Calculus Examples

$x^{5} \ln (x)$

Step 1

Write $x^{5} \ln (x)$ as a function.

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

Step 2

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

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{5}$ and $g (x) = \ln (x)$ .

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

Step 2.1.1.2

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

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

Step 2.1.1.3

Differentiate using the Power Rule.

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Step 2.1.1.3.1

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

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

Step 2.1.1.3.2

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

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Step 2.1.1.3.2.1

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

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

Step 2.1.1.3.2.2

Cancel the common factors.

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Step 2.1.1.3.2.2.1

Raise $x$ to the power of $1$ .

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

Step 2.1.1.3.2.2.2

Factor $x$ out of $x^{1}$ .

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

Step 2.1.1.3.2.2.3

Cancel the common factor.

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

Step 2.1.1.3.2.2.4

Rewrite the expression.

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

Step 2.1.1.3.2.2.5

Divide $x^{4}$ by $1$ .

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

Step 2.1.1.3.3

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

$x^{4} + \ln (x) (5 x^{4})$

Step 2.1.1.3.4

Reorder terms.

$f' (x) = x^{4} + 5 x^{4} \ln (x)$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

Differentiate.

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Step 2.1.2.1.1

By the Sum Rule, the derivative of $x^{4} + 5 x^{4} \ln (x)$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [5 x^{4} \ln (x)]$ .

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

Step 2.1.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 2.1.2.2

Evaluate $\frac{d}{d x} [5 x^{4} \ln (x)]$ .

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Step 2.1.2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{4} \ln (x)$ with respect to $x$ is $5 \frac{d}{d x} [x^{4} \ln (x)]$ .

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

Step 2.1.2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{4}$ and $g (x) = \ln (x)$ .

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

Step 2.1.2.2.3

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

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

Step 2.1.2.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + 5 (x^{4} \frac{1}{x} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.5

Combine $x^{4}$ and $\frac{1}{x}$ .

$4 x^{3} + 5 (\frac{x^{4}}{x} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6

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

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Step 2.1.2.2.6.1

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

$4 x^{3} + 5 (\frac{x \cdot x^{3}}{x} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6.2

Cancel the common factors.

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Step 2.1.2.2.6.2.1

Raise $x$ to the power of $1$ .

$4 x^{3} + 5 (\frac{x \cdot x^{3}}{x^{1}} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6.2.2

Factor $x$ out of $x^{1}$ .

$4 x^{3} + 5 (\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6.2.3

Cancel the common factor.

$4 x^{3} + 5 (\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6.2.4

Rewrite the expression.

$4 x^{3} + 5 (\frac{x^{3}}{1} + \ln (x) (4 x^{3}))$

Step 2.1.2.2.6.2.5

Divide $x^{3}$ by $1$ .

$4 x^{3} + 5 (x^{3} + \ln (x) (4 x^{3}))$

Step 2.1.2.3

Simplify.

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Step 2.1.2.3.1

Apply the distributive property.

$4 x^{3} + 5 x^{3} + 5 (\ln (x) (4 x^{3}))$

Step 2.1.2.3.2

Combine terms.

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Step 2.1.2.3.2.1

Multiply $4$ by $5$ .

$4 x^{3} + 5 x^{3} + 20 (\ln (x) (x^{3}))$

Step 2.1.2.3.2.2

Add $4 x^{3}$ and $5 x^{3}$ .

$9 x^{3} + 20 \ln (x) x^{3}$

Step 2.1.2.3.3

Reorder terms.

$f'' (x) = 9 x^{3} + 20 x^{3} \ln (x)$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $9 x^{3} + 20 x^{3} \ln (x)$ .

$9 x^{3} + 20 x^{3} \ln (x)$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $9 x^{3} + 20 x^{3} \ln (x) = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$9 x^{3} + 20 x^{3} \ln (x) = 0$

Step 2.2.2

Subtract $9 x^{3}$ from both sides of the equation.

$20 x^{3} \ln (x) = - 9 x^{3}$

Step 2.2.3

Divide each term in $20 x^{3} \ln (x) = - 9 x^{3}$ by $20 x^{3}$ and simplify.

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Step 2.2.3.1

Divide each term in $20 x^{3} \ln (x) = - 9 x^{3}$ by $20 x^{3}$ .

$\frac{20 x^{3} \ln (x)}{20 x^{3}} = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.2

Simplify the left side.

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Step 2.2.3.2.1

Cancel the common factor of $20$ .

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Step 2.2.3.2.1.1

Cancel the common factor.

$\frac{20 x^{3} \ln (x)}{20 x^{3}} = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.2.1.2

Rewrite the expression.

$\frac{x^{3} \ln (x)}{x^{3}} = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.2.2

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

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Step 2.2.3.2.2.1

Cancel the common factor.

$\frac{x^{3} \ln (x)}{x^{3}} = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.2.2.2

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

$\ln (x) = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.3

Simplify the right side.

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Step 2.2.3.3.1

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

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Step 2.2.3.3.1.1

Cancel the common factor.

$\ln (x) = \frac{- 9 x^{3}}{20 x^{3}}$

Step 2.2.3.3.1.2

Rewrite the expression.

$\ln (x) = \frac{- 9}{20}$

Step 2.2.3.3.2

Move the negative in front of the fraction.

$\ln (x) = - \frac{9}{20}$

Step 2.2.4

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

$e^{\ln (x)} = e^{- \frac{9}{20}}$

Step 2.2.5

Rewrite $\ln (x) = - \frac{9}{20}$ 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 $b^{y} = x$ .

$e^{- \frac{9}{20}} = x$

Step 2.2.6

Solve for $x$ .

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Step 2.2.6.1

Rewrite the equation as $x = e^{- \frac{9}{20}}$ .

$x = e^{- \frac{9}{20}}$

Step 2.2.6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{e^{\frac{9}{20}}}$

Step 3

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

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Step 3.1

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

$x > 0$

Step 3.2

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 4

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

$(0, \frac{1}{e^{\frac{9}{20}}}) \cup (\frac{1}{e^{\frac{9}{20}}}, \infty)$

Step 5

Substitute any number from the interval $(0, \frac{1}{e^{\frac{9}{20}}})$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $0.32$ in the expression.

$f'' (0.32) = 9 {(0.32)}^{3} + 20 {(0.32)}^{3} \ln (0.32)$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise $0.32$ to the power of $3$ .

$f'' (0.32) = 9 \cdot 0.032768 + 20 {(0.32)}^{3} \ln (0.32)$

Step 5.2.1.2

Multiply $9$ by $0.032768$ .

$f'' (0.32) = 0.294912 + 20 {(0.32)}^{3} \ln (0.32)$

Step 5.2.1.3

Raise $0.32$ to the power of $3$ .

$f'' (0.32) = 0.294912 + 20 \cdot (0.032768 \ln (0.32))$

Step 5.2.1.4

Multiply $20$ by $0.032768$ .

$f'' (0.32) = 0.294912 + 0.65536 \ln (0.32)$

Step 5.2.1.5

Simplify $0.65536 \ln (0.32)$ by moving $0.65536$ inside the logarithm.

$f'' (0.32) = 0.294912 + \ln ({0.32}^{0.65536})$

Step 5.2.1.6

Raise $0.32$ to the power of $0.65536$ .

$f'' (0.32) = 0.294912 + \ln (0.47390914)$

Step 5.2.2

Add $0.294912$ and $\ln (0.47390914)$ .

$f'' (0.32) = - 0.45182765$

Step 5.2.3

The final answer is $- 0.45182765$ .

$- 0.45182765$

Step 5.3

The graph is concave down on the interval $(0, \frac{1}{e^{\frac{9}{20}}})$ because $f'' (0.32)$ is negative.

Concave down on $(0, \frac{1}{e^{\frac{9}{20}}})$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

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

$f'' (3) = 9 {(3)}^{3} + 20 {(3)}^{3} \ln (3)$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

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

$f'' (3) = 9 \cdot 27 + 20 {(3)}^{3} \ln (3)$

Step 6.2.1.2

Multiply $9$ by $27$ .

$f'' (3) = 243 + 20 {(3)}^{3} \ln (3)$

Step 6.2.1.3

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

$f'' (3) = 243 + 20 \cdot (27 \ln (3))$

Step 6.2.1.4

Multiply $20$ by $27$ .

$f'' (3) = 243 + 540 \ln (3)$

Step 6.2.1.5

Simplify $540 \ln (3)$ by moving $540$ inside the logarithm.

$f'' (3) = 243 + \ln (3^{540})$

Step 6.2.2

The final answer is $243 + \ln (3^{540})$ .

$243 + \ln (3^{540})$

Step 6.3

The graph is concave up on the interval $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ because $f'' (3)$ is positive.

Concave up on $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ since $f'' (x)$ is positive

Step 7

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

Concave down on $(0, \frac{1}{e^{\frac{9}{20}}})$ since $f'' (x)$ is negative

Concave up on $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ since $f'' (x)$ is positive

Step 8