Calculus Examples

$f (x) = \sqrt{x^{2} + 4}$

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 4}$ as ${(x^{2} + 4)}^{\frac{1}{2}}$ .

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

Step 1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{2} + 4$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 4$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Replace all occurrences of $u$ with $x^{2} + 4$ .

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

Step 1.1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.1.4

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 1.1.5

Combine the numerators over the common denominator.

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

Step 1.1.6

Simplify the numerator.

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Step 1.1.6.1

Multiply $- 1$ by $2$ .

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

Step 1.1.6.2

Subtract $2$ from $1$ .

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

Step 1.1.7

Combine fractions.

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Step 1.1.7.1

Move the negative in front of the fraction.

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

Step 1.1.7.2

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

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

Step 1.1.7.3

Move ${(x^{2} + 4)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.8

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

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

Step 1.1.9

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

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

Step 1.1.10

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{2 {(x^{2} + 4)}^{\frac{1}{2}}} \cdot (2 x + 0)$

Step 1.1.11

Simplify terms.

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Step 1.1.11.1

Add $2 x$ and $0$ .

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

Step 1.1.11.2

Combine $2$ and $\frac{1}{2 {(x^{2} + 4)}^{\frac{1}{2}}}$ .

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

Step 1.1.11.3

Combine $\frac{2}{2 {(x^{2} + 4)}^{\frac{1}{2}}}$ and $x$ .

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

Step 1.1.11.4

Cancel the common factor.

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

Step 1.1.11.5

Rewrite the expression.

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x}{{(x^{2} + 4)}^{\frac{1}{2}}}$ .

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

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{x}{{(x^{2} + 4)}^{\frac{1}{2}}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$x = 0$

Step 3

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

$0$

Step 4

After finding the point that makes the derivative $f' (x) = \frac{x}{{(x^{2} + 4)}^{\frac{1}{2}}}$ equal to $0$ or undefined, the interval to check where $f (x) = \sqrt{x^{2} + 4}$ 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.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the denominator.

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Step 5.2.1.1

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

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

Step 5.2.1.2

Add $1$ and $4$ .

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

Step 5.2.2

Move the negative in front of the fraction.

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

Step 5.2.3

The final answer is $- \frac{1}{5^{\frac{1}{2}}}$ .

$- \frac{1}{5^{\frac{1}{2}}}$

Step 5.3

At $x = - 1$ the derivative is $- \frac{1}{5^{\frac{1}{2}}}$ . 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.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the denominator.

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Step 6.2.1.1

One to any power is one.

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

Step 6.2.1.2

Add $1$ and $4$ .

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

Step 6.2.2

The final answer is $\frac{1}{5^{\frac{1}{2}}}$ .

$\frac{1}{5^{\frac{1}{2}}}$

Step 6.3

At $x = 1$ the derivative is $\frac{1}{5^{\frac{1}{2}}}$ . 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