Calculus Examples

$f (x) = x^{7} \ln (x)$

Step 1

Find the first 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 $f (x) = x^{7}$ and $g (x) = \ln (x)$ .

$x^{7} \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.2

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

$x^{7} \frac{1}{x} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3

Differentiate using the Power Rule.

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Step 1.1.3.1

Combine $x^{7}$ and $\frac{1}{x}$ .

$\frac{x^{7}}{x} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3.2

Cancel the common factor of $x^{7}$ and $x$ .

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Step 1.1.3.2.1

Factor $x$ out of $x^{7}$ .

$\frac{x \cdot x^{6}}{x} + \ln (x) \frac{d}{d x} [x^{7}]$

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 $1$ .

$\frac{x \cdot x^{6}}{x^{1}} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3.2.2.2

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

$\frac{x \cdot x^{6}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3.2.2.3

Cancel the common factor.

$\frac{x \cdot x^{6}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3.2.2.4

Rewrite the expression.

$\frac{x^{6}}{1} + \ln (x) \frac{d}{d x} [x^{7}]$

Step 1.1.3.2.2.5

Divide $x^{6}$ by $1$ .

$x^{6} + \ln (x) \frac{d}{d x} [x^{7}]$

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 $n = 7$ .

$x^{6} + \ln (x) (7 x^{6})$

Step 1.1.3.4

Reorder terms.

$f' (x) = x^{6} + 7 x^{6} \ln (x)$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $x^{6} + 7 x^{6} \ln (x)$ .

$x^{6} + 7 x^{6} \ln (x)$

Step 2

Set the first derivative equal to $0$ then solve the equation $x^{6} + 7 x^{6} \ln (x) = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$x^{6} + 7 x^{6} \ln (x) = 0$

Step 2.2

Subtract $x^{6}$ from both sides of the equation.

$7 x^{6} \ln (x) = - x^{6}$

Step 2.3

Divide each term in $7 x^{6} \ln (x) = - x^{6}$ by $7 x^{6}$ and simplify.

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Step 2.3.1

Divide each term in $7 x^{6} \ln (x) = - x^{6}$ by $7 x^{6}$ .

$\frac{7 x^{6} \ln (x)}{7 x^{6}} = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $7$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{7 x^{6} \ln (x)}{7 x^{6}} = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.2.1.2

Rewrite the expression.

$\frac{x^{6} \ln (x)}{x^{6}} = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.2.2

Cancel the common factor of $x^{6}$ .

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Step 2.3.2.2.1

Cancel the common factor.

$\frac{x^{6} \ln (x)}{x^{6}} = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.2.2.2

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

$\ln (x) = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $x^{6}$ .

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Step 2.3.3.1.1

Cancel the common factor.

$\ln (x) = \frac{- x^{6}}{7 x^{6}}$

Step 2.3.3.1.2

Rewrite the expression.

$\ln (x) = \frac{- 1}{7}$

Step 2.3.3.2

Move the negative in front of the fraction.

$\ln (x) = - \frac{1}{7}$

Step 2.4

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

$e^{\ln (x)} = e^{- \frac{1}{7}}$

Step 2.5

Rewrite $\ln (x) = - \frac{1}{7}$ 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 $b^{y} = x$ .

$e^{- \frac{1}{7}} = x$

Step 2.6

Solve for $x$ .

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Step 2.6.1

Rewrite the equation as $x = e^{- \frac{1}{7}}$ .

$x = e^{- \frac{1}{7}}$

Step 2.6.2

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

$x = \frac{1}{e^{\frac{1}{7}}}$

Step 3

The values which make the derivative equal to $0$ are $\frac{1}{e^{\frac{1}{7}}}$ .

$\frac{1}{e^{\frac{1}{7}}}$

Step 4

Find where the derivative is undefined.

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Step 4.1

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

$x \leq 0$

Step 4.2

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 5

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

$(- \infty, 0) \cup (0, \frac{1}{e^{\frac{1}{7}}}) \cup (\frac{1}{e^{\frac{1}{7}}}, \infty)$

Step 6

Exclude the intervals that are not in the domain.

$(0, \frac{1}{e^{\frac{1}{7}}})$

Step 7

Substitute a value from the interval $(0, \frac{1}{e^{\frac{1}{7}}})$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.43343895$ in the expression.

$f' (0.43343895) = {(0.43343895)}^{6} + 7 {(0.43343895)}^{6} \ln (0.43343895)$

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 $0.43343895$ to the power of $6$ .

$f' (0.43343895) = 0.00663082 + 7 {(0.43343895)}^{6} \ln (0.43343895)$

Step 7.2.1.2

Raise $0.43343895$ to the power of $6$ .

$f' (0.43343895) = 0.00663082 + 7 \cdot (0.00663082 \ln (0.43343895))$

Step 7.2.1.3

Multiply $7$ by $0.00663082$ .

$f' (0.43343895) = 0.00663082 + 0.04641578 \ln (0.43343895)$

Step 7.2.1.4

Simplify $0.04641578 \ln (0.43343895)$ by moving $0.04641578$ inside the logarithm.

$f' (0.43343895) = 0.00663082 + \ln ({0.43343895}^{0.04641578})$

Step 7.2.1.5

Raise $0.43343895$ to the power of $0.04641578$ .

$f' (0.43343895) = 0.00663082 + \ln (0.96193943)$

Step 7.2.2

Add $0.00663082$ and $\ln (0.96193943)$ .

$f' (0.43343895) = - 0.03217296$

Step 7.2.3

The final answer is $- 0.03217296$ .

$- 0.03217296$

Step 7.3

At $x = 0.43343895$ the derivative is $- 0.03217296$ . Since this is negative, the function is decreasing on $(0, \frac{1}{e^{\frac{1}{7}}})$ .

Decreasing on $(0, \frac{1}{e^{\frac{1}{7}}})$ since $f' (x) < 0$

Step 8

Exclude the intervals that are not in the domain.

$(0, \frac{1}{e^{\frac{1}{7}}}), (\frac{1}{e^{\frac{1}{7}}}, \infty)$

Step 9

Substitute a value from the interval $(\frac{1}{e^{\frac{1}{7}}}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $1.8668779$ in the expression.

$f' (1.8668779) = {(1.8668779)}^{6} + 7 {(1.8668779)}^{6} \ln (1.8668779)$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify each term.

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Step 9.2.1.1

Raise $1.8668779$ to the power of $6$ .

$f' (1.8668779) = 42.33460261 + 7 {(1.8668779)}^{6} \ln (1.8668779)$

Step 9.2.1.2

Raise $1.8668779$ to the power of $6$ .

$f' (1.8668779) = 42.33460261 + 7 \cdot (42.33460261 \ln (1.8668779))$

Step 9.2.1.3

Multiply $7$ by $42.33460261$ .

$f' (1.8668779) = 42.33460261 + 296.34221828 \ln (1.8668779)$

Step 9.2.1.4

Simplify $296.34221828 \ln (1.8668779)$ by moving $296.34221828$ inside the logarithm.

$f' (1.8668779) = 42.33460261 + \ln ({1.8668779}^{296.34221828})$

Step 9.2.2

Add $42.33460261$ and $\ln ({1.8668779}^{296.34221828})$ .

$f' (1.8668779) = 227.33140751$

Step 9.2.3

The final answer is $227.33140751$ .

$227.33140751$

Step 9.3

At $x = 1.8668779$ the derivative is $227.33140751$ . Since this is positive, the function is increasing on $(\frac{1}{e^{\frac{1}{7}}}, \infty)$ .

Increasing on $(\frac{1}{e^{\frac{1}{7}}}, \infty)$ since $f' (x) > 0$

Step 10

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

Increasing on: $(\frac{1}{e^{\frac{1}{7}}}, \infty)$

Decreasing on: $(0, \frac{1}{e^{\frac{1}{7}}})$

Step 11