Calculus Examples

$\sqrt{x + y} + x y = 21$

Step 1

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

${(x + y)}^{\frac{1}{2}} + x y = 21$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} ({(x + y)}^{\frac{1}{2}} + x y) = \frac{d}{d x} (21)$

Step 3

Differentiate the left side of the equation.

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Step 3.1

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

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

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 $u$ as $x + y$ .

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

Combine the numerators over the common denominator.

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

Step 3.2.8

Simplify the numerator.

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Step 3.2.8.1

Multiply $- 1$ by $2$ .

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

Step 3.2.8.2

Subtract $2$ from $1$ .

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

Step 3.2.9

Move the negative in front of the fraction.

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

Step 3.2.10

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

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

Step 3.2.11

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

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

Step 3.3

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

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Step 3.3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $y$ by $1$ .

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

Step 3.4

Reorder terms.

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

Step 4

Since $21$ is constant with respect to $x$ , the derivative of $21$ with respect to $x$ is $0$ .

$0$

Step 5

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

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

Step 6

Solve for $y'$ .

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

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

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

Step 6.1.3

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

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

Step 6.1.4

Combine the numerators over the common denominator.

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

Step 6.1.5

Rewrite using the commutative property of multiplication.

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

Step 6.1.6

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

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

Step 6.1.7

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

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

Step 6.1.8

Combine the numerators over the common denominator.

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

Step 6.1.9

Rewrite using the commutative property of multiplication.

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

Step 6.2

Set the numerator equal to zero.

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

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 $1$ from both sides of the equation.

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

Step 6.3.1.2

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

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

Step 6.3.2

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

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Step 6.3.2.1

Factor $y'$ out of ${y'}^{1}$ .

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

Step 6.3.2.2

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

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

Step 6.3.2.3

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

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

Step 6.3.3

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

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Step 6.3.3.1

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

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

Step 6.3.3.2

Simplify the left side.

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Step 6.3.3.2.1

Cancel the common factor.

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

Step 6.3.3.2.2

Divide $y'$ by $1$ .

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

Step 6.3.3.3

Simplify the right side.

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Step 6.3.3.3.1

Combine the numerators over the common denominator.

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

Step 6.3.3.3.2

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 6.3.3.3.3

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

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

Step 6.3.3.3.4

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

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

Step 6.3.3.3.5

Move the negative in front of the fraction.

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

Step 7

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

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