Calculus Examples

$4 y^{3} + \sqrt{x^{2} + y^{2}}$

Step 1

By the Sum Rule, the derivative of $4 y^{3} + \sqrt{x^{2} + y^{2}}$ with respect to $y$ is $\frac{d}{d y} [4 y^{3}] + \frac{d}{d y} [\sqrt{x^{2} + y^{2}}]$ .

$\frac{d}{d y} [4 y^{3}] + \frac{d}{d y} [\sqrt{x^{2} + y^{2}}]$

Step 2

Evaluate $\frac{d}{d y} [4 y^{3}]$ .

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Step 2.1

Since $4$ is constant with respect to $y$ , the derivative of $4 y^{3}$ with respect to $y$ is $4 \frac{d}{d y} [y^{3}]$ .

$4 \frac{d}{d y} [y^{3}] + \frac{d}{d y} [\sqrt{x^{2} + y^{2}}]$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

$4 (3 y^{2}) + \frac{d}{d y} [\sqrt{x^{2} + y^{2}}]$

Step 2.3

Multiply $3$ by $4$ .

$12 y^{2} + \frac{d}{d y} [\sqrt{x^{2} + y^{2}}]$

Step 3

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

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Step 3.1

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

$12 y^{2} + \frac{d}{d y} [{(x^{2} + y^{2})}^{\frac{1}{2}}]$

Step 3.2

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

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Step 3.2.1

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

$12 y^{2} + \frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d y} [x^{2} + y^{2}]$

Step 3.2.2

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

$12 y^{2} + \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d y} [x^{2} + y^{2}]$

Step 3.2.3

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

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1} \frac{d}{d y} [x^{2} + y^{2}]$

Step 3.3

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

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1} (\frac{d}{d y} [x^{2}] + \frac{d}{d y} [y^{2}])$

Step 3.4

Since $x^{2}$ is constant with respect to $y$ , the derivative of $x^{2}$ with respect to $y$ is $0$ .

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1} (0 + \frac{d}{d y} [y^{2}])$

Step 3.5

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

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1} (0 + 2 y)$

Step 3.6

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

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (0 + 2 y)$

Step 3.7

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

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (0 + 2 y)$

Step 3.8

Combine the numerators over the common denominator.

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1 - 1 \cdot 2}{2}} (0 + 2 y)$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply $- 1$ by $2$ .

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{1 - 2}{2}} (0 + 2 y)$

Step 3.9.2

Subtract $2$ from $1$ .

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{\frac{- 1}{2}} (0 + 2 y)$

Step 3.10

Move the negative in front of the fraction.

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{- \frac{1}{2}} (0 + 2 y)$

Step 3.11

Add $0$ and $2 y$ .

$12 y^{2} + \frac{1}{2} {(x^{2} + y^{2})}^{- \frac{1}{2}} (2 y)$

Step 3.12

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

$12 y^{2} + \frac{{(x^{2} + y^{2})}^{- \frac{1}{2}}}{2} (2 y)$

Step 3.13

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

$12 y^{2} + \frac{2 {(x^{2} + y^{2})}^{- \frac{1}{2}}}{2} y$

Step 3.14

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

$12 y^{2} + \frac{2 {(x^{2} + y^{2})}^{- \frac{1}{2}} y}{2}$

Step 3.15

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

$12 y^{2} + \frac{2 y}{2 {(x^{2} + y^{2})}^{\frac{1}{2}}}$

Step 3.16

Cancel the common factor.

$12 y^{2} + \frac{2 y}{2 {(x^{2} + y^{2})}^{\frac{1}{2}}}$

Step 3.17

Rewrite the expression.

$12 y^{2} + \frac{y}{{(x^{2} + y^{2})}^{\frac{1}{2}}}$