Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $1$ .

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 2 x - 3 x^{2} (2 y)$

Step 1.4.3

Multiply $2$ by $- 3$ .

$\frac{\partial M}{\partial y} = 2 x - 6 x^{2} y$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = - 6 x^{2} y + 2 x$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 2 x - 2 y (3 x^{2})$

Step 2.3.3

Multiply $3$ by $- 2$ .

$\frac{\partial N}{\partial x} = 2 x - 6 y x^{2}$

Step 2.4

Reorder terms.

$\frac{\partial N}{\partial x} = - 6 x^{2} y + 2 x$

Step 3

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

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Step 3.1

Substitute $- 6 x^{2} y + 2 x$ for $\frac{\partial M}{\partial y}$ and $- 6 x^{2} y + 2 x$ for $\frac{\partial N}{\partial x}$ .

$- 6 x^{2} y + 2 x = - 6 x^{2} y + 2 x$

Step 3.2

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

$- 6 x^{2} y + 2 x = - 6 x^{2} y + 2 x$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Since $- 3 y^{2}$ is constant with respect to $x$ , move $- 3 y^{2}$ out of the integral.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 5.7

Simplify.

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Step 5.7.1

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

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

Step 5.7.2

Cancel the common factor of $2$ .

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Step 5.7.2.1

Cancel the common factor.

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

Step 5.7.2.2

Rewrite the expression.

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

Step 5.7.3

Multiply $y$ by $1$ .

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

Step 5.7.4

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

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

Step 5.7.5

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

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

Step 5.7.6

Combine $y^{2}$ and $\frac{- 3 x^{3}}{3}$ .

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

Step 5.7.7

Cancel the common factor of $- 3$ and $3$ .

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Step 5.7.7.1

Factor $3$ out of $y^{2} (- 3 x^{3})$ .

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

Step 5.7.7.2

Cancel the common factors.

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Step 5.7.7.2.1

Factor $3$ out of $3$ .

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

Step 5.7.7.2.2

Cancel the common factor.

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

Step 5.7.7.2.3

Rewrite the expression.

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

Step 5.7.7.2.4

Divide $y^{2} (- x^{3})$ by $1$ .

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $2$ by $- 1$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

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.

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

Step 9.1.2

Add $2 x^{3} y$ to both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

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

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

Step 10

Find the antiderivative of $0$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 10.2

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

$g (y) = \int 0 d y$

Step 10.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 10.4

Add $0$ and $C$ .

$g (y) = C$

Step 11

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

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

Step 12

Rewrite using the commutative property of multiplication.

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