Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

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

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

Step 1.5

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

$\frac{\partial M}{\partial y} = 0 + 2 y + 1$

Step 1.6

Add $0$ and $2 y$ .

$\frac{\partial M}{\partial y} = 2 y + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $1$ .

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

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 $n = 1$ .

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

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 2 y + 1 + 0$

Step 2.4.3

Add $2 y + 1$ and $0$ .

$\frac{\partial N}{\partial x} = 2 y + 1$

Step 3

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

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Step 3.1

Substitute $2 y + 1$ for $\frac{\partial M}{\partial y}$ and $2 y + 1$ for $\frac{\partial N}{\partial x}$ .

$2 y + 1 = 2 y + 1$

Step 3.2

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

$2 y + 1 = 2 y + 1$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Apply the constant rule.

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

Step 5.5

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

By the Sum Rule, the derivative of $\frac{1}{3} x^{3} + y^{2} x + y x + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{3} x^{3}] + \frac{d}{d y} [y^{2} x] + \frac{d}{d y} [y x] + \frac{d}{d y} [g (y)]$ .

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

Step 8.2.2

Since $\frac{1}{3} x^{3}$ is constant with respect to $y$ , the derivative of $\frac{1}{3} x^{3}$ with respect to $y$ is $0$ .

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Move $2$ to the left of $x$ .

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $x$ by $1$ .

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

Step 8.5

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

$0 + 2 x y + x + \frac{d g}{d y} = 2 x y + x + e^{y}$

Step 8.6

Simplify.

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Step 8.6.1

Add $0$ and $2 x y$ .

$2 x y + x + \frac{d g}{d y} = 2 x y + x + e^{y}$

Step 8.6.2

Reorder terms.

$\frac{d g}{d y} + 2 x y + x = 2 x y + x + e^{y}$

Step 9

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

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Step 9.1

Move all terms not containing $\frac{d g}{d y}$ to the right side of the equation.

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Step 9.1.1

Subtract $2 x y$ from both sides of the equation.

$\frac{d g}{d y} + x = 2 x y + x + e^{y} - 2 x y$

Step 9.1.2

Subtract $x$ from both sides of the equation.

$\frac{d g}{d y} = 2 x y + x + e^{y} - 2 x y - x$

Step 9.1.3

Combine the opposite terms in $2 x y + x + e^{y} - 2 x y - x$ .

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Step 9.1.3.1

Subtract $2 x y$ from $2 x y$ .

$\frac{d g}{d y} = x + e^{y} + 0 - x$

Step 9.1.3.2

Add $x + e^{y}$ and $0$ .

$\frac{d g}{d y} = x + e^{y} - x$

Step 9.1.3.3

Subtract $x$ from $x$ .

$\frac{d g}{d y} = 0 + e^{y}$

Step 9.1.3.4

Add $0$ and $e^{y}$ .

$\frac{d g}{d y} = e^{y}$

Step 10

Find the antiderivative of $e^{y}$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = e^{y}$ .

$\int \frac{d g}{d y} d y = \int e^{y} d y$

Step 10.2

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

$g (y) = \int e^{y} d y$

Step 10.3

The integral of $e^{y}$ with respect to $y$ is $e^{y}$ .

$g (y) = e^{y} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = \frac{1}{3} x^{3} + y^{2} x + y x + g (y)$ .

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

Step 12

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

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