Calculus Examples

$\sqrt{4 x + 8}$

Step 1

Rewrite $\frac{d}{d x} [\sqrt{4 x + 8}]$ as $\frac{d}{d x} [2 \sqrt{x + 2}]$ .

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Step 1.1

Rewrite $4 x + 8$ as $2^{2} (x + 2)$ .

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Step 1.1.1

Factor $4$ out of $4 x$ .

$\frac{d}{d x} [\sqrt{4 (x) + 8}]$

Step 1.1.2

Factor $4$ out of $8$ .

$\frac{d}{d x} [\sqrt{4 (x) + 4 (2)}]$

Step 1.1.3

Factor $4$ out of $4 (x) + 4 (2)$ .

$\frac{d}{d x} [\sqrt{4 (x + 2)}]$

Step 1.1.4

Rewrite $4$ as $2^{2}$ .

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

Step 1.2

Pull terms out from under the radical.

$\frac{d}{d x} [2 \sqrt{x + 2}]$

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Replace all occurrences of $u$ with $x + 2$ .

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

Step 4

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

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

Step 5

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

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

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

Step 7.2

Subtract $2$ from $1$ .

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

Step 8

Simplify terms.

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Step 8.1

Move the negative in front of the fraction.

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

Step 8.2

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

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

Cancel the common factor.

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

Step 8.6

Rewrite the expression.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

Step 12

Simplify the expression.

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Step 12.1

Add $1$ and $0$ .

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

Step 12.2

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

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