Calculus Examples

$f (x) = x^{2} \ln (x)$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 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 $f (x) = x^{2}$ and $g (x) = \ln (x)$ .

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

Step 1.1.2

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

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

Step 1.1.3

Differentiate using the Power Rule.

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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Step 1.1.3.2.1

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

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

Step 1.1.3.2.2

Cancel the common factors.

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Step 1.1.3.2.2.1

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

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

Step 1.1.3.2.2.2

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

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

Step 1.1.3.2.2.3

Cancel the common factor.

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

Step 1.1.3.2.2.4

Rewrite the expression.

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

Step 1.1.3.2.2.5

Divide $x$ by $1$ .

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

Step 1.1.3.3

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

$x + \ln (x) (2 x)$

Step 1.1.3.4

Reorder terms.

$f' (x) = 2 x \ln (x) + x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $2 x \ln (x) + x$ with respect to $x$ is $\frac{d}{d x} [2 x \ln (x)] + \frac{d}{d x} [x]$ .

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

Step 1.2.2

Evaluate $\frac{d}{d x} [2 x \ln (x)]$ .

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Step 1.2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x \ln (x)$ with respect to $x$ is $2 \frac{d}{d x} [x \ln (x)]$ .

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

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

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

Step 1.2.2.3

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

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

Step 1.2.2.4

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

$2 (x \frac{1}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [x]$

Step 1.2.2.5

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

$2 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [x]$

Step 1.2.2.6

Cancel the common factor of $x$ .

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Step 1.2.2.6.1

Cancel the common factor.

$2 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [x]$

Step 1.2.2.6.2

Rewrite the expression.

$2 (1 + \ln (x) \cdot 1) + \frac{d}{d x} [x]$

Step 1.2.2.7

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

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

Step 1.2.3

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

$2 (1 + \ln (x)) + 1$

Step 1.2.4

Simplify.

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Step 1.2.4.1

Apply the distributive property.

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

Step 1.2.4.2

Combine terms.

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Step 1.2.4.2.1

Multiply $2$ by $1$ .

$2 + 2 \ln (x) + 1$

Step 1.2.4.2.2

Add $2$ and $1$ .

$f'' (x) = 2 \ln (x) + 3$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $2 \ln (x) + 3$ .

$2 \ln (x) + 3$

Step 2

Set the second derivative equal to $0$ then solve the equation $2 \ln (x) + 3 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$2 \ln (x) + 3 = 0$

Step 2.2

Subtract $3$ from both sides of the equation.

$2 \ln (x) = - 3$

Step 2.3

Divide each term in $2 \ln (x) = - 3$ by $2$ and simplify.

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Step 2.3.1

Divide each term in $2 \ln (x) = - 3$ by $2$ .

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

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

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Move the negative in front of the fraction.

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

Step 2.4

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

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

Step 2.5

Rewrite $\ln (x) = - \frac{3}{2}$ 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 $b^{y} = x$ .

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

Step 2.6

Solve for $x$ .

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Step 2.6.1

Rewrite the equation as $x = e^{- \frac{3}{2}}$ .

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

Step 2.6.2

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

$x = \frac{1}{e^{\frac{3}{2}}}$

Step 3

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

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Step 3.1

Substitute $\frac{1}{e^{\frac{3}{2}}}$ in $f (x) = x^{2} \ln (x)$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $\frac{1}{e^{\frac{3}{2}}}$ in the expression.

$f (\frac{1}{e^{\frac{3}{2}}}) = {(\frac{1}{e^{\frac{3}{2}}})}^{2} \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Apply the product rule to $\frac{1}{e^{\frac{3}{2}}}$ .

$f (\frac{1}{e^{\frac{3}{2}}}) = \frac{1^{2}}{{(e^{\frac{3}{2}})}^{2}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2.2

One to any power is one.

$f (\frac{1}{e^{\frac{3}{2}}}) = \frac{1}{{(e^{\frac{3}{2}})}^{2}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2.3

Multiply the exponents in ${(e^{\frac{3}{2}})}^{2}$ .

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Step 3.1.2.3.1

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

$f (\frac{1}{e^{\frac{3}{2}}}) = \frac{1}{e^{\frac{3}{2} \cdot 2}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2.3.2

Cancel the common factor of $2$ .

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Step 3.1.2.3.2.1

Cancel the common factor.

$f (\frac{1}{e^{\frac{3}{2}}}) = \frac{1}{e^{\frac{3}{2} \cdot 2}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2.3.2.2

Rewrite the expression.

$f (\frac{1}{e^{\frac{3}{2}}}) = \frac{1}{e^{3}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 3.1.2.4

Move $e^{\frac{3}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

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

Step 3.1.2.5

Expand $\ln (e^{- \frac{3}{2}})$ by moving $- \frac{3}{2}$ outside the logarithm.

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

Step 3.1.2.6

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

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

Step 3.1.2.7

Multiply $- 1$ by $1$ .

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

Step 3.1.2.8

Multiply $\frac{1}{e^{3}}$ by $\frac{3}{2}$ .

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

Step 3.1.2.9

Move $2$ to the left of $e^{3}$ .

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

Step 3.1.2.10

The final answer is $- \frac{3}{2 e^{3}}$ .

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

Step 3.2

The point found by substituting $\frac{1}{e^{\frac{3}{2}}}$ in $f (x) = x^{2} \ln (x)$ is $(\frac{1}{e^{\frac{3}{2}}}, - \frac{3}{2 e^{3}})$ . This point can be an inflection point.

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

Step 4

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

$(- \infty, \frac{1}{e^{\frac{3}{2}}}) \cup (\frac{1}{e^{\frac{3}{2}}}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{1}{e^{\frac{3}{2}}})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $0.12313016$ in the expression.

$f'' (0.12313016) = 2 \ln (0.12313016) + 3$

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

Simplify $2 \ln (0.12313016)$ by moving $2$ inside the logarithm.

$f'' (0.12313016) = \ln ({0.12313016}^{2}) + 3$

Step 5.2.1.2

Raise $0.12313016$ to the power of $2$ .

$f'' (0.12313016) = \ln (0.01516103) + 3$

Step 5.2.2

The final answer is $\ln (0.01516103) + 3$ .

$\ln (0.01516103) + 3$

Step 5.3

At $0.12313016$ , the second derivative is $- 1.18902654$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{e^{\frac{3}{2}}})$

Decreasing on $(- \infty, \frac{1}{e^{\frac{3}{2}}})$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.32313016$ in the expression.

$f'' (0.32313016) = 2 \ln (0.32313016) + 3$

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

Simplify $2 \ln (0.32313016)$ by moving $2$ inside the logarithm.

$f'' (0.32313016) = \ln ({0.32313016}^{2}) + 3$

Step 6.2.1.2

Raise $0.32313016$ to the power of $2$ .

$f'' (0.32313016) = \ln (0.1044131) + 3$

Step 6.2.2

The final answer is $\ln (0.1044131) + 3$ .

$\ln (0.1044131) + 3$

Step 6.3

At $0.32313016$ , the second derivative is $0.74059987$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ .

Increasing on $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ since $f'' (x) > 0$

Step 7

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

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

Step 8