Calculus Examples

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

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 Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x^{2} + 2}$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

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

Step 1.1.4

Differentiate.

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Step 1.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 1.1.4.2

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

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

Step 1.1.4.3

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

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

Step 1.1.4.4

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

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

Step 1.1.4.5

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 1.1.4.6

Simplify the expression.

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Step 1.1.4.6.1

Add $2 x$ and $0$ .

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

Step 1.1.4.6.2

Move $2$ to the left of ${(x^{2} + 2)}^{- 2}$ .

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

Step 1.1.4.7

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

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

Step 1.1.4.8

Simplify the expression.

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Step 1.1.4.8.1

Multiply $\frac{1}{x^{2} + 2}$ by $0$ .

$2 {(x^{2} + 2)}^{- 2} x + 0$

Step 1.1.4.8.2

Add $2 {(x^{2} + 2)}^{- 2} x$ and $0$ .

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

Step 1.1.5

Simplify.

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Step 1.1.5.1

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

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

Step 1.1.5.2

Combine terms.

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Step 1.1.5.2.1

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

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

Step 1.1.5.2.2

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

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$2 x = 0$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide $x$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $2$ .

$x = 0$

Step 3

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

$0$

Step 4

After finding the point that makes the derivative $f' (x) = \frac{2 x}{{(x^{2} + 2)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = - \frac{1}{x^{2} + 2}$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

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

Step 5

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $2$ by $- 1$ .

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

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Raise $- 1$ to the power of $2$ .

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

Step 5.2.2.2

Add $1$ and $2$ .

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

Step 5.2.2.3

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

$f' (- 1) = \frac{- 2}{9}$

Step 5.2.3

Move the negative in front of the fraction.

$f' (- 1) = - \frac{2}{9}$

Step 5.2.4

The final answer is $- \frac{2}{9}$ .

$- \frac{2}{9}$

Step 5.3

At $x = - 1$ the derivative is $- \frac{2}{9}$ . Since this is negative, the function is decreasing on $(- \infty, 0)$ .

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

Step 6

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $1$ .

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

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

One to any power is one.

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

Step 6.2.2.2

Add $1$ and $2$ .

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

Step 6.2.2.3

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

$f' (1) = \frac{2}{9}$

Step 6.2.3

The final answer is $\frac{2}{9}$ .

$\frac{2}{9}$

Step 6.3

At $x = 1$ the derivative is $\frac{2}{9}$ . Since this is positive, the function is increasing on $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f' (x) > 0$

Step 7

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

Increasing on: $(0, \infty)$

Decreasing on: $(- \infty, 0)$

Step 8