Calculus Examples

$f (x) = \ln (x^{4} + 27)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{4} + 27$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $x^{4} + 27$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{4} + 27]$

Step 1.1.2

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

$\frac{1}{u} \frac{d}{d x} [x^{4} + 27]$

Step 1.1.3

Replace all occurrences of $u$ with $x^{4} + 27$ .

$\frac{1}{x^{4} + 27} \frac{d}{d x} [x^{4} + 27]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{4} + 27$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [27]$ .

$\frac{1}{x^{4} + 27} (\frac{d}{d x} [x^{4}] + \frac{d}{d x} [27])$

Step 1.2.2

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

$\frac{1}{x^{4} + 27} (4 x^{3} + \frac{d}{d x} [27])$

Step 1.2.3

Since $27$ is constant with respect to $x$ , the derivative of $27$ with respect to $x$ is $0$ .

$\frac{1}{x^{4} + 27} (4 x^{3} + 0)$

Step 1.2.4

Combine fractions.

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Step 1.2.4.1

Add $4 x^{3}$ and $0$ .

$\frac{1}{x^{4} + 27} (4 x^{3})$

Step 1.2.4.2

Combine $4$ and $\frac{1}{x^{4} + 27}$ .

$\frac{4}{x^{4} + 27} x^{3}$

Step 1.2.4.3

Combine $\frac{4}{x^{4} + 27}$ and $x^{3}$ .

$\frac{4 x^{3}}{x^{4} + 27}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $4$ is constant with respect to $x$ , the derivative of $\frac{4 x^{3}}{x^{4} + 27}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{x^{3}}{x^{4} + 27}]$ .

$f'' (x) = 4 \frac{d}{d x} (\frac{x^{3}}{x^{4} + 27})$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{3}$ and $g (x) = x^{4} + 27$ .

$f'' (x) = 4 (\frac{(x^{4} + 27) \frac{d}{d x} (x^{3}) - x^{3} \frac{d}{d x} (x^{4} + 27)}{{(x^{4} + 27)}^{2}})$

Step 2.3

Differentiate.

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Step 2.3.1

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

$f'' (x) = 4 (\frac{(x^{4} + 27) (3 x^{2}) - x^{3} \frac{d}{d x} (x^{4} + 27)}{{(x^{4} + 27)}^{2}})$

Step 2.3.2

Move $3$ to the left of $x^{4} + 27$ .

$f'' (x) = 4 (\frac{3 \cdot ((x^{4} + 27) x^{2}) - x^{3} \frac{d}{d x} (x^{4} + 27)}{{(x^{4} + 27)}^{2}})$

Step 2.3.3

By the Sum Rule, the derivative of $x^{4} + 27$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [27]$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - x^{3} (\frac{d}{d x} (x^{4}) + \frac{d}{d x} (27))}{{(x^{4} + 27)}^{2}})$

Step 2.3.4

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

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - x^{3} (4 x^{3} + \frac{d}{d x} (27))}{{(x^{4} + 27)}^{2}})$

Step 2.3.5

Since $27$ is constant with respect to $x$ , the derivative of $27$ with respect to $x$ is $0$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - x^{3} (4 x^{3} + 0)}{{(x^{4} + 27)}^{2}})$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $4 x^{3}$ and $0$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - x^{3} (4 x^{3})}{{(x^{4} + 27)}^{2}})$

Step 2.3.6.2

Multiply $4$ by $- 1$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - 4 x^{3} x^{3}}{{(x^{4} + 27)}^{2}})$

Step 2.4

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

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Step 2.4.1

Move $x^{3}$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - 4 (x^{3} x^{3})}{{(x^{4} + 27)}^{2}})$

Step 2.4.2

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

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - 4 x^{3 + 3}}{{(x^{4} + 27)}^{2}})$

Step 2.4.3

Add $3$ and $3$ .

$f'' (x) = 4 (\frac{3 (x^{4} + 27) x^{2} - 4 x^{6}}{{(x^{4} + 27)}^{2}})$

Step 2.5

Combine $4$ and $\frac{3 (x^{4} + 27) x^{2} - 4 x^{6}}{{(x^{4} + 27)}^{2}}$ .

$f'' (x) = \frac{4 (3 (x^{4} + 27) x^{2} - 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$f'' (x) = \frac{4 ((3 x^{4} + 3 \cdot 27) x^{2} - 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.2

Apply the distributive property.

$f'' (x) = \frac{4 (3 x^{4} x^{2} + 3 \cdot (27 x^{2}) - 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.3

Apply the distributive property.

$f'' (x) = \frac{4 (3 x^{4} x^{2}) + 4 (3 \cdot (27 x^{2})) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4

Simplify the numerator.

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Step 2.6.4.1

Simplify each term.

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Step 2.6.4.1.1

Multiply $x^{4}$ by $x^{2}$ by adding the exponents.

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Step 2.6.4.1.1.1

Move $x^{2}$ .

$f'' (x) = \frac{4 (3 (x^{2} x^{4})) + 4 (3 \cdot (27 x^{2})) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.1.2

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

$f'' (x) = \frac{4 (3 x^{2 + 4}) + 4 (3 \cdot (27 x^{2})) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.1.3

Add $2$ and $4$ .

$f'' (x) = \frac{4 (3 x^{6}) + 4 (3 \cdot (27 x^{2})) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.2

Multiply $3$ by $4$ .

$f'' (x) = \frac{12 x^{6} + 4 (3 \cdot (27 x^{2})) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.3

Multiply $3$ by $27$ .

$f'' (x) = \frac{12 x^{6} + 4 (81 x^{2}) + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.4

Multiply $81$ by $4$ .

$f'' (x) = \frac{12 x^{6} + 324 x^{2} + 4 (- 4 x^{6})}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.1.5

Multiply $- 4$ by $4$ .

$f'' (x) = \frac{12 x^{6} + 324 x^{2} - 16 x^{6}}{{(x^{4} + 27)}^{2}}$

Step 2.6.4.2

Subtract $16 x^{6}$ from $12 x^{6}$ .

$f'' (x) = \frac{- 4 x^{6} + 324 x^{2}}{{(x^{4} + 27)}^{2}}$

Step 3

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

$\frac{4 x^{3}}{x^{4} + 27} = 0$

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 using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{4} + 27$ .

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Step 4.1.1.1

To apply the Chain Rule, set $u$ as $x^{4} + 27$ .

$f' (x) = \frac{d}{d u} (\ln (u)) \frac{d}{d x} (x^{4} + 27)$

Step 4.1.1.2

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

$f' (x) = \frac{1}{u} \cdot \frac{d}{d x} (x^{4} + 27)$

Step 4.1.1.3

Replace all occurrences of $u$ with $x^{4} + 27$ .

$f' (x) = \frac{1}{x^{4} + 27} \cdot \frac{d}{d x} (x^{4} + 27)$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $x^{4} + 27$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [27]$ .

$f' (x) = \frac{1}{x^{4} + 27} \cdot (\frac{d}{d x} (x^{4}) + \frac{d}{d x} (27))$

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 $n = 4$ .

$f' (x) = \frac{1}{x^{4} + 27} \cdot (4 x^{3} + \frac{d}{d x} (27))$

Step 4.1.2.3

Since $27$ is constant with respect to $x$ , the derivative of $27$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{x^{4} + 27} \cdot (4 x^{3} + 0)$

Step 4.1.2.4

Combine fractions.

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Step 4.1.2.4.1

Add $4 x^{3}$ and $0$ .

$f' (x) = \frac{1}{x^{4} + 27} \cdot (4 x^{3})$

Step 4.1.2.4.2

Combine $4$ and $\frac{1}{x^{4} + 27}$ .

$f' (x) = \frac{4}{x^{4} + 27} \cdot x^{3}$

Step 4.1.2.4.3

Combine $\frac{4}{x^{4} + 27}$ and $x^{3}$ .

$f' (x) = \frac{4 x^{3}}{x^{4} + 27}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{4 x^{3}}{x^{4} + 27}$ .

$\frac{4 x^{3}}{x^{4} + 27}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{4 x^{3}}{x^{4} + 27} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{4 x^{3}}{x^{4} + 27} = 0$

Step 5.2

Set the numerator equal to zero.

$4 x^{3} = 0$

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

Divide each term in $4 x^{3} = 0$ by $4$ and simplify.

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Step 5.3.1.1

Divide each term in $4 x^{3} = 0$ by $4$ .

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

Step 5.3.1.2

Simplify the left side.

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Step 5.3.1.2.1

Cancel the common factor of $4$ .

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Step 5.3.1.2.1.1

Cancel the common factor.

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

Step 5.3.1.2.1.2

Divide $x^{3}$ by $1$ .

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

Step 5.3.1.3

Simplify the right side.

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Step 5.3.1.3.1

Divide $0$ by $4$ .

$x^{3} = 0$

Step 5.3.2

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

$x = \sqrt[3]{0}$

Step 5.3.3

Simplify $\sqrt[3]{0}$ .

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Step 5.3.3.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 5.3.3.2

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

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{- 4 {(0)}^{6} + 324 {(0)}^{2}}{{({(0)}^{4} + 27)}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Raising $0$ to any positive power yields $0$ .

$\frac{- 4 \cdot 0 + 324 \cdot 0^{2}}{{(0^{4} + 27)}^{2}}$

Step 9.1.2

Multiply $- 4$ by $0$ .

$\frac{0 + 324 \cdot 0^{2}}{{(0^{4} + 27)}^{2}}$

Step 9.1.3

Raising $0$ to any positive power yields $0$ .

$\frac{0 + 324 \cdot 0}{{(0^{4} + 27)}^{2}}$

Step 9.1.4

Multiply $324$ by $0$ .

$\frac{0 + 0}{{(0^{4} + 27)}^{2}}$

Step 9.1.5

Add $0$ and $0$ .

$\frac{0}{{(0^{4} + 27)}^{2}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{0}{{(0 + 27)}^{2}}$

Step 9.2.2

Add $0$ and $27$ .

$\frac{0}{27^{2}}$

Step 9.2.3

Raise $27$ to the power of $2$ .

$\frac{0}{729}$

Step 9.3

Divide $0$ by $729$ .

$0$

Step 10

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 10.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, \infty)$

Step 10.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $\frac{4 x^{3}}{x^{4} + 27}$ to check if the result is negative or positive.

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Step 10.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = \frac{4 {(- 2)}^{3}}{{(- 2)}^{4} + 27}$

Step 10.2.2

Simplify the result.

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Step 10.2.2.1

Raise $- 2$ to the power of $3$ .

$f' (- 2) = \frac{4 \cdot - 8}{{(- 2)}^{4} + 27}$

Step 10.2.2.2

Simplify the denominator.

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Step 10.2.2.2.1

Raise $- 2$ to the power of $4$ .

$f' (- 2) = \frac{4 \cdot - 8}{16 + 27}$

Step 10.2.2.2.2

Add $16$ and $27$ .

$f' (- 2) = \frac{4 \cdot - 8}{43}$

Step 10.2.2.3

Simplify the expression.

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Step 10.2.2.3.1

Multiply $4$ by $- 8$ .

$f' (- 2) = \frac{- 32}{43}$

Step 10.2.2.3.2

Move the negative in front of the fraction.

$f' (- 2) = - \frac{32}{43}$

Step 10.2.2.4

The final answer is $- \frac{32}{43}$ .

$- \frac{32}{43}$

Step 10.3

Substitute any number, such as $2$ , from the interval $(0, \infty)$ in the first derivative $\frac{4 x^{3}}{x^{4} + 27}$ to check if the result is negative or positive.

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Step 10.3.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = \frac{4 {(2)}^{3}}{{(2)}^{4} + 27}$

Step 10.3.2

Simplify the result.

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Step 10.3.2.1

Simplify the numerator.

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Step 10.3.2.1.1

Rewrite $4$ as $2^{2}$ .

$f' (2) = \frac{2^{2} \cdot 2^{3}}{2^{4} + 27}$

Step 10.3.2.1.2

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

$f' (2) = \frac{2^{2 + 3}}{2^{4} + 27}$

Step 10.3.2.1.3

Add $2$ and $3$ .

$f' (2) = \frac{2^{5}}{2^{4} + 27}$

Step 10.3.2.2

Simplify the denominator.

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Step 10.3.2.2.1

Raise $2$ to the power of $4$ .

$f' (2) = \frac{2^{5}}{16 + 27}$

Step 10.3.2.2.2

Add $16$ and $27$ .

$f' (2) = \frac{2^{5}}{43}$

Step 10.3.2.3

Raise $2$ to the power of $5$ .

$f' (2) = \frac{32}{43}$

Step 10.3.2.4

The final answer is $\frac{32}{43}$ .

$\frac{32}{43}$

Step 10.4

Since the first derivative changed signs from negative to positive around $x = 0$ , then $x = 0$ is a local minimum.

$x = 0$ is a local minimum

Step 11