Calculus Examples

$y = \frac{1}{1 + x^{2}}$

Step 1

Write $y = \frac{1}{1 + x^{2}}$ as a function.

$f (x) = \frac{1}{1 + x^{2}}$

Step 2

Find the first derivative of the function.

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Step 2.1

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

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

Step 2.2

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 $g (x) = 1 + x^{2}$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $1 + x^{2}$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $1 + x^{2}$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

By the Sum Rule, the derivative of $1 + x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{2}]$ .

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

Step 2.3.2

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

$- {(1 + x^{2})}^{- 2} (0 + \frac{d}{d x} [x^{2}])$

Step 2.3.3

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

Step 2.3.4

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

$- {(1 + x^{2})}^{- 2} (2 x)$

Step 2.3.5

Multiply $2$ by $- 1$ .

$- 2 {(1 + x^{2})}^{- 2} x$

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 2$ and $\frac{1}{{(1 + x^{2})}^{2}}$ .

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

Step 2.4.2.2

Move the negative in front of the fraction.

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

Step 2.4.2.3

Combine $x$ and $\frac{2}{{(1 + x^{2})}^{2}}$ .

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

Step 2.4.2.4

Move $2$ to the left of $x$ .

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

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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 $f (x) = x$ and $g (x) = {(1 + x^{2})}^{2}$ .

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

Step 3.3

Differentiate using the Power Rule.

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Step 3.3.1

Multiply the exponents in ${({(1 + x^{2})}^{2})}^{2}$ .

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Step 3.3.1.1

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

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

Step 3.3.1.2

Multiply $2$ by $2$ .

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

Step 3.3.2

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

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

Step 3.3.3

Multiply ${(1 + x^{2})}^{2}$ by $1$ .

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

Step 3.4

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

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Step 3.4.1

To apply the Chain Rule, set $u$ as $1 + x^{2}$ .

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

Step 3.4.2

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

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

Step 3.4.3

Replace all occurrences of $u$ with $1 + x^{2}$ .

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

Step 3.5

Simplify with factoring out.

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Step 3.5.1

Multiply $2$ by $- 1$ .

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

Step 3.5.2

Factor $1 + x^{2}$ out of ${(1 + x^{2})}^{2} - 2 x ((1 + x^{2}) \frac{d}{d x} [1 + x^{2}])$ .

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Step 3.5.2.1

Factor $1 + x^{2}$ out of ${(1 + x^{2})}^{2}$ .

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

Step 3.5.2.2

Factor $1 + x^{2}$ out of $- 2 x ((1 + x^{2}) \frac{d}{d x} [1 + x^{2}])$ .

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

Step 3.5.2.3

Factor $1 + x^{2}$ out of $(1 + x^{2}) (1 + x^{2}) + (1 + x^{2}) (- 2 x (\frac{d}{d x} [1 + x^{2}]))$ .

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

Step 3.6

Cancel the common factors.

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Step 3.6.1

Factor $1 + x^{2}$ out of ${(1 + x^{2})}^{4}$ .

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

Step 3.6.2

Cancel the common factor.

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

Step 3.6.3

Rewrite the expression.

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

Step 3.7

By the Sum Rule, the derivative of $1 + x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{2}]$ .

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

Step 3.8

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

$f'' (x) = - 2 \frac{1 + x^{2} - 2 x (0 + \frac{d}{d x} (x^{2}))}{{(1 + x^{2})}^{3}}$

Step 3.9

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

Step 3.10

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

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

Step 3.11

Multiply $2$ by $- 2$ .

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

Step 3.12

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

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

Step 3.13

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

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

Step 3.14

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

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

Step 3.15

Add $1$ and $1$ .

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

Step 3.16

Subtract $4 x^{2}$ from $x^{2}$ .

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

Step 3.17

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

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

Step 3.18

Move the negative in front of the fraction.

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

Step 3.19

Simplify.

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Step 3.19.1

Apply the distributive property.

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

Step 3.19.2

Simplify each term.

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Step 3.19.2.1

Multiply $2$ by $1$ .

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

Step 3.19.2.2

Multiply $- 3$ by $2$ .

$f'' (x) = - \frac{2 - 6 x^{2}}{{(1 + x^{2})}^{3}}$

Step 3.19.3

Factor $2$ out of $2 - 6 x^{2}$ .

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Step 3.19.3.1

Factor $2$ out of $2$ .

$f'' (x) = - \frac{2 (1) - 6 x^{2}}{{(1 + x^{2})}^{3}}$

Step 3.19.3.2

Factor $2$ out of $- 6 x^{2}$ .

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

Step 3.19.3.3

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

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

Step 4

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

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

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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

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

Step 5.1.2

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 $g (x) = 1 + x^{2}$ .

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Step 5.1.2.1

To apply the Chain Rule, set $u$ as $1 + x^{2}$ .

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Replace all occurrences of $u$ with $1 + x^{2}$ .

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

Step 5.1.3

Differentiate.

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Step 5.1.3.1

By the Sum Rule, the derivative of $1 + x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{2}]$ .

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

Step 5.1.3.2

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

$f' (x) = - {(1 + x^{2})}^{- 2} (0 + \frac{d}{d x} (x^{2}))$

Step 5.1.3.3

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

Step 5.1.3.4

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

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

Step 5.1.3.5

Multiply $2$ by $- 1$ .

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

Step 5.1.4

Simplify.

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Step 5.1.4.1

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

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

Step 5.1.4.2

Combine terms.

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Step 5.1.4.2.1

Combine $- 2$ and $\frac{1}{{(1 + x^{2})}^{2}}$ .

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

Step 5.1.4.2.2

Move the negative in front of the fraction.

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

Step 5.1.4.2.3

Combine $x$ and $\frac{2}{{(1 + x^{2})}^{2}}$ .

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

Step 5.1.4.2.4

Move $2$ to the left of $x$ .

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

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{2 x}{{(1 + x^{2})}^{2}}$ .

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

Step 6

Set the first derivative equal to $0$ then solve the equation $- \frac{2 x}{{(1 + x^{2})}^{2}} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

Set the numerator equal to zero.

$2 x = 0$

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Cancel the common factor of $2$ .

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Step 6.3.2.1.1

Cancel the common factor.

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

Step 6.3.2.1.2

Divide $x$ by $1$ .

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

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Divide $0$ by $2$ .

$x = 0$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 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.

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

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify the numerator.

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Step 10.1.1

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

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

Step 10.1.2

Multiply $- 3$ by $0$ .

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

Step 10.1.3

Add $1$ and $0$ .

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

Step 10.2

Simplify the denominator.

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Step 10.2.1

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

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

Step 10.2.2

Add $1$ and $0$ .

$- \frac{2 \cdot 1}{1^{3}}$

Step 10.2.3

One to any power is one.

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

Step 10.3

Simplify the expression.

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Step 10.3.1

Multiply $2$ by $1$ .

$- \frac{2}{1}$

Step 10.3.2

Divide $2$ by $1$ .

$- 1 \cdot 2$

Step 10.3.3

Multiply $- 1$ by $2$ .

$- 2$

Step 11

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 12

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

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Step 12.1

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

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

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify the denominator.

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Step 12.2.1.1

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

$f (0) = \frac{1}{1 + 0}$

Step 12.2.1.2

Add $1$ and $0$ .

$f (0) = \frac{1}{1}$

Step 12.2.2

Divide $1$ by $1$ .

$f (0) = 1$

Step 12.2.3

The final answer is $1$ .

$y = 1$

Step 13

These are the local extrema for $f (x) = \frac{1}{1 + x^{2}}$ .

$(0, 1)$ is a local maxima

Step 14