Calculus Examples

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

Step 1

Find the first derivative of the function.

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

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

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

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

Differentiate using the Power Rule.

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

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

$1 + \ln (x)$

Step 2

Find the second derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $1 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = \frac{d}{d x} (1) + \frac{d}{d x} (\ln (x))$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (\ln (x))$

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

$f'' (x) = 0 + \frac{1}{x}$

Add $0$ and $\frac{1}{x}$ .

$f'' (x) = \frac{1}{x}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$1 + \ln (x) = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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

$f' (x) = x \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x)$

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

$f' (x) = x (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x)$

Differentiate using the Power Rule.

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

$f' (x) = \frac{x}{x} + \ln (x) \frac{d}{d x} (x)$

Cancel the common factor of $x$ .

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Cancel the common factor.

$f' (x) = \frac{x}{x} + \ln (x) \frac{d}{d x} (x)$

Rewrite the expression.

$f' (x) = 1 + \ln (x) \frac{d}{d x} (x)$

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

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

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

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

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

$1 + \ln (x)$

Step 5

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

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Set the first derivative equal to $0$ .

$1 + \ln (x) = 0$

Subtract $1$ from both sides of the equation.

$\ln (x) = - 1$

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

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

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 $b^{y} = x$ .

$e^{- 1} = x$

Solve for $x$ .

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

$x = e^{- 1}$

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

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

Step 6

Find the values where the derivative is undefined.

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

$x \leq 0$

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 7

Critical points to evaluate.

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

Step 8

Evaluate the second derivative at $x = \frac{1}{e}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{1}{\frac{1}{e}}$

Step 9

Evaluate the second derivative.

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Multiply the numerator by the reciprocal of the denominator.

$1 e$

Multiply $e$ by $1$ .

$e$

Step 10

$x = \frac{1}{e}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{e}$ is a local minimum

Step 11

Find the y-value when $x = \frac{1}{e}$ .

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

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

Simplify the result.

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Rewrite $\frac{1}{e}$ as $1 e^{- 1}$ .

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

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

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

Use logarithm rules to move $- 1$ out of the exponent.

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

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

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

Multiply $- 1$ by $1$ .

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

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

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

Subtract $1$ from $0$ .

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

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

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

Move the negative in front of the fraction.

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

The final answer is $- \frac{1}{e}$ .

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

Step 12

These are the local extrema for $f (x) = x \ln (x)$ .

$(\frac{1}{e}, - \frac{1}{e})$ is a local minima

Step 13