Calculus Examples

$h (y) = \arctan (y^{2})$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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Step 1.1.1

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

$\frac{d}{d u} [\arctan (u)] \frac{d}{d y} [y^{2}]$

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.2

Differentiate using the Power Rule.

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Step 1.2.1

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

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Step 1.2.1.1

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

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

Step 1.2.1.2

Multiply $2$ by $2$ .

$\frac{1}{1 + y^{4}} \frac{d}{d y} [y^{2}]$

Step 1.2.2

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

$\frac{1}{1 + y^{4}} (2 y)$

Step 1.2.3

Combine fractions.

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Step 1.2.3.1

Combine $2$ and $\frac{1}{1 + y^{4}}$ .

$\frac{2}{1 + y^{4}} y$

Step 1.2.3.2

Combine $\frac{2}{1 + y^{4}}$ and $y$ .

$\frac{2 y}{1 + y^{4}}$

Step 1.2.3.3

Reorder terms.

$\frac{2 y}{y^{4} + 1}$

Step 2

Find the second derivative of the function.

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Step 2.1

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

$h'' (y) = 2 \frac{d}{d y} (\frac{y}{y^{4} + 1})$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d y} [\frac{f (y)}{g (y)}]$ is $\frac{g (y) \frac{d}{d y} [f (y)] - f (y) \frac{d}{d y} [g (y)]}{{g (y)}^{2}}$ where $f (y) = y$ and $g (y) = y^{4} + 1$ .

$h'' (y) = 2 (\frac{(y^{4} + 1) \frac{d}{d y} (y) - y \frac{d}{d y} (y^{4} + 1)}{{(y^{4} + 1)}^{2}})$

Step 2.3

Differentiate.

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Step 2.3.1

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

$h'' (y) = 2 (\frac{(y^{4} + 1) \cdot 1 - y \frac{d}{d y} (y^{4} + 1)}{{(y^{4} + 1)}^{2}})$

Step 2.3.2

Multiply $y^{4} + 1$ by $1$ .

$h'' (y) = 2 (\frac{y^{4} + 1 - y \frac{d}{d y} (y^{4} + 1)}{{(y^{4} + 1)}^{2}})$

Step 2.3.3

By the Sum Rule, the derivative of $y^{4} + 1$ with respect to $y$ is $\frac{d}{d y} [y^{4}] + \frac{d}{d y} [1]$ .

$h'' (y) = 2 (\frac{y^{4} + 1 - y (\frac{d}{d y} (y^{4}) + \frac{d}{d y} (1))}{{(y^{4} + 1)}^{2}})$

Step 2.3.4

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

$h'' (y) = 2 (\frac{y^{4} + 1 - y (4 y^{3} + \frac{d}{d y} (1))}{{(y^{4} + 1)}^{2}})$

Step 2.3.5

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

$h'' (y) = 2 (\frac{y^{4} + 1 - y (4 y^{3} + 0)}{{(y^{4} + 1)}^{2}})$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

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

$h'' (y) = 2 (\frac{y^{4} + 1 - y (4 y^{3})}{{(y^{4} + 1)}^{2}})$

Step 2.3.6.2

Multiply $4$ by $- 1$ .

$h'' (y) = 2 (\frac{y^{4} + 1 - 4 y \cdot y^{3}}{{(y^{4} + 1)}^{2}})$

Step 2.4

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

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Step 2.4.1

Move $y^{3}$ .

$h'' (y) = 2 (\frac{y^{4} + 1 - 4 (y^{3} y)}{{(y^{4} + 1)}^{2}})$

Step 2.4.2

Multiply $y^{3}$ by $y$ .

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Step 2.4.2.1

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

$h'' (y) = 2 (\frac{y^{4} + 1 - 4 (y^{3} y)}{{(y^{4} + 1)}^{2}})$

Step 2.4.2.2

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

$h'' (y) = 2 (\frac{y^{4} + 1 - 4 y^{3 + 1}}{{(y^{4} + 1)}^{2}})$

Step 2.4.3

Add $3$ and $1$ .

$h'' (y) = 2 (\frac{y^{4} + 1 - 4 y^{4}}{{(y^{4} + 1)}^{2}})$

Step 2.5

Subtract $4 y^{4}$ from $y^{4}$ .

$h'' (y) = 2 (\frac{- 3 y^{4} + 1}{{(y^{4} + 1)}^{2}})$

Step 2.6

Combine $2$ and $\frac{- 3 y^{4} + 1}{{(y^{4} + 1)}^{2}}$ .

$h'' (y) = \frac{2 (- 3 y^{4} + 1)}{{(y^{4} + 1)}^{2}}$

Step 2.7

Simplify.

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Step 2.7.1

Apply the distributive property.

$h'' (y) = \frac{2 (- 3 y^{4}) + 2 \cdot 1}{{(y^{4} + 1)}^{2}}$

Step 2.7.2

Simplify each term.

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Step 2.7.2.1

Multiply $- 3$ by $2$ .

$h'' (y) = \frac{- 6 y^{4} + 2 \cdot 1}{{(y^{4} + 1)}^{2}}$

Step 2.7.2.2

Multiply $2$ by $1$ .

$h'' (y) = \frac{- 6 y^{4} + 2}{{(y^{4} + 1)}^{2}}$

Step 2.7.3

Factor $2$ out of $- 6 y^{4} + 2$ .

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Step 2.7.3.1

Factor $2$ out of $- 6 y^{4}$ .

$h'' (y) = \frac{2 (- 3 y^{4}) + 2}{{(y^{4} + 1)}^{2}}$

Step 2.7.3.2

Factor $2$ out of $2$ .

$h'' (y) = \frac{2 (- 3 y^{4}) + 2 (1)}{{(y^{4} + 1)}^{2}}$

Step 2.7.3.3

Factor $2$ out of $2 (- 3 y^{4}) + 2 (1)$ .

$h'' (y) = \frac{2 (- 3 y^{4} + 1)}{{(y^{4} + 1)}^{2}}$

Step 2.7.4

Factor $- 1$ out of $- 3 y^{4}$ .

$h'' (y) = \frac{2 (- (3 y^{4}) + 1)}{{(y^{4} + 1)}^{2}}$

Step 2.7.5

Rewrite $1$ as $- 1 (- 1)$ .

$h'' (y) = \frac{2 (- (3 y^{4}) - 1 \cdot - 1)}{{(y^{4} + 1)}^{2}}$

Step 2.7.6

Factor $- 1$ out of $- (3 y^{4}) - 1 (- 1)$ .

$h'' (y) = \frac{2 (- (3 y^{4} - 1))}{{(y^{4} + 1)}^{2}}$

Step 2.7.7

Rewrite $- (3 y^{4} - 1)$ as $- 1 (3 y^{4} - 1)$ .

$h'' (y) = \frac{2 (- 1 (3 y^{4} - 1))}{{(y^{4} + 1)}^{2}}$

Step 2.7.8

Move the negative in front of the fraction.

$h'' (y) = - \frac{2 (3 y^{4} - 1)}{{(y^{4} + 1)}^{2}}$

Step 3

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

$\frac{2 y}{y^{4} + 1} = 0$

Step 4

Set the numerator equal to zero.

$2 y = 0$

Step 5

Divide each term in $2 y = 0$ by $2$ and simplify.

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Step 5.1

Divide each term in $2 y = 0$ by $2$ .

$\frac{2 y}{2} = \frac{0}{2}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $2$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{0}{2}$

Step 5.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $0$ by $2$ .

$y = 0$

Step 6

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

$- \frac{2 (3 {(0)}^{4} - 1)}{{({(0)}^{4} + 1)}^{2}}$

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 $0$ .

$- \frac{2 (3 \cdot 0 - 1)}{{(0^{4} + 1)}^{2}}$

Step 7.1.2

Multiply $3$ by $0$ .

$- \frac{2 (0 - 1)}{{(0^{4} + 1)}^{2}}$

Step 7.1.3

Subtract $1$ from $0$ .

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

Step 7.2

Simplify the denominator.

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Step 7.2.1

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

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

Step 7.2.2

Add $0$ and $1$ .

$- \frac{2 \cdot - 1}{1^{2}}$

Step 7.2.3

One to any power is one.

$- \frac{2 \cdot - 1}{1}$

Step 7.3

Simplify the expression.

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Step 7.3.1

Multiply $2$ by $- 1$ .

$- \frac{- 2}{1}$

Step 7.3.2

Divide $- 2$ by $1$ .

$- - 2$

Step 7.3.3

Multiply $- 1$ by $- 2$ .

$2$

Step 8

$y = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$y = 0$ is a local minimum

Step 9

Find the y-value when $y = 0$ .

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Step 9.1

Replace the variable $y$ with $0$ in the expression.

$h (0) = \arctan ({(0)}^{2})$

Step 9.2

Simplify the result.

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Step 9.2.1

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

$h (0) = \arctan (0)$

Step 9.2.2

The exact value of $\arctan (0)$ is $0$ .

$h (0) = 0$

Step 9.2.3

The final answer is $0$ .

$y = 0$

Step 10

These are the local extrema for $h (y) = \arctan (y^{2})$ .

$(0, 0)$ is a local minima

Step 11