Calculus Examples

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

Step 1

Write $x^{2} \ln (x)$ as a function.

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

Step 2

Find the first 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 $f (x) = x^{2}$ and $g (x) = \ln (x)$ .

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

Step 2.1.2

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 2.1.3

Differentiate using the Power Rule.

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

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Step 2.1.3.2.1

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

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

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 $1$ .

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

Step 2.1.3.2.2.2

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

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

Step 2.1.3.2.2.3

Cancel the common factor.

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

Step 2.1.3.2.2.4

Rewrite the expression.

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

Step 2.1.3.2.2.5

Divide $x$ by $1$ .

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

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 $n = 2$ .

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

Step 2.1.3.4

Reorder terms.

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

Step 2.2

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

$2 x \ln (x) + x$

Step 3

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

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Step 3.1

Set the first derivative equal to $0$ .

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

Step 3.2

Subtract $x$ from both sides of the equation.

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

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

Step 3.3.2.2

Cancel the common factor of $x$ .

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Step 3.3.2.2.1

Cancel the common factor.

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

Step 3.3.2.2.2

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

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Cancel the common factor of $x$ .

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Step 3.3.3.1.1

Cancel the common factor.

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

Step 3.3.3.1.2

Rewrite the expression.

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

Step 3.3.3.2

Move the negative in front of the fraction.

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

Step 3.4

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

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

Step 3.5

Rewrite $\ln (x) = - \frac{1}{2}$ 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 $b^{y} = x$ .

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

Step 3.6

Solve for $x$ .

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Step 3.6.1

Rewrite the equation as $x = e^{- \frac{1}{2}}$ .

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

Step 3.6.2

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

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

Step 4

The values which make the derivative equal to $0$ are $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{1}{e^{\frac{1}{2}}}$

Step 5

Find where the derivative is undefined.

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Step 5.1

Set the argument in $\ln (x)$ less than or equal to $0$ to find where the expression is undefined.

$x \leq 0$

Step 5.2

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 6

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, 0) \cup (0, \frac{1}{e^{\frac{1}{2}}}) \cup (\frac{1}{e^{\frac{1}{2}}}, \infty)$

Step 7

Exclude the intervals that are not in the domain.

$(0, \frac{1}{e^{\frac{1}{2}}})$

Step 8

Substitute a value from the interval $(0, \frac{1}{e^{\frac{1}{2}}})$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $0.30326533$ in the expression.

$f' (0.30326533) = 2 (0.30326533) \ln (0.30326533) + 0.30326533$

Step 8.2

Simplify the result.

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Step 8.2.1

Remove parentheses.

$f' (0.30326533) = 2 (0.30326533) \ln (0.30326533) + 0.30326533$

Step 8.2.2

Simplify each term.

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Step 8.2.2.1

Multiply $2$ by $0.30326533$ .

$f' (0.30326533) = 0.60653067 \ln (0.30326533) + 0.30326533$

Step 8.2.2.2

Simplify $0.60653067 \ln (0.30326533)$ by moving $0.60653067$ inside the logarithm.

$f' (0.30326533) = \ln ({0.30326533}^{0.60653067}) + 0.30326533$

Step 8.2.2.3

Raise $0.30326533$ to the power of $0.60653067$ .

$f' (0.30326533) = \ln (0.48496413) + 0.30326533$

Step 8.2.3

Add $\ln (0.48496413)$ and $0.30326533$ .

$f' (0.30326533) = - 0.42041501$

Step 8.2.4

The final answer is $- 0.42041501$ .

$- 0.42041501$

Step 8.3

At $x = 0.30326533$ the derivative is $- 0.42041501$ . Since this is negative, the function is decreasing on $(0, \frac{1}{e^{\frac{1}{2}}})$ .

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

Step 9

Exclude the intervals that are not in the domain.

$(0, \frac{1}{e^{\frac{1}{2}}}), (\frac{1}{e^{\frac{1}{2}}}, \infty)$

Step 10

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

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Step 10.1

Replace the variable $x$ with $1.60653067$ in the expression.

$f' (1.60653067) = 2 (1.60653067) \ln (1.60653067) + 1.60653067$

Step 10.2

Simplify the result.

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Step 10.2.1

Remove parentheses.

$f' (1.60653067) = 2 (1.60653067) \ln (1.60653067) + 1.60653067$

Step 10.2.2

Simplify each term.

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Step 10.2.2.1

Multiply $2$ by $1.60653067$ .

$f' (1.60653067) = 3.21306134 \ln (1.60653067) + 1.60653067$

Step 10.2.2.2

Simplify $3.21306134 \ln (1.60653067)$ by moving $3.21306134$ inside the logarithm.

$f' (1.60653067) = \ln ({1.60653067}^{3.21306134}) + 1.60653067$

Step 10.2.2.3

Raise $1.60653067$ to the power of $3.21306134$ .

$f' (1.60653067) = \ln (4.58705612) + 1.60653067$

Step 10.2.3

Add $\ln (4.58705612)$ and $1.60653067$ .

$f' (1.60653067) = 3.12976912$

Step 10.2.4

The final answer is $3.12976912$ .

$3.12976912$

Step 10.3

At $x = 1.60653067$ the derivative is $3.12976912$ . Since this is positive, the function is increasing on $(\frac{1}{e^{\frac{1}{2}}}, \infty)$ .

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

Step 11

List the intervals on which the function is increasing and decreasing.

Increasing on: $(\frac{1}{e^{\frac{1}{2}}}, \infty)$

Decreasing on: $(0, \frac{1}{e^{\frac{1}{2}}})$

Step 12