Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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 1.3.2

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 1.3.3

Multiply $2$ by $1$ .

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

Step 1.4

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $1$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

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

Step 2.4.2

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

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

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 $M (x, y)$ .

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Multiply $x^{2}$ by $1$ .

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

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

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

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 $x^{2}$ from both sides of the equation.

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

Step 9.1.2

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

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

Step 9.1.3

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

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Step 9.1.3.1

Subtract $x^{2}$ from $x^{2}$ .

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

Step 9.1.3.2

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

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

Step 9.1.3.3

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

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

Step 9.1.3.4

Add $- y$ and $0$ .

$\frac{d g}{d y} = - y$

Step 10

Find the antiderivative of $- y$ to find $g (y)$ .

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Step 10.1

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

$\int \frac{d g}{d y} d y = \int - y d y$

Step 10.2

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

$g (y) = \int - y d y$

Step 10.3

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

$g (y) = - \int y d y$

Step 10.4

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

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

Step 10.5

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

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

Step 11

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

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

Step 12

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

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Step 12.1

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

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

Step 12.2

Subtract $\frac{y^{2}}{2}$ from $y^{2} x$ .

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Step 12.2.1

Reorder $y^{2}$ and $x$ .

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

Step 12.2.2

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

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

Step 12.2.3

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

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

Step 12.2.4

Combine the numerators over the common denominator.

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

Step 12.3

Simplify the numerator.

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Step 12.3.1

Factor $y^{2}$ out of $x y^{2} \cdot 2 - y^{2}$ .

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Step 12.3.1.1

Factor $y^{2}$ out of $x y^{2} \cdot 2$ .

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

Step 12.3.1.2

Factor $y^{2}$ out of $- y^{2}$ .

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

Step 12.3.1.3

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

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

Step 12.3.2

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

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

Step 12.4

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

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

Step 12.5

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

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

Step 12.6

Combine the numerators over the common denominator.

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

Step 12.7

Simplify the numerator.

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Step 12.7.1

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

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Step 12.7.1.1

Factor $y$ out of $y x^{2} \cdot 2$ .

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

Step 12.7.1.2

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

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

Step 12.7.1.3

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

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

Step 12.7.2

Move $2$ to the left of $x^{2}$ .

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

Step 12.7.3

Apply the distributive property.

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

Step 12.7.4

Rewrite using the commutative property of multiplication.

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

Step 12.7.5

Move $- 1$ to the left of $y$ .

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

Step 12.7.6

Simplify each term.

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