Calculus Examples

$f (x) = \frac{(5 \ln (x))}{x}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

Tap for more steps...

Step 1.1

Find the second derivative.

Tap for more steps...

Step 1.1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1.1

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

$5 \frac{d}{d x} [\frac{\ln (x)}{x}]$

Step 1.1.1.2

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 $g (x) = x$ .

$5 \frac{x \frac{d}{d x} [\ln (x)] - \ln (x) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.1.3

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

$5 \frac{x \frac{1}{x} - \ln (x) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.1.4

Differentiate using the Power Rule.

Tap for more steps...

Step 1.1.1.4.1

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

$5 \frac{\frac{x}{x} - \ln (x) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.1.4.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.1.4.2.1

Cancel the common factor.

$5 \frac{\frac{x}{x} - \ln (x) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.1.4.2.2

Rewrite the expression.

$5 \frac{1 - \ln (x) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.1.4.3

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

$5 \frac{1 - \ln (x) \cdot 1}{x^{2}}$

Step 1.1.1.4.4

Combine fractions.

Tap for more steps...

Step 1.1.1.4.4.1

Multiply $- 1$ by $1$ .

$5 \frac{1 - \ln (x)}{x^{2}}$

Step 1.1.1.4.4.2

Combine $5$ and $\frac{1 - \ln (x)}{x^{2}}$ .

$\frac{5 (1 - \ln (x))}{x^{2}}$

Step 1.1.1.5

Simplify.

Tap for more steps...

Step 1.1.1.5.1

Apply the distributive property.

$\frac{5 \cdot 1 + 5 (- \ln (x))}{x^{2}}$

Step 1.1.1.5.2

Simplify each term.

Tap for more steps...

Step 1.1.1.5.2.1

Multiply $5$ by $1$ .

$\frac{5 + 5 (- \ln (x))}{x^{2}}$

Step 1.1.1.5.2.2

Multiply $5 (- \ln (x))$ .

Tap for more steps...

Step 1.1.1.5.2.2.1

Multiply $- 1$ by $5$ .

$\frac{5 - 5 \ln (x)}{x^{2}}$

Step 1.1.1.5.2.2.2

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

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

Step 1.1.2

Find the second derivative.

Tap for more steps...

Step 1.1.2.1

$\frac{x^{2} \frac{d}{d x} [5 - \ln (x^{5})] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{{(x^{2})}^{2}}$

Step 1.1.2.2

Differentiate.

Tap for more steps...

Step 1.1.2.2.1

Multiply the exponents in ${(x^{2})}^{2}$ .

Tap for more steps...

Step 1.1.2.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{x^{2} \frac{d}{d x} [5 - \ln (x^{5})] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{2 \cdot 2}}$

Step 1.1.2.2.1.2

Multiply $2$ by $2$ .

$\frac{x^{2} \frac{d}{d x} [5 - \ln (x^{5})] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.2.2

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

$\frac{x^{2} (\frac{d}{d x} [5] + \frac{d}{d x} [- \ln (x^{5})]) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.2.3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{x^{2} (0 + \frac{d}{d x} [- \ln (x^{5})]) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.2.4

Add $0$ and $\frac{d}{d x} [- \ln (x^{5})]$ .

$\frac{x^{2} \frac{d}{d x} [- \ln (x^{5})] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.2.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- \ln (x^{5})$ with respect to $x$ is $- \frac{d}{d x} [\ln (x^{5})]$ .

$\frac{x^{2} (- \frac{d}{d x} [\ln (x^{5})]) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.3

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 $g (x) = x^{5}$ .

Tap for more steps...

Step 1.1.2.3.1

To apply the Chain Rule, set $u$ as $x^{5}$ .

$\frac{x^{2} (- (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{5}])) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.3.2

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

$\frac{x^{2} (- (\frac{1}{u} \frac{d}{d x} [x^{5}])) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.3.3

Replace all occurrences of $u$ with $x^{5}$ .

$\frac{x^{2} (- (\frac{1}{x^{5}} \frac{d}{d x} [x^{5}])) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4

Differentiate using the Power Rule.

Tap for more steps...

Step 1.1.2.4.1

Combine $x^{2}$ and $\frac{1}{x^{5}}$ .

$\frac{- \frac{x^{2}}{x^{5}} \frac{d}{d x} [x^{5}] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.2

Cancel the common factor of $x^{2}$ and $x^{5}$ .

Tap for more steps...

Step 1.1.2.4.2.1

Multiply by $1$ .

$\frac{- \frac{x^{2} \cdot 1}{x^{5}} \frac{d}{d x} [x^{5}] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.2.2

Cancel the common factors.

Tap for more steps...

Step 1.1.2.4.2.2.1

Factor $x^{2}$ out of $x^{5}$ .

$\frac{- \frac{x^{2} \cdot 1}{x^{2} x^{3}} \frac{d}{d x} [x^{5}] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.2.2.2

Cancel the common factor.

$\frac{- \frac{x^{2} \cdot 1}{x^{2} x^{3}} \frac{d}{d x} [x^{5}] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.2.2.3

Rewrite the expression.

$\frac{- \frac{1}{x^{3}} \frac{d}{d x} [x^{5}] - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{- \frac{1}{x^{3}} (5 x^{4}) - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4

Simplify terms.

Tap for more steps...

Step 1.1.2.4.4.1

Multiply $5$ by $- 1$ .

$\frac{- 5 \frac{1}{x^{3}} x^{4} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.2

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

$\frac{\frac{- 5}{x^{3}} x^{4} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.3

Combine $\frac{- 5}{x^{3}}$ and $x^{4}$ .

$\frac{\frac{- 5 x^{4}}{x^{3}} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.4

Cancel the common factor of $x^{4}$ and $x^{3}$ .

Tap for more steps...

Step 1.1.2.4.4.4.1

Factor $x^{3}$ out of $- 5 x^{4}$ .

$\frac{\frac{x^{3} (- 5 x)}{x^{3}} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.4.2

Cancel the common factors.

Tap for more steps...

Step 1.1.2.4.4.4.2.1

Multiply by $1$ .

$\frac{\frac{x^{3} (- 5 x)}{x^{3} \cdot 1} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.4.2.2

Cancel the common factor.

$\frac{\frac{x^{3} (- 5 x)}{x^{3} \cdot 1} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.4.2.3

Rewrite the expression.

$\frac{\frac{- 5 x}{1} - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.4.4.2.4

Divide $- 5 x$ by $1$ .

$\frac{- 5 x - (5 - \ln (x^{5})) \frac{d}{d x} [x^{2}]}{x^{4}}$

Step 1.1.2.4.5

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

$\frac{- 5 x - (5 - \ln (x^{5})) (2 x)}{x^{4}}$

Step 1.1.2.4.6

Simplify with factoring out.

Tap for more steps...

Step 1.1.2.4.6.1

Multiply $2$ by $- 1$ .

$\frac{- 5 x - 2 (5 - \ln (x^{5})) x}{x^{4}}$

Step 1.1.2.4.6.2

Factor $x$ out of $- 5 x - 2 (5 - \ln (x^{5})) x$ .

Tap for more steps...

Step 1.1.2.4.6.2.1

Factor $x$ out of $- 5 x$ .

$\frac{x \cdot - 5 - 2 (5 - \ln (x^{5})) x}{x^{4}}$

Step 1.1.2.4.6.2.2

Factor $x$ out of $- 2 (5 - \ln (x^{5})) x$ .

$\frac{x \cdot - 5 + x (- 2 (5 - \ln (x^{5})))}{x^{4}}$

Step 1.1.2.4.6.2.3

Factor $x$ out of $x \cdot - 5 + x (- 2 (5 - \ln (x^{5})))$ .

$\frac{x (- 5 - 2 (5 - \ln (x^{5})))}{x^{4}}$

Step 1.1.2.5

Cancel the common factors.

Tap for more steps...

Step 1.1.2.5.1

Factor $x$ out of $x^{4}$ .

$\frac{x (- 5 - 2 (5 - \ln (x^{5})))}{x \cdot x^{3}}$

Step 1.1.2.5.2

Cancel the common factor.

$\frac{x (- 5 - 2 (5 - \ln (x^{5})))}{x \cdot x^{3}}$

Step 1.1.2.5.3

Rewrite the expression.

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

Step 1.1.2.6

Simplify.

Tap for more steps...

Step 1.1.2.6.1

Apply the distributive property.

$\frac{- 5 - 2 \cdot 5 - 2 (- \ln (x^{5}))}{x^{3}}$

Step 1.1.2.6.2

Simplify the numerator.

Tap for more steps...

Step 1.1.2.6.2.1

Simplify each term.

Tap for more steps...

Step 1.1.2.6.2.1.1

Multiply $- 2$ by $5$ .

$\frac{- 5 - 10 - 2 (- \ln (x^{5}))}{x^{3}}$

Step 1.1.2.6.2.1.2

Multiply $- 2 (- \ln (x^{5}))$ .

Tap for more steps...

Step 1.1.2.6.2.1.2.1

Multiply $- 1$ by $- 2$ .

$\frac{- 5 - 10 + 2 \ln (x^{5})}{x^{3}}$

Step 1.1.2.6.2.1.2.2

Simplify $2 \ln (x^{5})$ by moving $2$ inside the logarithm.

$\frac{- 5 - 10 + \ln ({(x^{5})}^{2})}{x^{3}}$

Step 1.1.2.6.2.1.3

Multiply the exponents in ${(x^{5})}^{2}$ .

Tap for more steps...

Step 1.1.2.6.2.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{- 5 - 10 + \ln (x^{5 \cdot 2})}{x^{3}}$

Step 1.1.2.6.2.1.3.2

Multiply $5$ by $2$ .

$\frac{- 5 - 10 + \ln (x^{10})}{x^{3}}$

Step 1.1.2.6.2.2

Subtract $10$ from $- 5$ .

$\frac{- 15 + \ln (x^{10})}{x^{3}}$

Step 1.1.2.6.3

Rewrite $- 15$ as $- 1 (15)$ .

$\frac{- 1 (15) + \ln (x^{10})}{x^{3}}$

Step 1.1.2.6.4

Factor $- 1$ out of $\ln (x^{10})$ .

$\frac{- 1 (15) - 1 (- \ln (x^{10}))}{x^{3}}$

Step 1.1.2.6.5

Factor $- 1$ out of $- 1 (15) - 1 (- \ln (x^{10}))$ .

$\frac{- 1 (15 - \ln (x^{10}))}{x^{3}}$

Step 1.1.2.6.6

Move the negative in front of the fraction.

$f'' (x) = - \frac{15 - \ln (x^{10})}{x^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{15 - \ln (x^{10})}{x^{3}}$ .

$- \frac{15 - \ln (x^{10})}{x^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{15 - \ln (x^{10})}{x^{3}} = 0$ .

Tap for more steps...

Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{15 - \ln (x^{10})}{x^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$15 - \ln (x^{10}) = 0$

Step 1.2.3

Solve the equation for $x$ .

Tap for more steps...

Step 1.2.3.1

Subtract $15$ from both sides of the equation.

$- \ln (x^{10}) = - 15$

Step 1.2.3.2

Divide each term in $- \ln (x^{10}) = - 15$ by $- 1$ and simplify.

Tap for more steps...

Step 1.2.3.2.1

Divide each term in $- \ln (x^{10}) = - 15$ by $- 1$ .

$\frac{- \ln (x^{10})}{- 1} = \frac{- 15}{- 1}$

Step 1.2.3.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (x^{10})}{1} = \frac{- 15}{- 1}$

Step 1.2.3.2.2.2

Divide $\ln (x^{10})$ by $1$ .

$\ln (x^{10}) = \frac{- 15}{- 1}$

Step 1.2.3.2.3

Simplify the right side.

Tap for more steps...

Step 1.2.3.2.3.1

Divide $- 15$ by $- 1$ .

$\ln (x^{10}) = 15$

Step 1.2.3.3

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

$e^{\ln (x^{10})} = e^{15}$

Step 1.2.3.4

Rewrite $\ln (x^{10}) = 15$ 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 $b^{y} = x$ .

$e^{15} = x^{10}$

Step 1.2.3.5

Solve for $x$ .

Tap for more steps...

Step 1.2.3.5.1

Rewrite the equation as $x^{10} = e^{15}$ .

$x^{10} = e^{15}$

Step 1.2.3.5.2

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

$x = \pm \sqrt[10]{e^{15}}$

Step 1.2.3.5.3

Simplify $\pm \sqrt[10]{e^{15}}$ .

Tap for more steps...

Step 1.2.3.5.3.1

Factor out $e^{10}$ .

$x = \pm \sqrt[10]{e^{10} e^{5}}$

Step 1.2.3.5.3.2

Pull terms out from under the radical.

$x = \pm e \sqrt[10]{e^{5}}$

Step 1.2.3.5.3.3

Rewrite $\sqrt[10]{e^{5}}$ as $\sqrt{\sqrt[5]{e^{5}}}$ .

$x = \pm e \sqrt{\sqrt[5]{e^{5}}}$

Step 1.2.3.5.3.4

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

$x = \pm e \sqrt{e}$

Step 1.2.3.5.4

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

Tap for more steps...

Step 1.2.3.5.4.1

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

$x = e \sqrt{e}$

Step 1.2.3.5.4.2

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

$x = - e \sqrt{e}$

Step 1.2.3.5.4.3

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

$x = e \sqrt{e}, - e \sqrt{e}$

Step 2

Find the domain of $f (x) = \frac{5 \ln (x)}{x}$ .

Tap for more steps...

Step 2.1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 2.2

Set the denominator in $\frac{5 \ln (x)}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 2.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(0, e \sqrt{e}) \cup (e \sqrt{e}, \infty)$

Step 4

Substitute any number from the interval $(0, e \sqrt{e})$ into the second derivative and evaluate to determine the concavity.

Tap for more steps...

Step 4.1

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

$f'' (2) = - \frac{15 - \ln ({(2)}^{10})}{{(2)}^{3}}$

Step 4.2

Simplify the result.

Tap for more steps...

Step 4.2.1

Raise $2$ to the power of $10$ .

$f'' (2) = - \frac{15 - \ln (1024)}{2^{3}}$

Step 4.2.2

Raise $2$ to the power of $3$ .

$f'' (2) = - \frac{15 - \ln (1024)}{8}$

Step 4.2.3

The final answer is $- \frac{15 - \ln (1024)}{8}$ .

$- \frac{15 - \ln (1024)}{8}$

Step 4.3

The graph is concave down on the interval $(0, e \sqrt{e})$ because $f'' (2)$ is negative.

Concave down on $(0, e \sqrt{e})$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(e \sqrt{e}, \infty)$ into the second derivative and evaluate to determine the concavity.

Tap for more steps...

Step 5.1

Replace the variable $x$ with $7$ in the expression.

$f'' (7) = - \frac{15 - \ln ({(7)}^{10})}{{(7)}^{3}}$

Step 5.2

Simplify the result.

Tap for more steps...

Step 5.2.1

Raise $7$ to the power of $10$ .

$f'' (7) = - \frac{15 - \ln (282475249)}{7^{3}}$

Step 5.2.2

Raise $7$ to the power of $3$ .

$f'' (7) = - \frac{15 - \ln (282475249)}{343}$

Step 5.2.3

The final answer is $- \frac{15 - \ln (282475249)}{343}$ .

$- \frac{15 - \ln (282475249)}{343}$

Step 5.3

The graph is concave up on the interval $(e \sqrt{e}, \infty)$ because $f'' (7)$ is positive.

Concave up on $(e \sqrt{e}, \infty)$ since $f'' (x)$ is positive

Step 6

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(0, e \sqrt{e})$ since $f'' (x)$ is negative

Concave up on $(e \sqrt{e}, \infty)$ since $f'' (x)$ is positive

Step 7