Calculus Examples

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

Step 1

Find the derivative.

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

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

Step 1.2

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 $g (x) = x^{2}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

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

Step 1.2.2

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

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

Step 1.2.3

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

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

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

Step 1.3.2

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

Cancel the common factors.

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Step 1.3.2.3.1

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

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

Step 1.3.2.3.2

Cancel the common factor.

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

Step 1.3.2.3.3

Rewrite the expression.

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

Step 1.3.3

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

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

Step 1.3.4

Simplify terms.

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Step 1.3.4.1

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

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

Step 1.3.4.2

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

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

Step 1.3.4.3

Cancel the common factor of $x$ .

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Step 1.3.4.3.1

Cancel the common factor.

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

Step 1.3.4.3.2

Divide $2$ by $1$ .

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $\ln (x^{2})$ by $1$ .

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

Step 2

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

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Step 2.1

Subtract $2$ from both sides of the equation.

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

Step 2.2

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

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

Step 2.3

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

$e^{- 2} = x^{2}$

Step 2.4

Solve for $x$ .

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Step 2.4.1

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

$x^{2} = e^{- 2}$

Step 2.4.2

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

$x = \pm \sqrt{e^{- 2}}$

Step 2.4.3

Simplify $\pm \sqrt{e^{- 2}}$ .

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Step 2.4.3.1

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

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

Step 2.4.3.2

Rewrite $\sqrt{\frac{1}{e^{2}}}$ as $\frac{\sqrt{1}}{\sqrt{e^{2}}}$ .

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

Step 2.4.3.3

Any root of $1$ is $1$ .

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

Step 2.4.3.4

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

$x = \pm \frac{1}{e}$

Step 2.4.4

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

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Step 2.4.4.1

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

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

Step 2.4.4.2

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

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

Step 2.4.4.3

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

$x = \frac{1}{e}, - \frac{1}{e}$

Step 3

Solve the original function $f (x) = x \ln (x^{2})$ at $x = \frac{1}{e}$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Rewrite $\ln (1^{2} e^{- 2})$ as $\ln (1^{2}) + \ln (e^{- 2})$ .

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

Step 3.2.4

Use logarithm rules to move $- 2$ out of the exponent.

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

Step 3.2.5

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

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

Step 3.2.6

Multiply $- 2$ by $1$ .

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

Step 3.2.7

Simplify each term.

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Step 3.2.7.1

Expand $\ln (1^{2})$ by moving $2$ outside the logarithm.

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

Step 3.2.7.2

The natural logarithm of $1$ is $0$ .

$f (\frac{1}{e}) = \frac{1}{e} \cdot (2 \cdot 0 - 2)$

Step 3.2.7.3

Multiply $2$ by $0$ .

$f (\frac{1}{e}) = \frac{1}{e} \cdot (0 - 2)$

Step 3.2.8

Combine fractions.

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Step 3.2.8.1

Subtract $2$ from $0$ .

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

Step 3.2.8.2

Combine $\frac{1}{e}$ and $- 2$ .

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

Step 3.2.8.3

Move the negative in front of the fraction.

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

Step 3.2.9

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

$- \frac{2}{e}$

Step 4

Solve the original function $f (x) = x \ln (x^{2})$ at $x = - \frac{1}{e}$ .

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Step 4.1

Replace the variable $x$ with $- \frac{1}{e}$ in the expression.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.2.1.1

Apply the product rule to $- \frac{1}{e}$ .

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

Step 4.2.1.2

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

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

Step 4.2.2

Raise $- 1$ to the power of $2$ .

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Rewrite $\ln (1^{2} e^{- 2})$ as $\ln (1^{2}) + \ln (e^{- 2})$ .

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

Step 4.2.6

Use logarithm rules to move $- 2$ out of the exponent.

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

Step 4.2.7

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

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

Step 4.2.8

Multiply $- 2$ by $1$ .

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

Step 4.2.9

Simplify each term.

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Step 4.2.9.1

Expand $\ln (1^{2})$ by moving $2$ outside the logarithm.

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

Step 4.2.9.2

The natural logarithm of $1$ is $0$ .

$f (- \frac{1}{e}) = - \frac{1}{e} \cdot (2 \cdot 0 - 2)$

Step 4.2.9.3

Multiply $2$ by $0$ .

$f (- \frac{1}{e}) = - \frac{1}{e} \cdot (0 - 2)$

Step 4.2.10

Subtract $2$ from $0$ .

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

Step 4.2.11

Multiply $- \frac{1}{e} \cdot - 2$ .

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Step 4.2.11.1

Multiply $- 2$ by $- 1$ .

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

Step 4.2.11.2

Combine $2$ and $\frac{1}{e}$ .

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

Step 4.2.12

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

$\frac{2}{e}$

Step 5

The horizontal tangent lines on function $f (x) = x \ln (x^{2})$ are $y = - \frac{2}{e}, y = \frac{2}{e}$ .

$y = - \frac{2}{e}, y = \frac{2}{e}$

Step 6