Calculus Examples

$y = x + 4 \sqrt{x + 2}$

Step 1

Differentiate.

Tap for more steps...

Step 1.1

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

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

Step 1.2

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

$1 + \frac{d}{d x} [4 \sqrt{x + 2}]$

Step 2

Evaluate $\frac{d}{d x} [4 \sqrt{x + 2}]$ .

Tap for more steps...

Step 2.1

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

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

Step 2.2

Since $4$ is constant with respect to $x$ , the derivative of $4 {(x + 2)}^{\frac{1}{2}}$ with respect to $x$ is $4 \frac{d}{d x} [{(x + 2)}^{\frac{1}{2}}]$ .

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

Step 2.3

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 $g (x) = x + 2$ .

Tap for more steps...

Step 2.3.1

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

$1 + 4 (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x + 2])$

Step 2.3.2

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

$1 + 4 (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x + 2])$

Step 2.3.3

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

$1 + 4 (\frac{1}{2} {(x + 2)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 2])$

Step 2.4

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

$1 + 4 (\frac{1}{2} {(x + 2)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [2]))$

Step 2.5

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

$1 + 4 (\frac{1}{2} {(x + 2)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [2]))$

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

Combine the numerators over the common denominator.

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

Step 2.10

Simplify the numerator.

Tap for more steps...

Step 2.10.1

Multiply $- 1$ by $2$ .

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

Step 2.10.2

Subtract $2$ from $1$ .

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

Step 2.11

Move the negative in front of the fraction.

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

Step 2.12

Add $1$ and $0$ .

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

Step 2.13

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

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

Step 2.14

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

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

Step 2.15

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

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

Step 2.16

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

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

Step 2.17

Factor $2$ out of $4$ .

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

Step 2.18

Cancel the common factors.

Tap for more steps...

Step 2.18.1

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

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

Step 2.18.2

Cancel the common factor.

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

Step 2.18.3

Rewrite the expression.

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