Calculus Examples

$f (x) = \arctan (x^{6})$

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 $f (x) = \arctan (x)$ and $g (x) = x^{6}$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $x^{6}$ .

$\frac{d}{d u} [\arctan (u)] \frac{d}{d x} [x^{6}]$

Step 1.1.2

The derivative of $\arctan (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

$\frac{1}{1 + u^{2}} \frac{d}{d x} [x^{6}]$

Step 1.1.3

Replace all occurrences of $u$ with $x^{6}$ .

$\frac{1}{1 + {(x^{6})}^{2}} \frac{d}{d x} [x^{6}]$

Step 1.2

Differentiate using the Power Rule.

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Step 1.2.1

Multiply the exponents in ${(x^{6})}^{2}$ .

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Step 1.2.1.1

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

$\frac{1}{1 + x^{6 \cdot 2}} \frac{d}{d x} [x^{6}]$

Step 1.2.1.2

Multiply $6$ by $2$ .

$\frac{1}{1 + x^{12}} \frac{d}{d x} [x^{6}]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 6$ .

$\frac{1}{1 + x^{12}} (6 x^{5})$

Step 1.2.3

Combine fractions.

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Step 1.2.3.1

Combine $6$ and $\frac{1}{1 + x^{12}}$ .

$\frac{6}{1 + x^{12}} x^{5}$

Step 1.2.3.2

Combine $\frac{6}{1 + x^{12}}$ and $x^{5}$ .

$\frac{6 x^{5}}{1 + x^{12}}$

Step 1.2.3.3

Reorder terms.

$\frac{6 x^{5}}{x^{12} + 1}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $6$ is constant with respect to $x$ , the derivative of $\frac{6 x^{5}}{x^{12} + 1}$ with respect to $x$ is $6 \frac{d}{d x} [\frac{x^{5}}{x^{12} + 1}]$ .

$f'' (x) = 6 \frac{d}{d x} (\frac{x^{5}}{x^{12} + 1})$

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 $f (x) = x^{5}$ and $g (x) = x^{12} + 1$ .

$f'' (x) = 6 (\frac{(x^{12} + 1) \frac{d}{d x} (x^{5}) - x^{5} \frac{d}{d x} (x^{12} + 1)}{{(x^{12} + 1)}^{2}})$

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 $n = 5$ .

$f'' (x) = 6 (\frac{(x^{12} + 1) (5 x^{4}) - x^{5} \frac{d}{d x} (x^{12} + 1)}{{(x^{12} + 1)}^{2}})$

Step 2.3.2

Move $5$ to the left of $x^{12} + 1$ .

$f'' (x) = 6 (\frac{5 \cdot ((x^{12} + 1) x^{4}) - x^{5} \frac{d}{d x} (x^{12} + 1)}{{(x^{12} + 1)}^{2}})$

Step 2.3.3

By the Sum Rule, the derivative of $x^{12} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{12}] + \frac{d}{d x} [1]$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - x^{5} (\frac{d}{d x} (x^{12}) + \frac{d}{d x} (1))}{{(x^{12} + 1)}^{2}})$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 12$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - x^{5} (12 x^{11} + \frac{d}{d x} (1))}{{(x^{12} + 1)}^{2}})$

Step 2.3.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - x^{5} (12 x^{11} + 0)}{{(x^{12} + 1)}^{2}})$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $12 x^{11}$ and $0$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - x^{5} (12 x^{11})}{{(x^{12} + 1)}^{2}})$

Step 2.3.6.2

Multiply $12$ by $- 1$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - 12 x^{5} x^{11}}{{(x^{12} + 1)}^{2}})$

Step 2.4

Multiply $x^{5}$ by $x^{11}$ by adding the exponents.

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Step 2.4.1

Move $x^{11}$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - 12 (x^{11} x^{5})}{{(x^{12} + 1)}^{2}})$

Step 2.4.2

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

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - 12 x^{11 + 5}}{{(x^{12} + 1)}^{2}})$

Step 2.4.3

Add $11$ and $5$ .

$f'' (x) = 6 (\frac{5 (x^{12} + 1) x^{4} - 12 x^{16}}{{(x^{12} + 1)}^{2}})$

Step 2.5

Combine $6$ and $\frac{5 (x^{12} + 1) x^{4} - 12 x^{16}}{{(x^{12} + 1)}^{2}}$ .

$f'' (x) = \frac{6 (5 (x^{12} + 1) x^{4} - 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$f'' (x) = \frac{6 ((5 x^{12} + 5 \cdot 1) x^{4} - 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.2

Apply the distributive property.

$f'' (x) = \frac{6 (5 x^{12} x^{4} + 5 \cdot (1 x^{4}) - 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.3

Apply the distributive property.

$f'' (x) = \frac{6 (5 x^{12} x^{4}) + 6 (5 \cdot (1 x^{4})) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

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 $x^{12}$ by $x^{4}$ by adding the exponents.

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Step 2.6.4.1.1.1

Move $x^{4}$ .

$f'' (x) = \frac{6 (5 (x^{4} x^{12})) + 6 (5 \cdot (1 x^{4})) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.1.2

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

$f'' (x) = \frac{6 (5 x^{4 + 12}) + 6 (5 \cdot (1 x^{4})) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.1.3

Add $4$ and $12$ .

$f'' (x) = \frac{6 (5 x^{16}) + 6 (5 \cdot (1 x^{4})) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.2

Multiply $5$ by $6$ .

$f'' (x) = \frac{30 x^{16} + 6 (5 \cdot (1 x^{4})) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.3

Multiply $5$ by $1$ .

$f'' (x) = \frac{30 x^{16} + 6 (5 x^{4}) + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.4

Multiply $5$ by $6$ .

$f'' (x) = \frac{30 x^{16} + 30 x^{4} + 6 (- 12 x^{16})}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.1.5

Multiply $- 12$ by $6$ .

$f'' (x) = \frac{30 x^{16} + 30 x^{4} - 72 x^{16}}{{(x^{12} + 1)}^{2}}$

Step 2.6.4.2

Subtract $72 x^{16}$ from $30 x^{16}$ .

$f'' (x) = \frac{- 42 x^{16} + 30 x^{4}}{{(x^{12} + 1)}^{2}}$

Step 3

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

$\frac{6 x^{5}}{x^{12} + 1} = 0$

Step 4

Set the numerator equal to zero.

$6 x^{5} = 0$

Step 5

Solve the equation for $x$ .

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Step 5.1

Divide each term in $6 x^{5} = 0$ by $6$ and simplify.

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Step 5.1.1

Divide each term in $6 x^{5} = 0$ by $6$ .

$\frac{6 x^{5}}{6} = \frac{0}{6}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $6$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{6 x^{5}}{6} = \frac{0}{6}$

Step 5.1.2.1.2

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

$x^{5} = \frac{0}{6}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Divide $0$ by $6$ .

$x^{5} = 0$

Step 5.2

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

$x = \sqrt[5]{0}$

Step 5.3

Simplify $\sqrt[5]{0}$ .

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Step 5.3.1

Rewrite $0$ as $0^{5}$ .

$x = \sqrt[5]{0^{5}}$

Step 5.3.2

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

$x = 0$

Step 6

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.

$\frac{- 42 {(0)}^{16} + 30 {(0)}^{4}}{{({(0)}^{12} + 1)}^{2}}$

Step 7

Evaluate the second derivative.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

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

$\frac{- 42 \cdot 0 + 30 \cdot 0^{4}}{{(0^{12} + 1)}^{2}}$

Step 7.1.2

Multiply $- 42$ by $0$ .

$\frac{0 + 30 \cdot 0^{4}}{{(0^{12} + 1)}^{2}}$

Step 7.1.3

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

$\frac{0 + 30 \cdot 0}{{(0^{12} + 1)}^{2}}$

Step 7.1.4

Multiply $30$ by $0$ .

$\frac{0 + 0}{{(0^{12} + 1)}^{2}}$

Step 7.1.5

Add $0$ and $0$ .

$\frac{0}{{(0^{12} + 1)}^{2}}$

Step 7.2

Simplify the denominator.

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Step 7.2.1

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

$\frac{0}{{(0 + 1)}^{2}}$

Step 7.2.2

Add $0$ and $1$ .

$\frac{0}{1^{2}}$

Step 7.2.3

One to any power is one.

$\frac{0}{1}$

Step 7.3

Divide $0$ by $1$ .

$0$

Step 8

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

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Step 8.1

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

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

Step 8.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $\frac{6 x^{5}}{x^{12} + 1}$ to check if the result is negative or positive.

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Step 8.2.1

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

$f' (- 2) = \frac{6 {(- 2)}^{5}}{{(- 2)}^{12} + 1}$

Step 8.2.2

Simplify the result.

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Step 8.2.2.1

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

$f' (- 2) = \frac{6 \cdot - 32}{{(- 2)}^{12} + 1}$

Step 8.2.2.2

Simplify the denominator.

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Step 8.2.2.2.1

Raise $- 2$ to the power of $12$ .

$f' (- 2) = \frac{6 \cdot - 32}{4096 + 1}$

Step 8.2.2.2.2

Add $4096$ and $1$ .

$f' (- 2) = \frac{6 \cdot - 32}{4097}$

Step 8.2.2.3

Simplify the expression.

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Step 8.2.2.3.1

Multiply $6$ by $- 32$ .

$f' (- 2) = \frac{- 192}{4097}$

Step 8.2.2.3.2

Move the negative in front of the fraction.

$f' (- 2) = - \frac{192}{4097}$

Step 8.2.2.4

The final answer is $- \frac{192}{4097}$ .

$- \frac{192}{4097}$

Step 8.3

Substitute any number, such as $2$ , from the interval $(0, \infty)$ in the first derivative $\frac{6 x^{5}}{x^{12} + 1}$ to check if the result is negative or positive.

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Step 8.3.1

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

$f' (2) = \frac{6 {(2)}^{5}}{{(2)}^{12} + 1}$

Step 8.3.2

Simplify the result.

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Step 8.3.2.1

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

$f' (2) = \frac{6 \cdot 32}{2^{12} + 1}$

Step 8.3.2.2

Simplify the denominator.

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Step 8.3.2.2.1

Raise $2$ to the power of $12$ .

$f' (2) = \frac{6 \cdot 32}{4096 + 1}$

Step 8.3.2.2.2

Add $4096$ and $1$ .

$f' (2) = \frac{6 \cdot 32}{4097}$

Step 8.3.2.3

Multiply $6$ by $32$ .

$f' (2) = \frac{192}{4097}$

Step 8.3.2.4

The final answer is $\frac{192}{4097}$ .

$\frac{192}{4097}$

Step 8.4

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 9