Calculus Examples

$f (x) = 6 x - 3 x \ln (x)$

Step 1

Find the first derivative of the function.

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By the Sum Rule, the derivative of $6 x - 3 x \ln (x)$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [- 3 x \ln (x)]$ .

$\frac{d}{d x} [6 x] + \frac{d}{d x} [- 3 x \ln (x)]$

Evaluate $\frac{d}{d x} [6 x]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x] + \frac{d}{d x} [- 3 x \ln (x)]$

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

$6 \cdot 1 + \frac{d}{d x} [- 3 x \ln (x)]$

Multiply $6$ by $1$ .

$6 + \frac{d}{d x} [- 3 x \ln (x)]$

Evaluate $\frac{d}{d x} [- 3 x \ln (x)]$ .

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x \ln (x)$ with respect to $x$ is $- 3 \frac{d}{d x} [x \ln (x)]$ .

$6 - 3 \frac{d}{d x} [x \ln (x)]$

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

$6 - 3 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])$

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

$6 - 3 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])$

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

$6 - 3 (x \frac{1}{x} + \ln (x) \cdot 1)$

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

$6 - 3 (\frac{x}{x} + \ln (x) \cdot 1)$

Cancel the common factor of $x$ .

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Cancel the common factor.

$6 - 3 (\frac{x}{x} + \ln (x) \cdot 1)$

Rewrite the expression.

$6 - 3 (1 + \ln (x) \cdot 1)$

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

$6 - 3 (1 + \ln (x))$

Simplify.

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Apply the distributive property.

$6 - 3 \cdot 1 - 3 \ln (x)$

Combine terms.

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Multiply $- 3$ by $1$ .

$6 - 3 - 3 \ln (x)$

Subtract $3$ from $6$ .

$3 - 3 \ln (x)$

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 $3 - 3 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- 3 \ln (x)]$ .

$f'' (x) = \frac{d}{d x} (3) + \frac{d}{d x} (- 3 \ln (x))$

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- 3 \ln (x))$

Evaluate $\frac{d}{d x} [- 3 \ln (x)]$ .

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 \ln (x)$ with respect to $x$ is $- 3 \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = 0 - 3 \frac{d}{d x} \ln (x)$

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

$f'' (x) = 0 - 3 \frac{1}{x}$

Combine $- 3$ and $\frac{1}{x}$ .

$f'' (x) = 0 + \frac{- 3}{x}$

Move the negative in front of the fraction.

$f'' (x) = 0 - \frac{3}{x}$

Subtract $\frac{3}{x}$ from $0$ .

$f'' (x) = - \frac{3}{x}$

Step 3

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

$3 - 3 \ln (x) = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $6 x - 3 x \ln (x)$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [- 3 x \ln (x)]$ .

$f' (x) = \frac{d}{d x} (6 x) + \frac{d}{d x} (- 3 x \ln (x))$

Evaluate $\frac{d}{d x} [6 x]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$f' (x) = 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 3 x \ln (x))$

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

$f' (x) = 6 \cdot 1 + \frac{d}{d x} (- 3 x \ln (x))$

Multiply $6$ by $1$ .

$f' (x) = 6 + \frac{d}{d x} (- 3 x \ln (x))$

Evaluate $\frac{d}{d x} [- 3 x \ln (x)]$ .

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x \ln (x)$ with respect to $x$ is $- 3 \frac{d}{d x} [x \ln (x)]$ .

$f' (x) = 6 - 3 \frac{d}{d x} (x \ln (x))$

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

$f' (x) = 6 - 3 (x \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x))$

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

$f' (x) = 6 - 3 (x (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x))$

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

$f' (x) = 6 - 3 (x (\frac{1}{x}) + \ln (x) \cdot 1)$

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

$f' (x) = 6 - 3 (\frac{x}{x} + \ln (x) \cdot 1)$

Cancel the common factor of $x$ .

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Cancel the common factor.

$f' (x) = 6 - 3 (\frac{x}{x} + \ln (x) \cdot 1)$

Rewrite the expression.

$f' (x) = 6 - 3 (1 + \ln (x) \cdot 1)$

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

$f' (x) = 6 - 3 (1 + \ln (x))$

Simplify.

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Apply the distributive property.

$f' (x) = 6 - 3 \cdot 1 - 3 \ln (x)$

Combine terms.

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Multiply $- 3$ by $1$ .

$f' (x) = 6 - 3 - 3 \ln (x)$

Subtract $3$ from $6$ .

$f' (x) = 3 - 3 \ln (x)$

The first derivative of $f (x)$ with respect to $x$ is $3 - 3 \ln (x)$ .

$3 - 3 \ln (x)$

Step 5

Set the first derivative equal to $0$ then solve the equation $3 - 3 \ln (x) = 0$ .

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Set the first derivative equal to $0$ .

$3 - 3 \ln (x) = 0$

Subtract $3$ from both sides of the equation.

$- 3 \ln (x) = - 3$

Divide each term in $- 3 \ln (x) = - 3$ by $- 3$ and simplify.

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Divide each term in $- 3 \ln (x) = - 3$ by $- 3$ .

$\frac{- 3 \ln (x)}{- 3} = \frac{- 3}{- 3}$

Simplify the left side.

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Cancel the common factor of $- 3$ .

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Cancel the common factor.

$\frac{- 3 \ln (x)}{- 3} = \frac{- 3}{- 3}$

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

$\ln (x) = \frac{- 3}{- 3}$

Simplify the right side.

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Divide $- 3$ by $- 3$ .

$\ln (x) = 1$

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 $b^{y} = x$ .

$e^{1} = x$

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

$x = e$

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.

$x = e$

Step 8

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

$- \frac{3}{e}$

Step 9

$x = e$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = e$ is a local maximum

Step 10

Find the y-value when $x = e$ .

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

$f (e) = 6 (e) - 3 e \ln (e)$

Simplify the result.

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Simplify each term.

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

$f (e) = 6 e - 3 e \cdot 1$

Multiply $- 3$ by $1$ .

$f (e) = 6 e - 3 e$

Subtract $3 e$ from $6 e$ .

$f (e) = 3 e$

The final answer is $3 e$ .

$y = 3 e$

Step 11

These are the local extrema for $f (x) = 6 x - 3 x \ln (x)$ .

$(e, 3 e)$ is a local maxima

Step 12