Calculus Examples

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

Step 1

Write $\ln (x^{4} + 27)$ as a function.

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

Step 2

Find the first derivative of the function.

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Combine fractions.

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

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

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

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

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

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

Step 3

Find the second derivative of the function.

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

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

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 $g (x) = x^{4} + 27$ .

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

Differentiate.

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

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

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

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

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

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

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

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

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

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

Simplify the expression.

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

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

Multiply $4$ by $- 1$ .

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

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

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Move $x^{3}$ .

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

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

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

Add $3$ and $3$ .

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

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

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

Simplify.

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify the numerator.

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Simplify each term.

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

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Move $x^{2}$ .

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

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

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

Add $2$ and $4$ .

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

Multiply $3$ by $4$ .

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

Multiply $3$ by $27$ .

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

Multiply $81$ by $4$ .

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

Multiply $- 4$ by $4$ .

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

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

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

Step 4

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 5

Find the first derivative.

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Find the first derivative.

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Combine fractions.

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

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

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

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

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

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

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 6

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

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Set the first derivative equal to $0$ .

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

Set the numerator equal to zero.

$4 x^{3} = 0$

Solve the equation for $x$ .

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

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

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

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

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

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

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

Simplify the right side.

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Divide $0$ by $4$ .

$x^{3} = 0$

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

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

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

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

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

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

$x = 0$

Step 7

Find the values where the derivative is undefined.

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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 8

Critical points to evaluate.

$x = 0$

Step 9

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 10

Evaluate the second derivative.

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Simplify the numerator.

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

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

Multiply $- 4$ by $0$ .

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

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

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

Multiply $324$ by $0$ .

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

Add $0$ and $0$ .

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

Simplify the denominator.

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

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

Add $0$ and $27$ .

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

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

$\frac{0}{729}$

Divide $0$ by $729$ .

$0$

Step 11

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

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

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

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

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

Simplify the result.

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

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

Simplify the denominator.

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

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

Add $16$ and $27$ .

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

Simplify the expression.

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Multiply $4$ by $- 8$ .

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

Move the negative in front of the fraction.

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

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

$- \frac{32}{43}$

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

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

Simplify the result.

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Simplify the numerator.

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

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

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

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

Add $2$ and $3$ .

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

Simplify the denominator.

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

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

Add $16$ and $27$ .

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

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

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

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

$\frac{32}{43}$

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 12