Calculus Examples

$2 x^{2} - 16 \ln (x)$

Step 1

Write $2 x^{2} - 16 \ln (x)$ as a function.

$f (x) = 2 x^{2} - 16 \ln (x)$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $2 x^{2} - 16 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 \ln (x)]$ .

$\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 \ln (x)]$

Step 2.1.2

Evaluate $\frac{d}{d x} [2 x^{2}]$ .

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Step 2.1.2.1

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

$2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16 \ln (x)]$

Step 2.1.2.2

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

$2 (2 x) + \frac{d}{d x} [- 16 \ln (x)]$

Step 2.1.2.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [- 16 \ln (x)]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 16 \ln (x)]$ .

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Step 2.1.3.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 \ln (x)$ with respect to $x$ is $- 16 \frac{d}{d x} [\ln (x)]$ .

$4 x - 16 \frac{d}{d x} [\ln (x)]$

Step 2.1.3.2

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

$4 x - 16 \frac{1}{x}$

Step 2.1.3.3

Combine $- 16$ and $\frac{1}{x}$ .

$4 x + \frac{- 16}{x}$

Step 2.1.3.4

Move the negative in front of the fraction.

$f' (x) = 4 x - \frac{16}{x}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $4 x - \frac{16}{x}$ .

$4 x - \frac{16}{x}$

Step 3

Set the first derivative equal to $0$ then solve the equation $4 x - \frac{16}{x} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$4 x - \frac{16}{x} = 0$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

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

$1, x, 1$

Step 3.2.2

The LCM of one and any expression is the expression.

$x$

Step 3.3

Multiply each term in $4 x - \frac{16}{x} = 0$ by $x$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $4 x - \frac{16}{x} = 0$ by $x$ .

$4 x \cdot x - \frac{16}{x} x = 0 x$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 3.3.2.1.1.1

Move $x$ .

$4 (x \cdot x) - \frac{16}{x} x = 0 x$

Step 3.3.2.1.1.2

Multiply $x$ by $x$ .

$4 x^{2} - \frac{16}{x} x = 0 x$

Step 3.3.2.1.2

Cancel the common factor of $x$ .

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Step 3.3.2.1.2.1

Move the leading negative in $- \frac{16}{x}$ into the numerator.

$4 x^{2} + \frac{- 16}{x} x = 0 x$

Step 3.3.2.1.2.2

Cancel the common factor.

$4 x^{2} + \frac{- 16}{x} x = 0 x$

Step 3.3.2.1.2.3

Rewrite the expression.

$4 x^{2} - 16 = 0 x$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Multiply $0$ by $x$ .

$4 x^{2} - 16 = 0$

Step 3.4

Solve the equation.

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Step 3.4.1

Add $16$ to both sides of the equation.

$4 x^{2} = 16$

Step 3.4.2

Divide each term in $4 x^{2} = 16$ by $4$ and simplify.

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Step 3.4.2.1

Divide each term in $4 x^{2} = 16$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{16}{4}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $4$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{16}{4}$

Step 3.4.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{16}{4}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Divide $16$ by $4$ .

$x^{2} = 4$

Step 3.4.3

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

$x = \pm \sqrt{4}$

Step 3.4.4

Simplify $\pm \sqrt{4}$ .

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Step 3.4.4.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 3.4.4.2

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

$x = \pm 2$

Step 3.4.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.4.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Step 3.4.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

Step 3.4.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 4

The values which make the derivative equal to $0$ are $2, - 2$ .

$2, - 2$

Step 5

Set the denominator in $\frac{16}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 6

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 0) \cup (0, 2) \cup (2, \infty)$

Step 7

Exclude the intervals that are not in the domain.

Step 8

Substitute a value from the interval $(- 2, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = 4 (- 1) - \frac{16}{- 1}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Multiply $4$ by $- 1$ .

$f' (- 1) = - 4 - \frac{16}{- 1}$

Step 8.2.1.2

Divide $16$ by $- 1$ .

$f' (- 1) = - 4 + 16$

Step 8.2.1.3

Multiply $- 1$ by $- 16$ .

$f' (- 1) = - 4 + 16$

Step 8.2.2

Add $- 4$ and $16$ .

$f' (- 1) = 12$

Step 8.2.3

The final answer is $12$ .

$12$

Step 9

Exclude the intervals that are not in the domain.

$(0, 2)$

Step 10

Substitute a value from the interval $(0, 2)$ into the derivative to determine if the function is increasing or decreasing.

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Step 10.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = 4 (1) - \frac{16}{1}$

Step 10.2

Simplify the result.

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Step 10.2.1

Simplify each term.

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Step 10.2.1.1

Multiply $4$ by $1$ .

$f' (1) = 4 - \frac{16}{1}$

Step 10.2.1.2

Divide $16$ by $1$ .

$f' (1) = 4 - 1 \cdot 16$

Step 10.2.1.3

Multiply $- 1$ by $16$ .

$f' (1) = 4 - 16$

Step 10.2.2

Subtract $16$ from $4$ .

$f' (1) = - 12$

Step 10.2.3

The final answer is $- 12$ .

$- 12$

Step 10.3

At $x = 1$ the derivative is $- 12$ . Since this is negative, the function is decreasing on $(0, 2)$ .

Decreasing on $(0, 2)$ since $f' (x) < 0$

Step 11

Exclude the intervals that are not in the domain.

$(0, 2), (2, \infty)$

Step 12

Substitute a value from the interval $(2, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 12.1

Replace the variable $x$ with $3$ in the expression.

$f' (3) = 4 (3) - \frac{16}{3}$

Step 12.2

Simplify the result.

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Step 12.2.1

Multiply $4$ by $3$ .

$f' (3) = 12 - \frac{16}{3}$

Step 12.2.2

To write $12$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f' (3) = 12 \cdot \frac{3}{3} - \frac{16}{3}$

Step 12.2.3

Combine $12$ and $\frac{3}{3}$ .

$f' (3) = \frac{12 \cdot 3}{3} - \frac{16}{3}$

Step 12.2.4

Combine the numerators over the common denominator.

$f' (3) = \frac{12 \cdot 3 - 16}{3}$

Step 12.2.5

Simplify the numerator.

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Step 12.2.5.1

Multiply $12$ by $3$ .

$f' (3) = \frac{36 - 16}{3}$

Step 12.2.5.2

Subtract $16$ from $36$ .

$f' (3) = \frac{20}{3}$

Step 12.2.6

The final answer is $\frac{20}{3}$ .

$\frac{20}{3}$

Step 12.3

At $x = 3$ the derivative is $\frac{20}{3}$ . Since this is positive, the function is increasing on $(2, \infty)$ .

Increasing on $(2, \infty)$ since $f' (x) > 0$

Step 13

List the intervals on which the function is increasing and decreasing.

Increasing on: $(2, \infty)$

Decreasing on: $(0, 2)$

Step 14