Calculus Examples

$f (x) = x \ln (x)$

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

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

Step 1.1.2

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

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

Step 1.1.3

Differentiate using the Power Rule.

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Step 1.1.3.1

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

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

Step 1.1.3.2

Cancel the common factor of $x$ .

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Step 1.1.3.2.1

Cancel the common factor.

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

Step 1.1.3.2.2

Rewrite the expression.

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

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 $n = 1$ .

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

Step 1.1.3.4

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

$f' (x) = 1 + \ln (x)$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $1 + \ln (x)$ .

$1 + \ln (x)$

Step 2

Set the first derivative equal to $0$ then solve the equation $1 + \ln (x) = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$1 + \ln (x) = 0$

Step 2.2

Subtract $1$ from both sides of the equation.

$\ln (x) = - 1$

Step 2.3

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

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

Step 2.4

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

$e^{- 1} = x$

Step 2.5

Solve for $x$ .

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Step 2.5.1

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

$x = e^{- 1}$

Step 2.5.2

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

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

Step 3

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

$\frac{1}{e}$

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.

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

Step 6

Exclude the intervals that are not in the domain.

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

Step 7

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

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Step 7.1

Replace the variable $x$ with $0.18393972$ in the expression.

$f' (0.18393972) = 1 + \ln (0.18393972)$

Step 7.2

The final answer is $1 + \ln (0.18393972)$ .

$1 + \ln (0.18393972)$

Step 7.3

Simplify.

$- 0.69314715$

Step 7.4

At $x = 0.18393972$ the derivative is $- 0.69314715$ . Since this is negative, the function is decreasing on $(0, \frac{1}{e})$ .

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

Step 8

Exclude the intervals that are not in the domain.

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

Step 9

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

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Step 9.1

Replace the variable $x$ with $1.36787945$ in the expression.

$f' (1.36787945) = 1 + \ln (1.36787945)$

Step 9.2

The final answer is $1 + \ln (1.36787945)$ .

$1 + \ln (1.36787945)$

Step 9.3

Simplify.

$1.31326169$

Step 9.4

At $x = 1.36787945$ the derivative is $1.31326169$ . Since this is positive, the function is increasing on $(\frac{1}{e}, \infty)$ .

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

Step 10

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

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

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

Step 11