Calculus Examples

$\frac{\ln (x)}{x}$

Step 1

Write $\frac{\ln (x)}{x}$ as a function.

$f (x) = \frac{\ln (x)}{x}$

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

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 $f (x) = \ln (x)$ and $g (x) = x$ .

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

Step 2.1.2

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

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

Step 2.1.3

Differentiate using the Power Rule.

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Step 2.1.3.1

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

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

Step 2.1.3.2

Cancel the common factor of $x$ .

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Step 2.1.3.2.1

Cancel the common factor.

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

Step 2.1.3.2.2

Rewrite the expression.

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

Step 2.1.3.3

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

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

Step 2.1.3.4

Multiply $- 1$ by $1$ .

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

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{1 - \ln (x)}{x^{2}}$ .

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

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{1 - \ln (x)}{x^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{1 - \ln (x)}{x^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$1 - \ln (x) = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

Subtract $1$ from both sides of the equation.

$- \ln (x) = - 1$

Step 3.3.2

Divide each term in $- \ln (x) = - 1$ by $- 1$ and simplify.

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Step 3.3.2.1

Divide each term in $- \ln (x) = - 1$ by $- 1$ .

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

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Dividing two negative values results in a positive value.

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

Step 3.3.2.2.2

Divide $\ln (x)$ by $1$ .

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

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Divide $- 1$ by $- 1$ .

$\ln (x) = 1$

Step 3.3.3

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

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

Step 3.3.4

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

$e^{1} = x$

Step 3.3.5

Rewrite the equation as $x = e^{1}$ .

$x = e$

Step 4

The values which make the derivative equal to $0$ are $e$ .

$e$

Step 5

Find where the derivative is undefined.

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Step 5.1

Set the denominator in $\frac{1 - \ln (x)}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 5.2

Solve for $x$ .

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Step 5.2.1

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

$x = \pm \sqrt{0}$

Step 5.2.2

Simplify $\pm \sqrt{0}$ .

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Step 5.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 5.2.2.2

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

$x = \pm 0$

Step 5.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 5.3

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

$x \leq 0$

Step 5.4

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 6

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

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

Step 7

Exclude the intervals that are not in the domain.

$(0, e)$

Step 8

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

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Step 8.1

Replace the variable $x$ with $1.35914085$ in the expression.

$f' (1.35914085) = \frac{1 - \ln (1.35914085)}{{(1.35914085)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Raise $1.35914085$ to the power of $2$ .

$f' (1.35914085) = \frac{1 - \ln (1.35914085)}{1.84726385}$

Step 8.2.2

Replace $e$ with an approximation.

$f' (1.35914085) = \frac{1 - \log_{2.71828182} (1.35914085)}{1.84726385}$

Step 8.2.3

Log base $2.71828182$ of $1.35914085$ is approximately $0.30685277$ .

$f' (1.35914085) = \frac{1 - 1 \cdot 0.30685277}{1.84726385}$

Step 8.2.4

Multiply $- 1$ by $0.30685277$ .

$f' (1.35914085) = \frac{1 - 0.30685277}{1.84726385}$

Step 8.2.5

Subtract $0.30685277$ from $1$ .

$f' (1.35914085) = \frac{0.69314722}{1.84726385}$

Step 8.2.6

Divide $0.69314722$ by $1.84726385$ .

$f' (1.35914085) = 0.37522914$

Step 8.2.7

The final answer is $0.37522914$ .

$0.37522914$

Step 8.3

At $x = 1.35914085$ the derivative is $0.37522914$ . Since this is positive, the function is increasing on $(0, e)$ .

Increasing on $(0, e)$ since $f' (x) > 0$

Step 9

Exclude the intervals that are not in the domain.

$(0, e), (e, \infty)$

Step 10

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

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Step 10.1

Replace the variable $x$ with $3.7182817$ in the expression.

$f' (3.7182817) = \frac{1 - \ln (3.7182817)}{{(3.7182817)}^{2}}$

Step 10.2

Simplify the result.

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Step 10.2.1

Raise $3.7182817$ to the power of $2$ .

$f' (3.7182817) = \frac{1 - \ln (3.7182817)}{13.8256188}$

Step 10.2.2

Replace $e$ with an approximation.

$f' (3.7182817) = \frac{1 - \log_{2.71828182} (3.7182817)}{13.8256188}$

Step 10.2.3

Log base $2.71828182$ of $3.7182817$ is approximately $1.31326165$ .

$f' (3.7182817) = \frac{1 - 1 \cdot 1.31326165}{13.8256188}$

Step 10.2.4

Multiply $- 1$ by $1.31326165$ .

$f' (3.7182817) = \frac{1 - 1.31326165}{13.8256188}$

Step 10.2.5

Subtract $1.31326165$ from $1$ .

$f' (3.7182817) = \frac{- 0.31326165}{13.8256188}$

Step 10.2.6

Divide $- 0.31326165$ by $13.8256188$ .

$f' (3.7182817) = - 0.02265805$

Step 10.2.7

The final answer is $- 0.02265805$ .

$- 0.02265805$

Step 10.3

At $x = 3.7182817$ the derivative is $- 0.02265805$ . Since this is negative, the function is decreasing on $(e, \infty)$ .

Decreasing on $(e, \infty)$ since $f' (x) < 0$

Step 11

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

Increasing on: $(0, e)$

Decreasing on: $(e, \infty)$

Step 12