Calculus Examples

\frac{ln (x)}{x}

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the Quotient Rule which states that

\frac{d}{d x} [\frac{f (x)}{g (x)}]

$\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is

\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}

$\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where

f (x) = ln (x)

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

g (x) = x

$g (x) = x$ .

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

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

Step 1.2

The derivative of

ln (x)

$\ln (x)$ with respect to

x

$x$ is

\frac{1}{x}

$\frac{1}{x}$ .

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

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

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

Combine

x

$x$ and

\frac{1}{x}

$\frac{1}{x}$ .

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

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

Step 1.3.2

Cancel the common factor of

x

$x$ .

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Step 1.3.2.1

Cancel the common factor.

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

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

Step 1.3.2.2

Rewrite the expression.

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

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

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

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

Step 1.3.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

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

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

Step 1.3.4

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

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

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

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

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

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)}]

$\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is

\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}

$\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where

f (x) = 1 - ln (x)

$f (x) = 1 - \ln (x)$ and

g (x) = x^{2}

$g (x) = x^{2}$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

Multiply the exponents in

${(x^{2})}^{2}$ .

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Step 2.2.1.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.2.1.2

Multiply

$2$ by

$2$ .

$f'' (x) = \frac{x^{2} \frac{d}{d x} (1 - \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.2.2

By the Sum Rule, the derivative of

$1 - \ln (x)$ with respect to

$x$ is

$\frac{d}{d x} [1] + \frac{d}{d x} [- \ln (x)]$ .

$f'' (x) = \frac{x^{2} (\frac{d}{d x} (1) + \frac{d}{d x} (- \ln (x))) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.2.3

Since

$1$ is constant with respect to

$x$ , the derivative of

$1$ with respect to

$x$ is

$0$ .

$f'' (x) = \frac{x^{2} (0 + \frac{d}{d x} (- \ln (x))) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.2.4

Add

$0$ and

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

$f'' (x) = \frac{x^{2} \frac{d}{d x} (- \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.2.5

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- \ln (x)$ with respect to

$x$ is

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

$f'' (x) = \frac{x^{2} (- \frac{d}{d x} \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.3

The derivative of

$\ln (x)$ with respect to

$x$ is

$\frac{1}{x}$ .

$f'' (x) = \frac{x^{2} (- \frac{1}{x}) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4

Differentiate using the Power Rule.

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Step 2.4.1

Combine

$x^{2}$ and

$\frac{1}{x}$ .

$f'' (x) = \frac{- \frac{x^{2}}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2

Cancel the common factor of

$x^{2}$ and

$x$ .

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Step 2.4.2.1

Factor

$x$ out of

$x^{2}$ .

$f'' (x) = \frac{- \frac{x \cdot x}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2.2

Cancel the common factors.

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Step 2.4.2.2.1

Raise

$x$ to the power of

$1$ .

$f'' (x) = \frac{- \frac{x \cdot x}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2.2.2

Factor

$x$ out of

$x^{1}$ .

$f'' (x) = \frac{- \frac{x \cdot x}{x \cdot 1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2.2.3

Cancel the common factor.

$f'' (x) = \frac{- \frac{x \cdot x}{x \cdot 1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2.2.4

Rewrite the expression.

$f'' (x) = \frac{- \frac{x}{1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.2.2.5

Divide

$x$ by

$1$ .

$f'' (x) = \frac{- x - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Step 2.4.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

$f'' (x) = \frac{- x - (1 - \ln (x)) (2 x)}{x^{4}}$

Step 2.4.4

Simplify with factoring out.

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Step 2.4.4.1

Multiply

$2$ by

$- 1$ .

$f'' (x) = \frac{- x - 2 (1 - \ln (x)) x}{x^{4}}$

Step 2.4.4.2

Factor

$x$ out of

$- x - 2 (1 - \ln (x)) x$ .

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Step 2.4.4.2.1

Factor

$x$ out of

$- x$ .

$f'' (x) = \frac{x \cdot - 1 - 2 (1 - \ln (x)) x}{x^{4}}$

Step 2.4.4.2.2

Factor

$x$ out of

$- 2 (1 - \ln (x)) x$ .

$f'' (x) = \frac{x \cdot - 1 + x (- 2 (1 - \ln (x)))}{x^{4}}$

Step 2.4.4.2.3

Factor

$x$ out of

$x \cdot - 1 + x (- 2 (1 - \ln (x)))$ .

$f'' (x) = \frac{x (- 1 - 2 (1 - \ln (x)))}{x^{4}}$

Step 2.5

Cancel the common factors.

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Step 2.5.1

Factor

$x$ out of

$x^{4}$ .

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

Step 2.5.2

Cancel the common factor.

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

Step 2.5.3

Rewrite the expression.

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

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

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

Step 2.6.2

Simplify the numerator.

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Step 2.6.2.1

Simplify each term.

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Step 2.6.2.1.1

Multiply

$- 2$ by

$1$ .

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

Step 2.6.2.1.2

Multiply

$- 2 (- \ln (x))$ .

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Step 2.6.2.1.2.1

Multiply

$- 1$ by

$- 2$ .

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

Step 2.6.2.1.2.2

Simplify

$2 \ln (x)$ by moving

$2$ inside the logarithm.

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

Step 2.6.2.2

Subtract

$2$ from

$- 1$ .

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

Step 2.6.3

Rewrite

$- 3$ as

$- 1 (3)$ .

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

Step 2.6.4

Factor

$- 1$ out of

$\ln (x^{2})$ .

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

Step 2.6.5

Factor

$- 1$ out of

$- 1 (3) - 1 (- \ln (x^{2}))$ .

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

Step 2.6.6

Move the negative in front of the fraction.

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

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

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 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 4.1.2

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 4.1.3

Differentiate using the Power Rule.

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Step 4.1.3.1

Combine

$x$ and

$\frac{1}{x}$ .

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

Step 4.1.3.2

Cancel the common factor of

$x$ .

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Step 4.1.3.2.1

Cancel the common factor.

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

Step 4.1.3.2.2

Rewrite the expression.

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

Step 4.1.3.3

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 4.1.3.4

Multiply

$- 1$ by

$1$ .

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

Step 4.2

The first derivative of

$f (x)$ with respect to

$x$ is

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

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

Step 5

Set the first derivative equal to

$0$ then solve the equation

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

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Step 5.1

Set the first derivative equal to

$0$ .

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

Step 5.2

Set the numerator equal to zero.

$1 - \ln (x) = 0$

Step 5.3

Solve the equation for

$x$ .

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Step 5.3.1

Subtract

$1$ from both sides of the equation.

$- \ln (x) = - 1$

Step 5.3.2

Divide each term in

$- \ln (x) = - 1$ by

$- 1$ and simplify.

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Step 5.3.2.1

Divide each term in

$- \ln (x) = - 1$ by

$- 1$ .

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

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.3.2.2.2

Divide

$\ln (x)$ by

$1$ .

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

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

Divide

$- 1$ by

$- 1$ .

$\ln (x) = 1$

Step 5.3.3

To solve for

$x$ , rewrite the equation using properties of logarithms.

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

Step 5.3.4

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 5.3.5

Rewrite the equation as

$x = e^{1}$ .

$x = e$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in

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

$0$ to find where the expression is undefined.

$x^{2} = 0$

Step 6.2

Solve for

$x$ .

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Step 6.2.1

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

$x = \pm \sqrt{0}$

Step 6.2.2

Simplify

$\pm \sqrt{0}$ .

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Step 6.2.2.1

Rewrite

$0$ as

$0^{2}$ .

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

Step 6.2.2.2

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

$x = \pm 0$

Step 6.2.2.3

Plus or minus

$0$ is

$0$ .

$x = 0$

Step 6.3

Set the argument in

$\ln (x)$ less than or equal to

$0$ to find where the expression is undefined.

$x \leq 0$

Step 6.4

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 7

Critical points to evaluate.

$x = e$

Step 8

Evaluate the second derivative at

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

$- \frac{3 - \ln ({(e)}^{2})}{{(e)}^{3}}$

Step 9

Simplify the numerator.

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Step 9.1

Use logarithm rules to move

$2$ out of the exponent.

$- \frac{3 - (2 \ln (e))}{e^{3}}$

Step 9.2

The natural logarithm of

$e$ is

$1$ .

$- \frac{3 - (2 \cdot 1)}{e^{3}}$

Step 9.3

Multiply

$2$ by

$1$ .

$- \frac{3 - 1 \cdot 2}{e^{3}}$

Step 9.4

Multiply

$- 1$ by

$2$ .

$- \frac{3 - 2}{e^{3}}$

Step 9.5

Subtract

$2$ from

$3$ .

$- \frac{1}{e^{3}}$

Step 10

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

$x = e$ is a local maximum

Step 11

Find the y-value when

$x = e$ .

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Step 11.1

Replace the variable

$x$ with

$e$ in the expression.

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

Step 11.2

Simplify the result.

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Step 11.2.1

The natural logarithm of

$e$ is

$1$ .

$f (e) = \frac{1}{e}$

Step 11.2.2

The final answer is

$\frac{1}{e}$ .

$y = \frac{1}{e}$

Step 12

These are the local extrema for

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

$(e, \frac{1}{e})$ is a local maxima

Step 13