}{x} x = 3 x$

Step 5.4.2.1.2

Rewrite the expression.

$6 = 3 x$

Step 5.5

Solve the equation.

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Step 5.5.1

Rewrite the equation as $3 x = 6$ .

$3 x = 6$

Step 5.5.2

Divide each term in $3 x = 6$ by $3$ and simplify.

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Step 5.5.2.1

Divide each term in $3 x = 6$ by $3$ .

$\frac{3 x}{3} = \frac{6}{3}$

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Cancel the common factor of $3$ .

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Step 5.5.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{6}{3}$

Step 5.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{6}{3}$

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

Divide $6$ by $3$ .

$x = 2$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{6}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 7

Critical points to evaluate.

$x = 2$

Step 8

Evaluate the second derivative at $x = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{6}{{(2)}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

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

$- \frac{6}{4}$

Step 9.2

Cancel the common factor of $6$ and $4$ .

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Step 9.2.1

Factor $2$ out of $6$ .

$- \frac{2 (3)}{4}$

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $2$ out of $4$ .

$- \frac{2 \cdot 3}{2 \cdot 2}$

Step 9.2.2.2

Cancel the common factor.

$- \frac{2 \cdot 3}{2 \cdot 2}$

Step 9.2.2.3

Rewrite the expression.

$- \frac{3}{2}$

Step 10

$x = 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 2$ is a local maximum

Step 11

Find the y-value when $x = 2$ .

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Step 11.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = - 3 \cdot 2 + 6 \ln (2 (2))$

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

Multiply $- 3$ by $2$ .

$f (2) = - 6 + 6 \ln (2 (2))$

Step 11.2.1.2

Multiply $2$ by $2$ .

$f (2) = - 6 + 6 \ln (4)$

Step 11.2.1.3

Simplify $6 \ln (4)$ by moving $6$ inside the logarithm.

$f (2) = - 6 + \ln (4^{6})$

Step 11.2.1.4

Raise $4$ to the power of $6$ .

$f (2) = - 6 + \ln (4096)$

Step 11.2.2

The final answer is $- 6 + \ln (4096)$ .

$y = - 6 + \ln (4096)$

Step 12

These are the local extrema for $f (x) = - 3 x + 6 \ln (2 x)$ .

$(2, - 6 + \ln (4096))$ is a local maxima

Step 13