Calculus Examples

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

Step 1

Rewrite the left side with rational exponents.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

Differentiate both sides of the equation.

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

Step 3

Differentiate the left side of the equation.

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Step 3.1

By the Sum Rule, the derivative of ${(x + y)}^{\frac{1}{2}} + {(x - y)}^{\frac{1}{2}}$ with respect to $x$ is $\frac{d}{d x} [{(x + y)}^{\frac{1}{2}}] + \frac{d}{d x} [{(x - y)}^{\frac{1}{2}}]$ .

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

Step 3.2

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

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Step 3.2.1

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 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

Combine the numerators over the common denominator.

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

Step 3.2.8

Simplify the numerator.

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Step 3.2.8.1

Multiply $- 1$ by $2$ .

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

Step 3.2.8.2

Subtract $2$ from $1$ .

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

Step 3.2.9

Move the negative in front of the fraction.

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

Step 3.2.10

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

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

Step 3.2.11

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

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

Step 3.3

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

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Step 3.3.1

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 3.3.1.1

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

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

Step 3.3.1.2

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

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

Step 3.3.1.3

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

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

Step 3.3.6

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

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

Step 3.3.7

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

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

Step 3.3.8

Combine the numerators over the common denominator.

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

Step 3.3.9

Simplify the numerator.

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Step 3.3.9.1

Multiply $- 1$ by $2$ .

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

Step 3.3.9.2

Subtract $2$ from $1$ .

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

Step 3.3.10

Move the negative in front of the fraction.

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

Step 3.3.11

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

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

Step 3.3.12

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

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

Step 3.4

Reorder terms.

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

Step 4

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

$0$

Step 5

Reform the equation by setting the left side equal to the right side.

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

Step 6

Solve for $y'$ .

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Step 6.1

Simplify $(1 + y') \frac{1}{2 {(x + y)}^{\frac{1}{2}}} + (1 - y') \frac{1}{2 {(x - y)}^{\frac{1}{2}}}$ .

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Step 6.1.1

Simplify each term.

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Step 6.1.1.1

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

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

Step 6.1.1.2

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

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

Step 6.1.2

To write $\frac{1 + y'}{2 {(x + y)}^{\frac{1}{2}}}$ as a fraction with a common denominator, multiply by $\frac{{(x - y)}^{\frac{1}{2}}}{{(x - y)}^{\frac{1}{2}}}$ .

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

Step 6.1.3

To write $\frac{1 - y'}{2 {(x - y)}^{\frac{1}{2}}}$ as a fraction with a common denominator, multiply by $\frac{{(x + y)}^{\frac{1}{2}}}{{(x + y)}^{\frac{1}{2}}}$ .

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

Step 6.1.4

Write each expression with a common denominator of $2 {(x + y)}^{\frac{1}{2}} {(x - y)}^{\frac{1}{2}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.1.4.1

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

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

Step 6.1.4.2

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

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

Step 6.1.4.3

Reorder the factors of $2 {(x - y)}^{\frac{1}{2}} {(x + y)}^{\frac{1}{2}}$ .

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

Step 6.1.5

Combine the numerators over the common denominator.

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

Step 6.1.6

Simplify the numerator.

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Step 6.1.6.1

Apply the distributive property.

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

Step 6.1.6.2

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

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

Step 6.1.6.3

Apply the distributive property.

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

Step 6.1.6.4

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

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

Step 6.2

Set the numerator equal to zero.

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

Step 6.3

Solve the equation for $y'$ .

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Step 6.3.1

Move all terms not containing $y'$ to the right side of the equation.

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Step 6.3.1.1

Subtract ${(x - y)}^{\frac{1}{2}}$ from both sides of the equation.

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

Step 6.3.1.2

Subtract ${(x + y)}^{\frac{1}{2}}$ from both sides of the equation.

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

Step 6.3.2

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

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Step 6.3.2.1

Factor $y'$ out of $y' {(x - y)}^{\frac{1}{2}}$ .

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

Step 6.3.2.2

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

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

Step 6.3.2.3

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

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

Step 6.3.3

Rewrite $- 1 {(x + y)}^{\frac{1}{2}}$ as $- {(x + y)}^{\frac{1}{2}}$ .

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

Step 6.3.4

Divide each term in $y' ({(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}) = - {(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}$ by ${(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}$ and simplify.

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Step 6.3.4.1

Divide each term in $y' ({(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}) = - {(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}$ by ${(x - y)}^{\frac{1}{2}} - {(x + y)}^{\frac{1}{2}}$ .

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

Step 6.3.4.2

Simplify the left side.

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Step 6.3.4.2.1

Cancel the common factor.

Step 6.3.4.2.2

Divide $y'$ by $1$ .

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

Step 6.3.4.3

Simplify the right side.

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Step 6.3.4.3.1

Combine the numerators over the common denominator.

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

Step 6.3.4.3.2

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

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Step 6.3.4.3.2.1

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

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

Step 6.3.4.3.2.2

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

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

Step 6.3.4.3.2.3

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

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

Step 6.3.4.3.3

Move the negative in front of the fraction.

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

Step 7

Replace $y'$ with $\frac{d y}{d x}$ .

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