Calculus Examples

$f (x) = 4 x \ln (x) - 7 x$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $4 x \ln (x) - 7 x$ with respect to $x$ is $\frac{d}{d x} [4 x \ln (x)] + \frac{d}{d x} [- 7 x]$ .

$\frac{d}{d x} [4 x \ln (x)] + \frac{d}{d x} [- 7 x]$

Step 1.2

Evaluate $\frac{d}{d x} [4 x \ln (x)]$ .

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Step 1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x \ln (x)$ with respect to $x$ is $4 \frac{d}{d x} [x \ln (x)]$ .

$4 \frac{d}{d x} [x \ln (x)] + \frac{d}{d x} [- 7 x]$

Step 1.2.2

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

$4 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 7 x]$

Step 1.2.3

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

$4 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 7 x]$

Step 1.2.4

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

$4 (x \frac{1}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 1.2.5

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

$4 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 1.2.6

Cancel the common factor of $x$ .

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Step 1.2.6.1

Cancel the common factor.

$4 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 1.2.6.2

Rewrite the expression.

$4 (1 + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 1.2.7

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

$4 (1 + \ln (x)) + \frac{d}{d x} [- 7 x]$

Step 1.3

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

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Step 1.3.1

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

$4 (1 + \ln (x)) - 7 \frac{d}{d x} [x]$

Step 1.3.2

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

$4 (1 + \ln (x)) - 7 \cdot 1$

Step 1.3.3

Multiply $- 7$ by $1$ .

$4 (1 + \ln (x)) - 7$

Step 1.4

Simplify.

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Step 1.4.1

Apply the distributive property.

$4 \cdot 1 + 4 \ln (x) - 7$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Multiply $4$ by $1$ .

$4 + 4 \ln (x) - 7$

Step 1.4.2.2

Subtract $7$ from $4$ .

$4 \ln (x) - 3$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $4 \ln (x) - 3$ with respect to $x$ is $\frac{d}{d x} [4 \ln (x)] + \frac{d}{d x} [- 3]$ .

$f'' (x) = \frac{d}{d x} (4 \ln (x)) + \frac{d}{d x} (- 3)$

Step 2.2

Evaluate $\frac{d}{d x} [4 \ln (x)]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 \ln (x)$ with respect to $x$ is $4 \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = 4 \frac{d}{d x} (\ln (x)) + \frac{d}{d x} (- 3)$

Step 2.2.2

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

$f'' (x) = 4 (\frac{1}{x}) + \frac{d}{d x} (- 3)$

Step 2.2.3

Combine $4$ and $\frac{1}{x}$ .

$f'' (x) = \frac{4}{x} + \frac{d}{d x} (- 3)$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

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

$f'' (x) = \frac{4}{x} + 0$

Step 2.3.2

Add $\frac{4}{x}$ and $0$ .

$f'' (x) = \frac{4}{x}$

Step 3

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

$4 \ln (x) - 3 = 0$

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 $4 x \ln (x) - 7 x$ with respect to $x$ is $\frac{d}{d x} [4 x \ln (x)] + \frac{d}{d x} [- 7 x]$ .

$\frac{d}{d x} [4 x \ln (x)] + \frac{d}{d x} [- 7 x]$

Step 4.1.2

Evaluate $\frac{d}{d x} [4 x \ln (x)]$ .

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Step 4.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x \ln (x)$ with respect to $x$ is $4 \frac{d}{d x} [x \ln (x)]$ .

$4 \frac{d}{d x} [x \ln (x)] + \frac{d}{d x} [- 7 x]$

Step 4.1.2.2

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

$4 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.3

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

$4 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x]) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.4

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

$4 (x \frac{1}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.5

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

$4 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.6

Cancel the common factor of $x$ .

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Step 4.1.2.6.1

Cancel the common factor.

$4 (\frac{x}{x} + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.6.2

Rewrite the expression.

$4 (1 + \ln (x) \cdot 1) + \frac{d}{d x} [- 7 x]$

Step 4.1.2.7

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

$4 (1 + \ln (x)) + \frac{d}{d x} [- 7 x]$

Step 4.1.3

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

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Step 4.1.3.1

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

$4 (1 + \ln (x)) - 7 \frac{d}{d x} [x]$

Step 4.1.3.2

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

$4 (1 + \ln (x)) - 7 \cdot 1$

Step 4.1.3.3

Multiply $- 7$ by $1$ .

$4 (1 + \ln (x)) - 7$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Apply the distributive property.

$4 \cdot 1 + 4 \ln (x) - 7$

Step 4.1.4.2

Combine terms.

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Step 4.1.4.2.1

Multiply $4$ by $1$ .

$4 + 4 \ln (x) - 7$

Step 4.1.4.2.2

Subtract $7$ from $4$ .

$f' (x) = 4 \ln (x) - 3$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $4 \ln (x) - 3$ .

$4 \ln (x) - 3$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 \ln (x) - 3 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$4 \ln (x) - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$4 \ln (x) = 3$

Step 5.3

Divide each term in $4 \ln (x) = 3$ by $4$ and simplify.

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Step 5.3.1

Divide each term in $4 \ln (x) = 3$ by $4$ .

$\frac{4 \ln (x)}{4} = \frac{3}{4}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{4 \ln (x)}{4} = \frac{3}{4}$

Step 5.3.2.1.2

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

$\ln (x) = \frac{3}{4}$

Step 5.4

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

$e^{\ln (x)} = e^{\frac{3}{4}}$

Step 5.5

Rewrite $\ln (x) = \frac{3}{4}$ 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 $b^{y} = x$ .

$e^{\frac{3}{4}} = x$

Step 5.6

Rewrite the equation as $x = e^{\frac{3}{4}}$ .

$x = e^{\frac{3}{4}}$

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.

$x = e^{\frac{3}{4}}$

Step 8

Evaluate the second derivative at $x = e^{\frac{3}{4}}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{4}{e^{\frac{3}{4}}}$

Step 9

$x = e^{\frac{3}{4}}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = e^{\frac{3}{4}}$ is a local minimum

Step 10

Find the y-value when $x = e^{\frac{3}{4}}$ .

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Step 10.1

Replace the variable $x$ with $e^{\frac{3}{4}}$ in the expression.

$f (e^{\frac{3}{4}}) = 4 (e^{\frac{3}{4}}) \ln (e^{\frac{3}{4}}) - 7 e^{\frac{3}{4}}$

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 $\frac{3}{4}$ out of the exponent.

$f (e^{\frac{3}{4}}) = 4 e^{\frac{3}{4}} (\frac{3}{4} \cdot \ln (e)) - 7 e^{\frac{3}{4}}$

Step 10.2.1.2

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

$f (e^{\frac{3}{4}}) = 4 e^{\frac{3}{4}} (\frac{3}{4} \cdot 1) - 7 e^{\frac{3}{4}}$

Step 10.2.1.3

Multiply $\frac{3}{4}$ by $1$ .

$f (e^{\frac{3}{4}}) = 4 e^{\frac{3}{4}} (\frac{3}{4}) - 7 e^{\frac{3}{4}}$

Step 10.2.1.4

Cancel the common factor of $4$ .

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Step 10.2.1.4.1

Factor $4$ out of $4 e^{\frac{3}{4}}$ .

$f (e^{\frac{3}{4}}) = 4 (e^{\frac{3}{4}}) (\frac{3}{4}) - 7 e^{\frac{3}{4}}$

Step 10.2.1.4.2

Cancel the common factor.

$f (e^{\frac{3}{4}}) = 4 e^{\frac{3}{4}} (\frac{3}{4}) - 7 e^{\frac{3}{4}}$

Step 10.2.1.4.3

Rewrite the expression.

$f (e^{\frac{3}{4}}) = e^{\frac{3}{4}} \cdot 3 - 7 e^{\frac{3}{4}}$

Step 10.2.1.5

Move $3$ to the left of $e^{\frac{3}{4}}$ .

$f (e^{\frac{3}{4}}) = 3 e^{\frac{3}{4}} - 7 e^{\frac{3}{4}}$

Step 10.2.2

Subtract $7 e^{\frac{3}{4}}$ from $3 e^{\frac{3}{4}}$ .

$f (e^{\frac{3}{4}}) = - 4 e^{\frac{3}{4}}$

Step 10.2.3

The final answer is $- 4 e^{\frac{3}{4}}$ .

$y = - 4 e^{\frac{3}{4}}$

Step 11

These are the local extrema for $f (x) = 4 x \ln (x) - 7 x$ .

$(e^{\frac{3}{4}}, - 4 e^{\frac{3}{4}})$ is a local minima

Step 12