Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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 $g (x) = 2 x + 4 y$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x + 4 y$ .

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

Step 2.2.2

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

$\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d x} [2 x + 4 y] + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $2 x + 4 y$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

$\frac{1}{2} {(2 x + 4 y)}^{\frac{1}{2} - 1} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.7

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

$\frac{1}{2} {(2 x + 4 y)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.8

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

$\frac{1}{2} {(2 x + 4 y)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.9

Combine the numerators over the common denominator.

$\frac{1}{2} {(2 x + 4 y)}^{\frac{1 - 1 \cdot 2}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.10

Simplify the numerator.

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Step 2.10.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(2 x + 4 y)}^{\frac{1 - 2}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.10.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(2 x + 4 y)}^{\frac{- 1}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.11

Move the negative in front of the fraction.

$\frac{1}{2} {(2 x + 4 y)}^{- \frac{1}{2}} (2 \cdot 1 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.12

Multiply $2$ by $1$ .

$\frac{1}{2} {(2 x + 4 y)}^{- \frac{1}{2}} (2 + 0) + \frac{d}{d x} [\sqrt{4 x y}]$

Step 2.13

Add $2$ and $0$ .

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

Step 2.14

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

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

Step 2.15

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

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

Step 2.16

Move $2$ to the left of ${(2 x + 4 y)}^{- \frac{1}{2}}$ .

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

Step 2.17

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

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

Step 2.18

Cancel the common factor.

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

Step 2.19

Rewrite the expression.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Factor $4$ out of $4 x y$ .

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

Step 3.3

Apply the product rule to $4 (x y)$ .

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

Step 3.4

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

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

Step 3.5

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.6

Cancel the common factor of $2$ .

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Step 3.6.1

Cancel the common factor.

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

Step 3.6.2

Rewrite the expression.

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

Step 3.7

Evaluate the exponent.

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

Step 3.8

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

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

Step 3.9

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 $g (x) = x y$ .

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Step 3.9.1

To apply the Chain Rule, set $u_{2}$ as $x y$ .

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

Step 3.9.2

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

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

Step 3.9.3

Replace all occurrences of $u_{2}$ with $x y$ .

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

Step 3.10

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

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

Step 3.11

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

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

Step 3.12

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

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

Step 3.13

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

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

Step 3.14

Combine the numerators over the common denominator.

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

Step 3.15

Simplify the numerator.

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Step 3.15.1

Multiply $- 1$ by $2$ .

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

Step 3.15.2

Subtract $2$ from $1$ .

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

Step 3.16

Move the negative in front of the fraction.

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

Step 3.17

Multiply $y$ by $1$ .

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

Step 3.18

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

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

Step 3.19

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

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

Step 3.20

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

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

Step 3.21

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

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

Step 3.22

Cancel the common factor.

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

Step 3.23

Rewrite the expression.

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

Step 4

Simplify.

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Step 4.1

Apply the product rule to $x y$ .

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

Step 4.2

Combine terms.

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Step 4.2.1

Move $y^{\frac{1}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

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

Step 4.2.2

Multiply $y$ by $y^{- \frac{1}{2}}$ by adding the exponents.

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Step 4.2.2.1

Multiply $y$ by $y^{- \frac{1}{2}}$ .

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Step 4.2.2.1.1

Raise $y$ to the power of $1$ .

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

Step 4.2.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.2.2.2

Write $1$ as a fraction with a common denominator.

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

Step 4.2.2.3

Combine the numerators over the common denominator.

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

Step 4.2.2.4

Subtract $1$ from $2$ .

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