Calculus Examples

${(x + y)}^{2} d x + (2 x y + x^{2} - 1) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = {(x + y)}^{2}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [{(x + y)}^{2}]$

Step 1.2

Rewrite ${(x + y)}^{2}$ as $(x + y) (x + y)$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [(x + y) (x + y)]$

Step 1.3

Expand $(x + y) (x + y)$ using the FOIL Method.

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Step 1.3.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x (x + y) + y (x + y)]$

Step 1.3.2

Apply the distributive property.

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x \cdot x + x y + y (x + y)]$

Step 1.3.3

Apply the distributive property.

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x \cdot x + x y + y x + y \cdot y]$

Step 1.4

Simplify and combine like terms.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply $x$ by $x$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{2} + x y + y x + y \cdot y]$

Step 1.4.1.2

Multiply $y$ by $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{2} + x y + y x + y^{2}]$

Step 1.4.2

Add $x y$ and $y x$ .

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Step 1.4.2.1

Reorder $y$ and $x$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{2} + x y + x y + y^{2}]$

Step 1.4.2.2

Add $x y$ and $x y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{2} + 2 x y + y^{2}]$

Step 1.5

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{2}] + \frac{d}{d y} [2 x y] + \frac{d}{d y} [y^{2}]$

Step 1.6

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [2 x y] + \frac{d}{d y} [y^{2}]$

Step 1.7

Add $0$ and $\frac{d}{d y} [2 x y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y] + \frac{d}{d y} [y^{2}]$

Step 1.8

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

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y] + \frac{d}{d y} [y^{2}]$

Step 1.9

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

$\frac{\partial M}{\partial y} = 2 x \cdot 1 + \frac{d}{d y} [y^{2}]$

Step 1.10

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x + \frac{d}{d y} [y^{2}]$

Step 1.11

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

$\frac{\partial M}{\partial y} = 2 x + 2 y$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 2 x y + x^{2} - 1$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 x y + x^{2} - 1]$

Step 2.2

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

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 x y] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 2 y \frac{d}{d x} [x] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 2 y \cdot 1 + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = 2 y + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$

Step 2.4

Differentiate.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 2 y + 2 x + \frac{d}{d x} [- 1]$

Step 2.4.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 2 y + 2 x + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $2 y + 2 x$ and $0$ .

$\frac{\partial N}{\partial x} = 2 y + 2 x$

Step 2.5.2

Reorder terms.

$\frac{\partial N}{\partial x} = 2 x + 2 y$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $2 x + 2 y$ for $\frac{\partial M}{\partial y}$ and $2 x + 2 y$ for $\frac{\partial N}{\partial x}$ .

$2 x + 2 y = 2 x + 2 y$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$2 x + 2 y = 2 x + 2 y$ is an identity.

Step 4

Set $f (x, y)$ equal to the integral of $M (x, y)$ .

$f (x, y) = \int {(x + y)}^{2} d x$

Step 5

Integrate $M (x, y) = {(x + y)}^{2}$ to find $f (x, y)$ .

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Step 5.1

Let $u = x + y$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1.1

Let $u = x + y$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1.1

Differentiate $x + y$ .

$\frac{d}{d x} [x + y]$

Step 5.1.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [y]$

Step 5.1.1.3

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

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

Step 5.1.1.4

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

$1 + 0$

Step 5.1.1.5

Add $1$ and $0$ .

$1$

Step 5.1.2

Rewrite the problem using $u$ and $d u$ .

$f (x, y) = \int u^{2} d u$

Step 5.2

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$f (x, y) = \frac{1}{3} u^{3} + C$

Step 5.3

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

$f (x, y) = \frac{1}{3} {(x + y)}^{3} + C$

Step 6

Since the integral of $g (y)$ will contain an integration constant, we can replace $C$ with $g (y)$ .

$f (x, y) = \frac{1}{3} {(x + y)}^{3} + g (y)$

Step 7

Set $\frac{\partial f}{\partial y} = N (x, y)$ .

$\frac{\partial f}{\partial y} = 2 x y + x^{2} - 1$

Step 8

Find $\frac{\partial f}{\partial y}$ .

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Step 8.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{1}{3} {(x + y)}^{3} + g (y)] = 2 x y + x^{2} - 1$

Step 8.2

By the Sum Rule, the derivative of $\frac{1}{3} {(x + y)}^{3} + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{3} {(x + y)}^{3}] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [\frac{1}{3} {(x + y)}^{3}] + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3

Evaluate $\frac{d}{d y} [\frac{1}{3} {(x + y)}^{3}]$ .

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Step 8.3.1

Since $\frac{1}{3}$ is constant with respect to $y$ , the derivative of $\frac{1}{3} {(x + y)}^{3}$ with respect to $y$ is $\frac{1}{3} \frac{d}{d y} [{(x + y)}^{3}]$ .

$\frac{1}{3} \frac{d}{d y} [{(x + y)}^{3}] + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{3}$ and $g (y) = x + y$ .

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Step 8.3.2.1

To apply the Chain Rule, set $u$ as $x + y$ .

$\frac{1}{3} (\frac{d}{d u} [u^{3}] \frac{d}{d y} [x + y]) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.2.2

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

$\frac{1}{3} (3 u^{2} \frac{d}{d y} [x + y]) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.2.3

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

$\frac{1}{3} (3 {(x + y)}^{2} \frac{d}{d y} [x + y]) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.3

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

$\frac{1}{3} (3 {(x + y)}^{2} (\frac{d}{d y} [x] + \frac{d}{d y} [y])) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.4

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

$\frac{1}{3} (3 {(x + y)}^{2} (0 + \frac{d}{d y} [y])) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.5

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

$\frac{1}{3} (3 {(x + y)}^{2} (0 + 1)) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.6

Add $0$ and $1$ .

$\frac{1}{3} (3 {(x + y)}^{2} \cdot 1) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.7

Multiply $3$ by $1$ .

$\frac{1}{3} (3 {(x + y)}^{2}) + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.8

Combine $3$ and $\frac{1}{3}$ .

$\frac{3}{3} {(x + y)}^{2} + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.9

Combine $\frac{3}{3}$ and ${(x + y)}^{2}$ .

$\frac{3 {(x + y)}^{2}}{3} + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.10

Cancel the common factor of $3$ .

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Step 8.3.10.1

Cancel the common factor.

$\frac{3 {(x + y)}^{2}}{3} + \frac{d}{d y} [g (y)] = 2 x y + x^{2} - 1$

Step 8.3.10.2

Divide ${(x + y)}^{2}$ by $1$ .

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

Step 8.4

Differentiate using the function rule which states that the derivative of $g (y)$ is $\frac{d g}{d y}$ .

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

Step 8.5

Reorder terms.

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

Step 9

Solve for $\frac{d g}{d y}$ .

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Step 9.1

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

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

Step 10

Find the antiderivative of $2 x y + x^{2} - 1 - {(x + y)}^{2}$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = 2 x y + x^{2} - 1 - {(x + y)}^{2}$ .

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

Step 10.2

Evaluate $\int \frac{d g}{d y} d y$ .

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

Step 10.3

Split the single integral into multiple integrals.

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

Step 10.4

Since $2 x$ is constant with respect to $y$ , move $2 x$ out of the integral.

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

Step 10.5

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

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

Step 10.6

Apply the constant rule.

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

Step 10.7

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

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

Step 10.8

Apply the constant rule.

$g (y) = 2 x (\frac{y^{2}}{2} + C) + x^{2} y + C - y + C + \int - {(x + y)}^{2} d y$

Step 10.9

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$g (y) = 2 x (\frac{y^{2}}{2} + C) + x^{2} y + C - y + C - \int {(x + y)}^{2} d y$

Step 10.10

Let $u = x + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.10.1

Let $u = x + y$ . Find $\frac{d u}{d y}$ .

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Step 10.10.1.1

Differentiate $x + y$ .

$\frac{d}{d y} [x + y]$

Step 10.10.1.2

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

$\frac{d}{d y} [x] + \frac{d}{d y} [y]$

Step 10.10.1.3

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

$0 + \frac{d}{d y} [y]$

Step 10.10.1.4

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

$0 + 1$

Step 10.10.1.5

Add $0$ and $1$ .

$1$

Step 10.10.2

Rewrite the problem using $u$ and $d u$ .

$g (y) = 2 x (\frac{y^{2}}{2} + C) + x^{2} y + C - y + C - \int u^{2} d u$

Step 10.11

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$g (y) = 2 x (\frac{y^{2}}{2} + C) + x^{2} y + C - y + C - (\frac{1}{3} u^{3} + C)$

Step 10.12

Combine $\frac{1}{3}$ and $u^{3}$ .

$g (y) = 2 x (\frac{y^{2}}{2} + C) + x^{2} y + C - y + C - (\frac{u^{3}}{3} + C)$

Step 10.13

Simplify.

$g (y) = x y^{2} + x^{2} y - y - \frac{u^{3}}{3} + C$

Step 10.14

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

$g (y) = x y^{2} + x^{2} y - y - \frac{{(x + y)}^{3}}{3} + C$

Step 10.15

Simplify.

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Step 10.15.1

Reorder terms.

$g (y) = x y^{2} + x^{2} y - y - (\frac{1}{3} {(x + y)}^{3}) + C$

Step 10.15.2

Remove parentheses.

$g (y) = x y^{2} + x^{2} y - y - \frac{1}{3} {(x + y)}^{3} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = \frac{1}{3} {(x + y)}^{3} + g (y)$ .

$f (x, y) = \frac{1}{3} {(x + y)}^{3} + x y^{2} + x^{2} y - y - \frac{1}{3} {(x + y)}^{3} + C$

Step 12

Combine the opposite terms in $\frac{1}{3} {(x + y)}^{3} + x y^{2} + x^{2} y - y - \frac{1}{3} {(x + y)}^{3} + C$ .

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Step 12.1

Subtract $\frac{1}{3} {(x + y)}^{3}$ from $\frac{1}{3} {(x + y)}^{3}$ .

$f (x, y) = x y^{2} + x^{2} y - y + 0 + C$

Step 12.2

Add $x y^{2} + x^{2} y - y$ and $0$ .

$f (x, y) = x y^{2} + x^{2} y - y + C$