Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $x^{4} + \frac{4}{x}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [\frac{4}{x}]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [\frac{4}{x}]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [\frac{4}{x}]$

Step 1.2

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

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Step 1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $\frac{4}{x}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{1}{x}]$ .

$4 x^{3} + 4 \frac{d}{d x} [\frac{1}{x}]$

Step 1.2.2

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

$4 x^{3} + 4 \frac{d}{d x} [x^{- 1}]$

Step 1.2.3

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

$4 x^{3} + 4 (- x^{- 2})$

Step 1.2.4

Multiply $- 1$ by $4$ .

$4 x^{3} - 4 x^{- 2}$

Step 1.3

Simplify.

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Step 1.3.1

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

$4 x^{3} - 4 \frac{1}{x^{2}}$

Step 1.3.2

Combine terms.

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Step 1.3.2.1

Combine $- 4$ and $\frac{1}{x^{2}}$ .

$4 x^{3} + \frac{- 4}{x^{2}}$

Step 1.3.2.2

Move the negative in front of the fraction.

$4 x^{3} - \frac{4}{x^{2}}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $4 x^{3} - \frac{4}{x^{2}}$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- \frac{4}{x^{2}}]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- \frac{4}{x^{2}})$

Step 2.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- \frac{4}{x^{2}})$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- \frac{4}{x^{2}})$

Step 2.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- \frac{4}{x^{2}})$

Step 2.3

Evaluate $\frac{d}{d x} [- \frac{4}{x^{2}}]$ .

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Step 2.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- \frac{4}{x^{2}}$ with respect to $x$ is $- 4 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = 12 x^{2} - 4 \frac{d}{d x} \frac{1}{x^{2}}$

Step 2.3.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f'' (x) = 12 x^{2} - 4 \frac{d}{d x} {(x^{2})}^{- 1}$

Step 2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 2.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$f'' (x) = 12 x^{2} - 4 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2}))$

Step 2.3.3.2

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

$f'' (x) = 12 x^{2} - 4 (- u^{- 2} \frac{d}{d x} x^{2})$

Step 2.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$f'' (x) = 12 x^{2} - 4 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2})$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 12 x^{2} - 4 (- {(x^{2})}^{- 2} (2 x))$

Step 2.3.5

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 2.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = 12 x^{2} - 4 (- x^{2 \cdot - 2} (2 x))$

Step 2.3.5.2

Multiply $2$ by $- 2$ .

$f'' (x) = 12 x^{2} - 4 (- x^{- 4} (2 x))$

Step 2.3.6

Multiply $2$ by $- 1$ .

$f'' (x) = 12 x^{2} - 4 (- 2 x^{- 4} x)$

Step 2.3.7

Raise $x$ to the power of $1$ .

$f'' (x) = 12 x^{2} - 4 (- 2 (x \cdot x^{- 4}))$

Step 2.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 12 x^{2} - 4 (- 2 x^{1 - 4})$

Step 2.3.9

Subtract $4$ from $1$ .

$f'' (x) = 12 x^{2} - 4 (- 2 x^{- 3})$

Step 2.3.10

Multiply $- 2$ by $- 4$ .

$f'' (x) = 12 x^{2} + 8 x^{- 3}$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = 12 x^{2} + 8 (\frac{1}{x^{3}})$

Step 2.4.2

Combine $8$ and $\frac{1}{x^{3}}$ .

$f'' (x) = 12 x^{2} + \frac{8}{x^{3}}$

Step 3

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

$4 x^{3} - \frac{4}{x^{2}} = 0$

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

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of $x^{4} + \frac{4}{x}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [\frac{4}{x}]$ .

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

Step 4.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 4.1.2

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

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Step 4.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $\frac{4}{x}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 4.1.2.2

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

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

Step 4.1.2.3

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

$f' (x) = 4 x^{3} + 4 (- x^{- 2})$

Step 4.1.2.4

Multiply $- 1$ by $4$ .

$f' (x) = 4 x^{3} - 4 x^{- 2}$

Step 4.1.3

Simplify.

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Step 4.1.3.1

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

$f' (x) = 4 x^{3} - 4 \frac{1}{x^{2}}$

Step 4.1.3.2

Combine terms.

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Step 4.1.3.2.1

Combine $- 4$ and $\frac{1}{x^{2}}$ .

$f' (x) = 4 x^{3} + \frac{- 4}{x^{2}}$

Step 4.1.3.2.2

Move the negative in front of the fraction.

$f' (x) = 4 x^{3} - \frac{4}{x^{2}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - \frac{4}{x^{2}}$ .

$4 x^{3} - \frac{4}{x^{2}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - \frac{4}{x^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$4 x^{3} - \frac{4}{x^{2}} = 0$

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x^{2}, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 5.3

Multiply each term in $4 x^{3} - \frac{4}{x^{2}} = 0$ by $x^{2}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $4 x^{3} - \frac{4}{x^{2}} = 0$ by $x^{2}$ .

$4 x^{3} x^{2} - \frac{4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply $x^{3}$ by $x^{2}$ by adding the exponents.

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Step 5.3.2.1.1.1

Move $x^{2}$ .

$4 (x^{2} x^{3}) - \frac{4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 x^{2 + 3} - \frac{4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2.1.1.3

Add $2$ and $3$ .

$4 x^{5} - \frac{4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2.1.2

Cancel the common factor of $x^{2}$ .

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Step 5.3.2.1.2.1

Move the leading negative in $- \frac{4}{x^{2}}$ into the numerator.

$4 x^{5} + \frac{- 4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2.1.2.2

Cancel the common factor.

$4 x^{5} + \frac{- 4}{x^{2}} x^{2} = 0 x^{2}$

Step 5.3.2.1.2.3

Rewrite the expression.

$4 x^{5} - 4 = 0 x^{2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $0$ by $x^{2}$ .

$4 x^{5} - 4 = 0$

Step 5.4

Solve the equation.

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Step 5.4.1

Add $4$ to both sides of the equation.

$4 x^{5} = 4$

Step 5.4.2

Divide each term in $4 x^{5} = 4$ by $4$ and simplify.

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Step 5.4.2.1

Divide each term in $4 x^{5} = 4$ by $4$ .

$\frac{4 x^{5}}{4} = \frac{4}{4}$

Step 5.4.2.2

Simplify the left side.

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Step 5.4.2.2.1

Cancel the common factor of $4$ .

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Step 5.4.2.2.1.1

Cancel the common factor.

$\frac{4 x^{5}}{4} = \frac{4}{4}$

Step 5.4.2.2.1.2

Divide $x^{5}$ by $1$ .

$x^{5} = \frac{4}{4}$

Step 5.4.2.3

Simplify the right side.

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Step 5.4.2.3.1

Divide $4$ by $4$ .

$x^{5} = 1$

Step 5.4.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[5]{1}$

Step 5.4.4

Any root of $1$ is $1$ .

$x = 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{4}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 6.2.2

Simplify $\pm \sqrt{0}$ .

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Step 6.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 6.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 6.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

Critical points to evaluate.

$x = 1$

Step 8

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

$12 {(1)}^{2} + \frac{8}{{(1)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

One to any power is one.

$12 \cdot 1 + \frac{8}{{(1)}^{3}}$

Step 9.1.2

Multiply $12$ by $1$ .

$12 + \frac{8}{{(1)}^{3}}$

Step 9.1.3

One to any power is one.

$12 + \frac{8}{1}$

Step 9.1.4

Divide $8$ by $1$ .

$12 + 8$

Step 9.2

Add $12$ and $8$ .

$20$

Step 10

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

$x = 1$ is a local minimum

Step 11

Find the y-value when $x = 1$ .

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Step 11.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{4} + \frac{4}{1}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

One to any power is one.

$f (1) = 1 + \frac{4}{1}$

Step 11.2.1.2

Divide $4$ by $1$ .

$f (1) = 1 + 4$

Step 11.2.2

Add $1$ and $4$ .

$f (1) = 5$

Step 11.2.3

The final answer is $5$ .

$y = 5$

Step 12

These are the local extrema for $f (x) = x^{4} + \frac{4}{x}$ .

$(1, 5)$ is a local minima

Step 13