Calculus Examples

$f (x) = 7 x - 2 x \ln (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $7 x - 2 x \ln (x)$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [- 2 x \ln (x)]$ .

$\frac{d}{d x} [7 x] + \frac{d}{d x} [- 2 x \ln (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [7 x]$ .

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Step 1.2.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 x \ln (x)]$

Step 1.2.2

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

$7 \cdot 1 + \frac{d}{d x} [- 2 x \ln (x)]$

Step 1.2.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d x} [- 2 x \ln (x)]$

Step 1.3

Evaluate $\frac{d}{d x} [- 2 x \ln (x)]$ .

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Step 1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x \ln (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [x \ln (x)]$ .

$7 - 2 \frac{d}{d x} [x \ln (x)]$

Step 1.3.2

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

$7 - 2 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])$

Step 1.3.3

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

$7 - 2 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])$

Step 1.3.4

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

$7 - 2 (x \frac{1}{x} + \ln (x) \cdot 1)$

Step 1.3.5

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

$7 - 2 (\frac{x}{x} + \ln (x) \cdot 1)$

Step 1.3.6

Cancel the common factor of $x$ .

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Step 1.3.6.1

Cancel the common factor.

$7 - 2 (\frac{x}{x} + \ln (x) \cdot 1)$

Step 1.3.6.2

Rewrite the expression.

$7 - 2 (1 + \ln (x) \cdot 1)$

Step 1.3.7

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

$7 - 2 (1 + \ln (x))$

Step 1.4

Simplify.

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Step 1.4.1

Apply the distributive property.

$7 - 2 \cdot 1 - 2 \ln (x)$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Multiply $- 2$ by $1$ .

$7 - 2 - 2 \ln (x)$

Step 1.4.2.2

Subtract $2$ from $7$ .

$5 - 2 \ln (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $5 - 2 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [5] + \frac{d}{d x} [- 2 \ln (x)]$ .

$f'' (x) = \frac{d}{d x} (5) + \frac{d}{d x} (- 2 \ln (x))$

Step 2.1.2

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- 2 \ln (x))$

Step 2.2

Evaluate $\frac{d}{d x} [- 2 \ln (x)]$ .

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Step 2.2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 \ln (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = 0 - 2 \frac{d}{d x} \ln (x)$

Step 2.2.2

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

$f'' (x) = 0 - 2 \frac{1}{x}$

Step 2.2.3

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

$f'' (x) = 0 + \frac{- 2}{x}$

Step 2.2.4

Move the negative in front of the fraction.

$f'' (x) = 0 - \frac{2}{x}$

Step 2.3

Subtract $\frac{2}{x}$ from $0$ .

$f'' (x) = - \frac{2}{x}$

Step 3

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

$5 - 2 \ln (x) = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $7 x - 2 x \ln (x)$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [- 2 x \ln (x)]$ .

$\frac{d}{d x} [7 x] + \frac{d}{d x} [- 2 x \ln (x)]$

Step 4.1.2

Evaluate $\frac{d}{d x} [7 x]$ .

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Step 4.1.2.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 x \ln (x)]$

Step 4.1.2.2

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

$7 \cdot 1 + \frac{d}{d x} [- 2 x \ln (x)]$

Step 4.1.2.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d x} [- 2 x \ln (x)]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- 2 x \ln (x)]$ .

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Step 4.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x \ln (x)$ with respect to $x$ is $- 2 \frac{d}{d x} [x \ln (x)]$ .

$7 - 2 \frac{d}{d x} [x \ln (x)]$

Step 4.1.3.2

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

$7 - 2 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])$

Step 4.1.3.3

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

$7 - 2 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])$

Step 4.1.3.4

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

$7 - 2 (x \frac{1}{x} + \ln (x) \cdot 1)$

Step 4.1.3.5

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

$7 - 2 (\frac{x}{x} + \ln (x) \cdot 1)$

Step 4.1.3.6

Cancel the common factor of $x$ .

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Step 4.1.3.6.1

Cancel the common factor.

$7 - 2 (\frac{x}{x} + \ln (x) \cdot 1)$

Step 4.1.3.6.2

Rewrite the expression.

$7 - 2 (1 + \ln (x) \cdot 1)$

Step 4.1.3.7

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

$7 - 2 (1 + \ln (x))$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Apply the distributive property.

$7 - 2 \cdot 1 - 2 \ln (x)$

Step 4.1.4.2

Combine terms.

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Step 4.1.4.2.1

Multiply $- 2$ by $1$ .

$7 - 2 - 2 \ln (x)$

Step 4.1.4.2.2

Subtract $2$ from $7$ .

$f' (x) = 5 - 2 \ln (x)$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $5 - 2 \ln (x)$ .

$5 - 2 \ln (x)$

Step 5

Set the first derivative equal to $0$ then solve the equation $5 - 2 \ln (x) = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$5 - 2 \ln (x) = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$- 2 \ln (x) = - 5$

Step 5.3

Divide each term in $- 2 \ln (x) = - 5$ by $- 2$ and simplify.

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Step 5.3.1

Divide each term in $- 2 \ln (x) = - 5$ by $- 2$ .

$\frac{- 2 \ln (x)}{- 2} = \frac{- 5}{- 2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $- 2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{- 2 \ln (x)}{- 2} = \frac{- 5}{- 2}$

Step 5.3.2.1.2

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

$\ln (x) = \frac{- 5}{- 2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Dividing two negative values results in a positive value.

$\ln (x) = \frac{5}{2}$

Step 5.4

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

$e^{\ln (x)} = e^{\frac{5}{2}}$

Step 5.5

Rewrite $\ln (x) = \frac{5}{2}$ 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 $b^{y} = x$ .

$e^{\frac{5}{2}} = x$

Step 5.6

Rewrite the equation as $x = e^{\frac{5}{2}}$ .

$x = e^{\frac{5}{2}}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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

$x \leq 0$

Step 6.2

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 7

Critical points to evaluate.

$x = e^{\frac{5}{2}}$

Step 8

Evaluate the second derivative at $x = e^{\frac{5}{2}}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2}{e^{\frac{5}{2}}}$

Step 9

$x = e^{\frac{5}{2}}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = e^{\frac{5}{2}}$ is a local maximum

Step 10

Find the y-value when $x = e^{\frac{5}{2}}$ .

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Step 10.1

Replace the variable $x$ with $e^{\frac{5}{2}}$ in the expression.

$f (e^{\frac{5}{2}}) = 7 (e^{\frac{5}{2}}) - 2 e^{\frac{5}{2}} \ln (e^{\frac{5}{2}})$

Step 10.2

Simplify the result.

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Step 10.2.1

Simplify each term.

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Step 10.2.1.1

Use logarithm rules to move $\frac{5}{2}$ out of the exponent.

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} - 2 e^{\frac{5}{2}} (\frac{5}{2} \cdot \ln (e))$

Step 10.2.1.2

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

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} - 2 e^{\frac{5}{2}} (\frac{5}{2} \cdot 1)$

Step 10.2.1.3

Multiply $\frac{5}{2}$ by $1$ .

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} - 2 e^{\frac{5}{2}} \frac{5}{2}$

Step 10.2.1.4

Cancel the common factor of $2$ .

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Step 10.2.1.4.1

Factor $2$ out of $- 2 e^{\frac{5}{2}}$ .

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} + 2 (- e^{\frac{5}{2}}) (\frac{5}{2})$

Step 10.2.1.4.2

Cancel the common factor.

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} + 2 (- e^{\frac{5}{2}}) (\frac{5}{2})$

Step 10.2.1.4.3

Rewrite the expression.

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} - e^{\frac{5}{2}} \cdot 5$

Step 10.2.1.5

Multiply $5$ by $- 1$ .

$f (e^{\frac{5}{2}}) = 7 e^{\frac{5}{2}} - 5 e^{\frac{5}{2}}$

Step 10.2.2

Subtract $5 e^{\frac{5}{2}}$ from $7 e^{\frac{5}{2}}$ .

$f (e^{\frac{5}{2}}) = 2 e^{\frac{5}{2}}$

Step 10.2.3

The final answer is $2 e^{\frac{5}{2}}$ .

$y = 2 e^{\frac{5}{2}}$

Step 11

These are the local extrema for $f (x) = 7 x - 2 x \ln (x)$ .

$(e^{\frac{5}{2}}, 2 e^{\frac{5}{2}})$ is a local maxima

Step 12