Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.4

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

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Step 1.4.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

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

Step 1.4.2

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

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

Step 1.4.3

Multiply $- 1$ by $1$ .

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

Step 1.5

Add $0$ and $2 x y$ .

$\frac{\partial M}{\partial y} = 2 x y - 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 + y (2 x) + \frac{d}{d x} [- x]$

Step 2.3.3

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

$\frac{\partial N}{\partial x} = 0 + 2 y x + \frac{d}{d x} [- x]$

Step 2.4

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

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Step 2.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = 0 + 2 y x - \frac{d}{d x} [x]$

Step 2.4.2

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

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

Step 2.4.3

Multiply $- 1$ by $1$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Add $0$ and $2 y x$ .

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

Step 2.5.2

Reorder terms.

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

Step 3

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

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Step 3.1

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

$2 x y - 1 = 2 x y - 1$

Step 3.2

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

$2 x y - 1 = 2 x y - 1$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$f (x, y) = \frac{1}{4} x^{4} + C + \int x y^{2} d x + \int - y d x$

Step 5.3

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

$f (x, y) = \frac{1}{4} x^{4} + C + y^{2} \int x d x + \int - y d x$

Step 5.4

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

$f (x, y) = \frac{1}{4} x^{4} + C + y^{2} (\frac{1}{2} x^{2} + C) + \int - y d x$

Step 5.5

Apply the constant rule.

$f (x, y) = \frac{1}{4} x^{4} + C + y^{2} (\frac{1}{2} x^{2} + C) - y x + C$

Step 5.6

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

$f (x, y) = \frac{1}{4} x^{4} + C + y^{2} (\frac{x^{2}}{2} + C) - y x + C$

Step 5.7

Simplify.

$f (x, y) = \frac{x^{4}}{4} + \frac{y^{2} x^{2}}{2} - y x + C$

Step 5.8

Reorder terms.

$f (x, y) = \frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + C$

Step 6

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

$f (x, y) = \frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + g (y)] = y^{3} + x^{2} y - x$

Step 8.2

Differentiate.

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Step 8.2.1

By the Sum Rule, the derivative of $\frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{4} x^{4}] + \frac{d}{d y} [\frac{1}{2} y^{2} x^{2}] + \frac{d}{d y} [- y x] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [\frac{1}{4} x^{4}] + \frac{d}{d y} [\frac{1}{2} y^{2} x^{2}] + \frac{d}{d y} [- y x] + \frac{d}{d y} [g (y)] = y^{3} + x^{2} y - x$

Step 8.2.2

Since $\frac{1}{4} x^{4}$ is constant with respect to $y$ , the derivative of $\frac{1}{4} x^{4}$ with respect to $y$ is $0$ .

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

Step 8.3

Evaluate $\frac{d}{d y} [\frac{1}{2} y^{2} x^{2}]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

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

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

Step 8.3.6

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

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

Step 8.3.7

Cancel the common factor of $2$ .

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Step 8.3.7.1

Cancel the common factor.

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

Step 8.3.7.2

Divide $x^{2} y$ by $1$ .

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

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $- 1$ by $1$ .

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

Step 8.5

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

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

Step 8.6

Simplify.

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Step 8.6.1

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

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

Step 8.6.2

Reorder terms.

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

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

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

Step 9.1.2

Add $x$ to both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Add $- x$ and $x$ .

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

Step 9.1.3.4

Add $y^{3}$ and $0$ .

$\frac{d g}{d y} = y^{3}$

Step 10

Find the antiderivative of $y^{3}$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = y^{3}$ .

$\int \frac{d g}{d y} d y = \int y^{3} d y$

Step 10.2

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

$g (y) = \int y^{3} d y$

Step 10.3

By the Power Rule, the integral of $y^{3}$ with respect to $y$ is $\frac{1}{4} y^{4}$ .

$g (y) = \frac{1}{4} y^{4} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = \frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + g (y)$ .

$f (x, y) = \frac{1}{4} x^{4} + \frac{1}{2} y^{2} x^{2} - y x + \frac{1}{4} y^{4} + C$

Step 12

Simplify each term.

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Step 12.1

Combine $\frac{1}{4}$ and $x^{4}$ .

$f (x, y) = \frac{x^{4}}{4} + \frac{1}{2} y^{2} x^{2} - y x + \frac{1}{4} y^{4} + C$

Step 12.2

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

$f (x, y) = \frac{x^{4}}{4} + \frac{y^{2}}{2} x^{2} - y x + \frac{1}{4} y^{4} + C$

Step 12.3

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

$f (x, y) = \frac{x^{4}}{4} + \frac{y^{2} x^{2}}{2} - y x + \frac{1}{4} y^{4} + C$

Step 12.4

Combine $\frac{1}{4}$ and $y^{4}$ .

$f (x, y) = \frac{x^{4}}{4} + \frac{y^{2} x^{2}}{2} - y x + \frac{y^{4}}{4} + C$