Calculus Examples

$(2 x y + y^{3}) d x + (x^{2} + 3 x y^{2} - 2 y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y + y^{3}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y + y^{3}]$

Step 1.2

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y] + \frac{d}{d y} [y^{3}]$

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

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y] + \frac{d}{d y} [y^{3}]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 2 x \cdot 1 + \frac{d}{d y} [y^{3}]$

Step 1.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x + \frac{d}{d y} [y^{3}]$

Step 1.4

Differentiate using the Power Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 2 x + 3 y^{2}$

Step 1.4.2

Reorder terms.

$\frac{\partial M}{\partial y} = 3 y^{2} + 2 x$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x^{2} + 3 x y^{2} - 2 y$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2} + 3 x y^{2} - 2 y]$

Step 2.2

Differentiate.

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Step 2.2.1

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

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x y^{2}] + \frac{d}{d x} [- 2 y]$

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 2 x + \frac{d}{d x} [3 x y^{2}] + \frac{d}{d x} [- 2 y]$

Step 2.3

Evaluate $\frac{d}{d x} [3 x y^{2}]$ .

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Step 2.3.1

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 x y^{2}$ with respect to $x$ is $3 y^{2} \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = 2 x + 3 y^{2} \frac{d}{d x} [x] + \frac{d}{d x} [- 2 y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 2 x + 3 y^{2} \cdot 1 + \frac{d}{d x} [- 2 y]$

Step 2.3.3

Multiply $3$ by $1$ .

$\frac{\partial N}{\partial x} = 2 x + 3 y^{2} + \frac{d}{d x} [- 2 y]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 2 x + 3 y^{2} + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $2 x + 3 y^{2}$ and $0$ .

$\frac{\partial N}{\partial x} = 2 x + 3 y^{2}$

Step 2.5.2

Reorder terms.

$\frac{\partial N}{\partial x} = 3 y^{2} + 2 x$

Step 3

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

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Step 3.1

Substitute $3 y^{2} + 2 x$ for $\frac{\partial M}{\partial y}$ and $3 y^{2} + 2 x$ for $\frac{\partial N}{\partial x}$ .

$3 y^{2} + 2 x = 3 y^{2} + 2 x$

Step 3.2

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

$3 y^{2} + 2 x = 3 y^{2} + 2 x$ is an identity.

Step 4

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

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

Step 5

Integrate $M (x, y) = 2 x y + y^{3}$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial y} = x^{2} + 3 x y^{2} - 2 y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.4

Evaluate $\frac{d}{d y} [y^{3} x]$ .

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Move $3$ to the left of $x$ .

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

Step 8.5

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

$x^{2} + 3 x y^{2} + \frac{d g}{d y} = x^{2} + 3 x y^{2} - 2 y$

Step 8.6

Reorder terms.

$\frac{d g}{d y} + x^{2} + 3 y^{2} x = x^{2} + 3 x y^{2} - 2 y$

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.

$\frac{d g}{d y} + 3 y^{2} x = x^{2} + 3 x y^{2} - 2 y - x^{2}$

Step 9.1.2

Subtract $3 y^{2} x$ from both sides of the equation.

$\frac{d g}{d y} = x^{2} + 3 x y^{2} - 2 y - x^{2} - 3 y^{2} x$

Step 9.1.3

Combine the opposite terms in $x^{2} + 3 x y^{2} - 2 y - x^{2} - 3 y^{2} x$ .

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Step 9.1.3.1

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

$\frac{d g}{d y} = 3 x y^{2} - 2 y + 0 - 3 y^{2} x$

Step 9.1.3.2

Add $3 x y^{2} - 2 y$ and $0$ .

$\frac{d g}{d y} = 3 x y^{2} - 2 y - 3 y^{2} x$

Step 9.1.3.3

Reorder the factors in the terms $3 x y^{2}$ and $- 3 y^{2} x$ .

$\frac{d g}{d y} = 3 y^{2} x - 2 y - 3 y^{2} x$

Step 9.1.3.4

Subtract $3 y^{2} x$ from $3 y^{2} x$ .

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

Step 9.1.3.5

Add $- 2 y$ and $0$ .

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

Step 10

Find the antiderivative of $- 2 y$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = - 2 y$ .

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

Step 10.2

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

$g (y) = \int - 2 y d y$

Step 10.3

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

$g (y) = - 2 \int y d y$

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

Step 10.5.2.2

Cancel the common factor of $- 2$ and $2$ .

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Step 10.5.2.2.1

Factor $2$ out of $- 2$ .

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

Step 10.5.2.2.2

Cancel the common factors.

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Step 10.5.2.2.2.1

Factor $2$ out of $2$ .

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

Step 10.5.2.2.2.2

Cancel the common factor.

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

Step 10.5.2.2.2.3

Rewrite the expression.

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

Step 10.5.2.2.2.4

Divide $- 1$ by $1$ .

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

Step 11

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

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