Calculus Examples

$h (w) = {(w^{2} + 2 w + 4)}^{\frac{5}{2}}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d w} [f (g (w))]$ is $f' (g (w)) g' (w)$ where $f (w) = w^{\frac{5}{2}}$ and $g (w) = w^{2} + 2 w + 4$ .

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Step 1.1

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

$\frac{d}{d u} [u^{\frac{5}{2}}] \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{5}{2}$ .

$\frac{5}{2} u^{\frac{5}{2} - 1} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 1.3

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

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{5}{2} - 1} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 2

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

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{5}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 3

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

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{5}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 4

Combine the numerators over the common denominator.

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{5 - 1 \cdot 2}{2}} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $2$ .

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{5 - 2}{2}} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 5.2

Subtract $2$ from $5$ .

$\frac{5}{2} {(w^{2} + 2 w + 4)}^{\frac{3}{2}} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 6

Combine $\frac{5}{2}$ and ${(w^{2} + 2 w + 4)}^{\frac{3}{2}}$ .

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} \frac{d}{d w} [w^{2} + 2 w + 4]$

Step 7

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

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (\frac{d}{d w} [w^{2}] + \frac{d}{d w} [2 w] + \frac{d}{d w} [4])$

Step 8

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

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + \frac{d}{d w} [2 w] + \frac{d}{d w} [4])$

Step 9

Since $2$ is constant with respect to $w$ , the derivative of $2 w$ with respect to $w$ is $2 \frac{d}{d w} [w]$ .

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2 \frac{d}{d w} [w] + \frac{d}{d w} [4])$

Step 10

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

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2 \cdot 1 + \frac{d}{d w} [4])$

Step 11

Multiply $2$ by $1$ .

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2 + \frac{d}{d w} [4])$

Step 12

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

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2 + 0)$

Step 13

Simplify the expression.

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Step 13.1

Add $2 w + 2$ and $0$ .

$\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2)$

Step 13.2

Reorder the factors of $\frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2} (2 w + 2)$ .

$(2 w + 2) \frac{5 {(w^{2} + 2 w + 4)}^{\frac{3}{2}}}{2}$