Calculus Examples

$f (x) = x - 5 \ln (3 x - 9)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $x - 5 \ln (3 x - 9)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5 \ln (3 x - 9)]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 5 \ln (3 x - 9)]$

Step 1.1.2

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

$1 + \frac{d}{d x} [- 5 \ln (3 x - 9)]$

Step 1.2

Evaluate $\frac{d}{d x} [- 5 \ln (3 x - 9)]$ .

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Step 1.2.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 \ln (3 x - 9)$ with respect to $x$ is $- 5 \frac{d}{d x} [\ln (3 x - 9)]$ .

$1 - 5 \frac{d}{d x} [\ln (3 x - 9)]$

Step 1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 3 x - 9$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $3 x - 9$ .

$1 - 5 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [3 x - 9])$

Step 1.2.2.2

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

$1 - 5 (\frac{1}{u} \frac{d}{d x} [3 x - 9])$

Step 1.2.2.3

Replace all occurrences of $u$ with $3 x - 9$ .

$1 - 5 (\frac{1}{3 x - 9} \frac{d}{d x} [3 x - 9])$

Step 1.2.3

By the Sum Rule, the derivative of $3 x - 9$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 9]$ .

$1 - 5 (\frac{1}{3 x - 9} (\frac{d}{d x} [3 x] + \frac{d}{d x} [- 9]))$

Step 1.2.4

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

$1 - 5 (\frac{1}{3 x - 9} (3 \frac{d}{d x} [x] + \frac{d}{d x} [- 9]))$

Step 1.2.5

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

$1 - 5 (\frac{1}{3 x - 9} (3 \cdot 1 + \frac{d}{d x} [- 9]))$

Step 1.2.6

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$1 - 5 (\frac{1}{3 x - 9} (3 \cdot 1 + 0))$

Step 1.2.7

Multiply $3$ by $1$ .

$1 - 5 (\frac{1}{3 x - 9} (3 + 0))$

Step 1.2.8

Add $3$ and $0$ .

$1 - 5 (\frac{1}{3 x - 9} \cdot 3)$

Step 1.2.9

Combine $\frac{1}{3 x - 9}$ and $3$ .

$1 - 5 \frac{3}{3 x - 9}$

Step 1.2.10

Cancel the common factor of $3$ and $3 x - 9$ .

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Step 1.2.10.1

Factor $3$ out of $3$ .

$1 - 5 \frac{3 \cdot 1}{3 x - 9}$

Step 1.2.10.2

Cancel the common factors.

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Step 1.2.10.2.1

Factor $3$ out of $3 x$ .

$1 - 5 \frac{3 \cdot 1}{3 (x) - 9}$

Step 1.2.10.2.2

Factor $3$ out of $- 9$ .

$1 - 5 \frac{3 \cdot 1}{3 (x) + 3 (- 3)}$

Step 1.2.10.2.3

Factor $3$ out of $3 (x) + 3 (- 3)$ .

$1 - 5 \frac{3 \cdot 1}{3 (x - 3)}$

Step 1.2.10.2.4

Cancel the common factor.

$1 - 5 \frac{3 \cdot 1}{3 (x - 3)}$

Step 1.2.10.2.5

Rewrite the expression.

$1 - 5 \frac{1}{x - 3}$

Step 1.2.11

Combine $- 5$ and $\frac{1}{x - 3}$ .

$1 + \frac{- 5}{x - 3}$

Step 1.2.12

Move the negative in front of the fraction.

$1 - \frac{5}{x - 3}$

Step 1.3

Combine terms.

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Step 1.3.1

Write $1$ as a fraction with a common denominator.

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

Step 1.3.2

Combine the numerators over the common denominator.

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

Step 1.3.3

Subtract $5$ from $- 3$ .

$\frac{x - 8}{x - 3}$

Step 2

Find the second derivative of the function.

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Step 2.1

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 $f (x) = x - 8$ and $g (x) = x - 3$ .

$f'' (x) = \frac{(x - 3) \frac{d}{d x} (x - 8) - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$f'' (x) = \frac{(x - 3) (\frac{d}{d x} (x) + \frac{d}{d x} (- 8)) - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2.2

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

$f'' (x) = \frac{(x - 3) (1 + \frac{d}{d x} (- 8)) - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2.3

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$f'' (x) = \frac{(x - 3) (1 + 0) - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{(x - 3) \cdot 1 - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2.4.2

Multiply $x - 3$ by $1$ .

$f'' (x) = \frac{x - 3 - (x - 8) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 2.2.5

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

$f'' (x) = \frac{x - 3 - (x - 8) (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))}{{(x - 3)}^{2}}$

Step 2.2.6

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

$f'' (x) = \frac{x - 3 - (x - 8) (1 + \frac{d}{d x} (- 3))}{{(x - 3)}^{2}}$

Step 2.2.7

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

$f'' (x) = \frac{x - 3 - (x - 8) (1 + 0)}{{(x - 3)}^{2}}$

Step 2.2.8

Simplify the expression.

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Step 2.2.8.1

Add $1$ and $0$ .

$f'' (x) = \frac{x - 3 - (x - 8) \cdot 1}{{(x - 3)}^{2}}$

Step 2.2.8.2

Multiply $- 1$ by $1$ .

$f'' (x) = \frac{x - 3 - (x - 8)}{{(x - 3)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$f'' (x) = \frac{x - 3 - x + 8}{{(x - 3)}^{2}}$

Step 2.3.2

Simplify the numerator.

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Step 2.3.2.1

Combine the opposite terms in $x - 3 - x - - 8$ .

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Step 2.3.2.1.1

Subtract $x$ from $x$ .

$f'' (x) = \frac{0 - 3 + 8}{{(x - 3)}^{2}}$

Step 2.3.2.1.2

Subtract $3$ from $0$ .

$f'' (x) = \frac{- 3 + 8}{{(x - 3)}^{2}}$

Step 2.3.2.2

Multiply $- 1$ by $- 8$ .

$f'' (x) = \frac{- 3 + 8}{{(x - 3)}^{2}}$

Step 2.3.2.3

Add $- 3$ and $8$ .

$f'' (x) = \frac{5}{{(x - 3)}^{2}}$

Step 3

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

$\frac{x - 8}{x - 3} = 0$

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

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of $x - 5 \ln (3 x - 9)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5 \ln (3 x - 9)]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 5 \ln (3 x - 9)]$

Step 4.1.1.2

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

$1 + \frac{d}{d x} [- 5 \ln (3 x - 9)]$

Step 4.1.2

Evaluate $\frac{d}{d x} [- 5 \ln (3 x - 9)]$ .

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Step 4.1.2.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 \ln (3 x - 9)$ with respect to $x$ is $- 5 \frac{d}{d x} [\ln (3 x - 9)]$ .

$1 - 5 \frac{d}{d x} [\ln (3 x - 9)]$

Step 4.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 3 x - 9$ .

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Step 4.1.2.2.1

To apply the Chain Rule, set $u$ as $3 x - 9$ .

$1 - 5 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [3 x - 9])$

Step 4.1.2.2.2

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

$1 - 5 (\frac{1}{u} \frac{d}{d x} [3 x - 9])$

Step 4.1.2.2.3

Replace all occurrences of $u$ with $3 x - 9$ .

$1 - 5 (\frac{1}{3 x - 9} \frac{d}{d x} [3 x - 9])$

Step 4.1.2.3

By the Sum Rule, the derivative of $3 x - 9$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 9]$ .

$1 - 5 (\frac{1}{3 x - 9} (\frac{d}{d x} [3 x] + \frac{d}{d x} [- 9]))$

Step 4.1.2.4

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

$1 - 5 (\frac{1}{3 x - 9} (3 \frac{d}{d x} [x] + \frac{d}{d x} [- 9]))$

Step 4.1.2.5

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

$1 - 5 (\frac{1}{3 x - 9} (3 \cdot 1 + \frac{d}{d x} [- 9]))$

Step 4.1.2.6

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$1 - 5 (\frac{1}{3 x - 9} (3 \cdot 1 + 0))$

Step 4.1.2.7

Multiply $3$ by $1$ .

$1 - 5 (\frac{1}{3 x - 9} (3 + 0))$

Step 4.1.2.8

Add $3$ and $0$ .

$1 - 5 (\frac{1}{3 x - 9} \cdot 3)$

Step 4.1.2.9

Combine $\frac{1}{3 x - 9}$ and $3$ .

$1 - 5 \frac{3}{3 x - 9}$

Step 4.1.2.10

Cancel the common factor of $3$ and $3 x - 9$ .

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Step 4.1.2.10.1

Factor $3$ out of $3$ .

$1 - 5 \frac{3 \cdot 1}{3 x - 9}$

Step 4.1.2.10.2

Cancel the common factors.

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Step 4.1.2.10.2.1

Factor $3$ out of $3 x$ .

$1 - 5 \frac{3 \cdot 1}{3 (x) - 9}$

Step 4.1.2.10.2.2

Factor $3$ out of $- 9$ .

$1 - 5 \frac{3 \cdot 1}{3 (x) + 3 (- 3)}$

Step 4.1.2.10.2.3

Factor $3$ out of $3 (x) + 3 (- 3)$ .

$1 - 5 \frac{3 \cdot 1}{3 (x - 3)}$

Step 4.1.2.10.2.4

Cancel the common factor.

$1 - 5 \frac{3 \cdot 1}{3 (x - 3)}$

Step 4.1.2.10.2.5

Rewrite the expression.

$1 - 5 \frac{1}{x - 3}$

Step 4.1.2.11

Combine $- 5$ and $\frac{1}{x - 3}$ .

$1 + \frac{- 5}{x - 3}$

Step 4.1.2.12

Move the negative in front of the fraction.

$1 - \frac{5}{x - 3}$

Step 4.1.3

Combine terms.

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Step 4.1.3.1

Write $1$ as a fraction with a common denominator.

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

Step 4.1.3.2

Combine the numerators over the common denominator.

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

Step 4.1.3.3

Subtract $5$ from $- 3$ .

$f' (x) = \frac{x - 8}{x - 3}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x - 8}{x - 3}$ .

$\frac{x - 8}{x - 3}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{x - 8}{x - 3} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{x - 8}{x - 3} = 0$

Step 5.2

Set the numerator equal to zero.

$x - 8 = 0$

Step 5.3

Add $8$ to both sides of the equation.

$x = 8$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{x - 8}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

Critical points to evaluate.

$x = 8$

Step 8

Evaluate the second derivative at $x = 8$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{5}{{((8) - 3)}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Subtract $3$ from $8$ .

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

Step 9.1.2

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

$\frac{5}{25}$

Step 9.2

Cancel the common factor of $5$ and $25$ .

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Step 9.2.1

Factor $5$ out of $5$ .

$\frac{5 (1)}{25}$

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $5$ out of $25$ .

$\frac{5 \cdot 1}{5 \cdot 5}$

Step 9.2.2.2

Cancel the common factor.

$\frac{5 \cdot 1}{5 \cdot 5}$

Step 9.2.2.3

Rewrite the expression.

$\frac{1}{5}$

Step 10

$x = 8$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 8$ is a local minimum

Step 11

Find the y-value when $x = 8$ .

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Step 11.1

Replace the variable $x$ with $8$ in the expression.

$f (8) = (8) - 5 \ln (3 (8) - 9)$

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 $8$ .

$f (8) = 8 - 5 \ln (24 - 9)$

Step 11.2.1.2

Subtract $9$ from $24$ .

$f (8) = 8 - 5 \ln (15)$

Step 11.2.1.3

Simplify $- 5 \ln (15)$ by moving $5$ inside the logarithm.

$f (8) = 8 - \ln (15^{5})$

Step 11.2.1.4

Raise $15$ to the power of $5$ .

$f (8) = 8 - \ln (759375)$

Step 11.2.2

The final answer is $8 - \ln (759375)$ .

$y = 8 - \ln (759375)$

Step 12

These are the local extrema for $f (x) = x - 5 \ln (3 x - 9)$ .

$(8, 8 - \ln (759375))$ is a local minima

Step 13