Calculus Examples

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

Step 1

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

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 2.1.2

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 2.1.3

Differentiate using the Power Rule.

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

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Step 2.1.3.2.1

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

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

Step 2.1.3.2.2

Cancel the common factors.

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Step 2.1.3.2.2.1

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

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

Step 2.1.3.2.2.2

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

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

Step 2.1.3.2.2.3

Cancel the common factor.

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

Step 2.1.3.2.2.4

Rewrite the expression.

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

Step 2.1.3.2.2.5

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

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

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 $n = 5$ .

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

Step 2.1.3.4

Reorder terms.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

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 2.2.1.2

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 2.2.2

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

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Step 2.2.2.1

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 2.2.2.2

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 2.2.2.3

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 2.2.2.4

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 2.2.2.5

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

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

Step 2.2.2.6

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

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Step 2.2.2.6.1

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

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

Step 2.2.2.6.2

Cancel the common factors.

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Step 2.2.2.6.2.1

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

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

Step 2.2.2.6.2.2

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

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

Step 2.2.2.6.2.3

Cancel the common factor.

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

Step 2.2.2.6.2.4

Rewrite the expression.

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

Step 2.2.2.6.2.5

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

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Apply the distributive property.

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

Step 2.2.3.2

Combine terms.

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Step 2.2.3.2.1

Multiply $4$ by $5$ .

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

Step 2.2.3.2.2

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

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

Step 2.2.3.3

Reorder terms.

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

Step 2.3

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 3

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 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $20$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

Cancel the common factor.

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

Step 3.3.2.2.2

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

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

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

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Step 3.3.3.1.1

Cancel the common factor.

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

Step 3.3.3.1.2

Rewrite the expression.

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

Step 3.3.3.2

Move the negative in front of the fraction.

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

Step 3.4

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

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

Step 3.5

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 3.6

Solve for $x$ .

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Step 3.6.1

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

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

Step 3.6.2

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

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

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $\frac{1}{e^{\frac{9}{20}}}$ in $f (x) = x^{5} \ln (x)$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\frac{1}{e^{\frac{9}{20}}}$ in the expression.

$f (\frac{1}{e^{\frac{9}{20}}}) = {(\frac{1}{e^{\frac{9}{20}}})}^{5} \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Apply the product rule to $\frac{1}{e^{\frac{9}{20}}}$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1^{5}}{{(e^{\frac{9}{20}})}^{5}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.2

One to any power is one.

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{{(e^{\frac{9}{20}})}^{5}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.3

Multiply the exponents in ${(e^{\frac{9}{20}})}^{5}$ .

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Step 4.1.2.3.1

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

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{20} \cdot 5}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.3.2

Cancel the common factor of $5$ .

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Step 4.1.2.3.2.1

Factor $5$ out of $20$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{5 (4)} \cdot 5}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.3.2.2

Cancel the common factor.

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{5 \cdot 4} \cdot 5}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.3.2.3

Rewrite the expression.

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{4}}} \cdot \ln (\frac{1}{e^{\frac{9}{20}}})$

Step 4.1.2.4

Move $e^{\frac{9}{20}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{4}}} \cdot \ln (e^{- \frac{9}{20}})$

Step 4.1.2.5

Expand $\ln (e^{- \frac{9}{20}})$ by moving $- \frac{9}{20}$ outside the logarithm.

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{4}}} \cdot (- \frac{9}{20} \cdot \ln (e))$

Step 4.1.2.6

The natural logarithm of $e$ is $1$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{4}}} \cdot (- \frac{9}{20} \cdot 1)$

Step 4.1.2.7

Multiply $- 1$ by $1$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = \frac{1}{e^{\frac{9}{4}}} \cdot (- \frac{9}{20})$

Step 4.1.2.8

Multiply $\frac{1}{e^{\frac{9}{4}}}$ by $\frac{9}{20}$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = - \frac{9}{e^{\frac{9}{4}} \cdot 20}$

Step 4.1.2.9

Move $20$ to the left of $e^{\frac{9}{4}}$ .

$f (\frac{1}{e^{\frac{9}{20}}}) = - \frac{9}{20 e^{\frac{9}{4}}}$

Step 4.1.2.10

The final answer is $- \frac{9}{20 e^{\frac{9}{4}}}$ .

$- \frac{9}{20 e^{\frac{9}{4}}}$

Step 4.2

The point found by substituting $\frac{1}{e^{\frac{9}{20}}}$ in $f (x) = x^{5} \ln (x)$ is $(\frac{1}{e^{\frac{9}{20}}}, - \frac{9}{20 e^{\frac{9}{4}}})$ . This point can be an inflection point.

$(\frac{1}{e^{\frac{9}{20}}}, - \frac{9}{20 e^{\frac{9}{4}}})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 6

Substitute a value from the interval $(- \infty, \frac{1}{e^{\frac{9}{20}}})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.53762815$ in the expression.

$f'' (0.53762815) = 9 {(0.53762815)}^{3} + 20 {(0.53762815)}^{3} \ln (0.53762815)$

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 $0.53762815$ to the power of $3$ .

$f'' (0.53762815) = 9 \cdot 0.1553982 + 20 {(0.53762815)}^{3} \ln (0.53762815)$

Step 6.2.1.2

Multiply $9$ by $0.1553982$ .

$f'' (0.53762815) = 1.39858386 + 20 {(0.53762815)}^{3} \ln (0.53762815)$

Step 6.2.1.3

Raise $0.53762815$ to the power of $3$ .

$f'' (0.53762815) = 1.39858386 + 20 \cdot (0.1553982 \ln (0.53762815))$

Step 6.2.1.4

Multiply $20$ by $0.1553982$ .

$f'' (0.53762815) = 1.39858386 + 3.10796414 \ln (0.53762815)$

Step 6.2.1.5

Simplify $3.10796414 \ln (0.53762815)$ by moving $3.10796414$ inside the logarithm.

$f'' (0.53762815) = 1.39858386 + \ln ({0.53762815}^{3.10796414})$

Step 6.2.1.6

Raise $0.53762815$ to the power of $3.10796414$ .

$f'' (0.53762815) = 1.39858386 + \ln (0.14532747)$

Step 6.2.2

Add $1.39858386$ and $\ln (0.14532747)$ .

$f'' (0.53762815) = - 0.53018177$

Step 6.2.3

The final answer is $- 0.53018177$ .

$- 0.53018177$

Step 6.3

At $0.53762815$ , the second derivative is $- 0.53018177$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{e^{\frac{9}{20}}})$

Decreasing on $(- \infty, \frac{1}{e^{\frac{9}{20}}})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.73762815$ in the expression.

$f'' (0.73762815) = 9 {(0.73762815)}^{3} + 20 {(0.73762815)}^{3} \ln (0.73762815)$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $0.73762815$ to the power of $3$ .

$f'' (0.73762815) = 9 \cdot 0.40134 + 20 {(0.73762815)}^{3} \ln (0.73762815)$

Step 7.2.1.2

Multiply $9$ by $0.40134$ .

$f'' (0.73762815) = 3.61206002 + 20 {(0.73762815)}^{3} \ln (0.73762815)$

Step 7.2.1.3

Raise $0.73762815$ to the power of $3$ .

$f'' (0.73762815) = 3.61206002 + 20 \cdot (0.40134 \ln (0.73762815))$

Step 7.2.1.4

Multiply $20$ by $0.40134$ .

$f'' (0.73762815) = 3.61206002 + 8.02680006 \ln (0.73762815)$

Step 7.2.1.5

Simplify $8.02680006 \ln (0.73762815)$ by moving $8.02680006$ inside the logarithm.

$f'' (0.73762815) = 3.61206002 + \ln ({0.73762815}^{8.02680006})$

Step 7.2.1.6

Raise $0.73762815$ to the power of $8.02680006$ .

$f'' (0.73762815) = 3.61206002 + \ln (0.08692764)$

Step 7.2.2

Add $3.61206002$ and $\ln (0.08692764)$ .

$f'' (0.73762815) = 1.16938082$

Step 7.2.3

The final answer is $1.16938082$ .

$1.16938082$

Step 7.3

At $0.73762815$ , the second derivative is $1.16938082$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ .

Increasing on $(\frac{1}{e^{\frac{9}{20}}}, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(\frac{1}{e^{\frac{9}{20}}}, - \frac{9}{20 e^{\frac{9}{4}}})$ .

$(\frac{1}{e^{\frac{9}{20}}}, - \frac{9}{20 e^{\frac{9}{4}}})$

Step 9