Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.2.4

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

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

Step 1.1.2.5

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

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

Step 1.1.2.6

Simplify the expression.

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Step 1.1.2.6.1

Add $2 x$ and $0$ .

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

Step 1.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 1.1.3

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

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

Add $1$ and $1$ .

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

Step 1.1.7

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

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$- x^{2} + 1 = 0$

Step 2.3

Solve the equation for $x$ .

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Step 2.3.1

Subtract $1$ from both sides of the equation.

$- x^{2} = - 1$

Step 2.3.2

Divide each term in $- x^{2} = - 1$ by $- 1$ and simplify.

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Step 2.3.2.1

Divide each term in $- x^{2} = - 1$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 1}{- 1}$

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 1}{- 1}$

Step 2.3.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 1}{- 1}$

Step 2.3.2.3

Simplify the right side.

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Step 2.3.2.3.1

Divide $- 1$ by $- 1$ .

$x^{2} = 1$

Step 2.3.3

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

$x = \pm \sqrt{1}$

Step 2.3.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.3.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.3.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 2.3.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 2.3.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 3

The values which make the derivative equal to $0$ are $1, - 1$ .

$1, - 1$

Step 4

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

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

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

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

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

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

Step 5.2.1.2

Multiply $- 1$ by $4$ .

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

Step 5.2.1.3

Add $- 4$ and $1$ .

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

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

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

Step 5.2.2.2

Add $4$ and $1$ .

$f' (- 2) = \frac{- 3}{5^{2}}$

Step 5.2.2.3

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

$f' (- 2) = \frac{- 3}{25}$

Step 5.2.3

Move the negative in front of the fraction.

$f' (- 2) = - \frac{3}{25}$

Step 5.2.4

The final answer is $- \frac{3}{25}$ .

$- \frac{3}{25}$

Step 5.3

At $x = - 2$ the derivative is $- \frac{3}{25}$ . Since this is negative, the function is decreasing on $(- \infty, - 1)$ .

Decreasing on $(- \infty, - 1)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- 1, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

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

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

Step 6.2.1.2

Multiply $- 1$ by $0$ .

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

Step 6.2.1.3

Add $0$ and $1$ .

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

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

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

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

Step 6.2.2.2

Add $0$ and $1$ .

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

Step 6.2.2.3

One to any power is one.

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

Step 6.2.3

Divide $1$ by $1$ .

$f' (0) = 1$

Step 6.2.4

The final answer is $1$ .

$1$

Step 6.3

At $x = 0$ the derivative is $1$ . Since this is positive, the function is increasing on $(- 1, 1)$ .

Increasing on $(- 1, 1)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(1, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the numerator.

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Step 7.2.1.1

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

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

Step 7.2.1.2

Multiply $- 1$ by $4$ .

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

Step 7.2.1.3

Add $- 4$ and $1$ .

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

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

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

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

Step 7.2.2.2

Add $4$ and $1$ .

$f' (2) = \frac{- 3}{5^{2}}$

Step 7.2.2.3

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

$f' (2) = \frac{- 3}{25}$

Step 7.2.3

Move the negative in front of the fraction.

$f' (2) = - \frac{3}{25}$

Step 7.2.4

The final answer is $- \frac{3}{25}$ .

$- \frac{3}{25}$

Step 7.3

At $x = 2$ the derivative is $- \frac{3}{25}$ . Since this is negative, the function is decreasing on $(1, \infty)$ .

Decreasing on $(1, \infty)$ since $f' (x) < 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 1, 1)$

Decreasing on: $(- \infty, - 1), (1, \infty)$

Step 9