Calculus Examples

$(4 x^{3} y^{3} - 2 x y) d x + (3 x^{4} y^{2} - x^{2}) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 4 x^{3} y^{3} - 2 x y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [4 x^{3} y^{3} - 2 x y]$

Step 1.2

By the Sum Rule, the derivative of $4 x^{3} y^{3} - 2 x y$ with respect to $y$ is $\frac{d}{d y} [4 x^{3} y^{3}] + \frac{d}{d y} [- 2 x y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [4 x^{3} y^{3}] + \frac{d}{d y} [- 2 x y]$

Step 1.3

Evaluate $\frac{d}{d y} [4 x^{3} y^{3}]$ .

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Step 1.3.1

Since $4 x^{3}$ is constant with respect to $y$ , the derivative of $4 x^{3} y^{3}$ with respect to $y$ is $4 x^{3} \frac{d}{d y} [y^{3}]$ .

$\frac{\partial M}{\partial y} = 4 x^{3} \frac{d}{d y} [y^{3}] + \frac{d}{d y} [- 2 x y]$

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $4$ .

$\frac{\partial M}{\partial y} = 12 x^{3} y^{2} + \frac{d}{d y} [- 2 x y]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 12 x^{3} y^{2} - 2 x \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 12 x^{3} y^{2} - 2 x \cdot 1$

Step 1.4.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 12 x^{3} y^{2} - 2 x$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 x^{4} y^{2} - x^{2}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{4} y^{2} - x^{2}]$

Step 2.2

By the Sum Rule, the derivative of $3 x^{4} y^{2} - x^{2}$ with respect to $x$ is $\frac{d}{d x} [3 x^{4} y^{2}] + \frac{d}{d x} [- x^{2}]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{4} y^{2}] + \frac{d}{d x} [- x^{2}]$

Step 2.3

Evaluate $\frac{d}{d x} [3 x^{4} y^{2}]$ .

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Step 2.3.1

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 x^{4} y^{2}$ with respect to $x$ is $3 y^{2} \frac{d}{d x} [x^{4}]$ .

$\frac{\partial N}{\partial x} = 3 y^{2} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- x^{2}]$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 2.3.3

Multiply $4$ by $3$ .

$\frac{\partial N}{\partial x} = 12 y^{2} x^{3} + \frac{d}{d x} [- x^{2}]$

Step 2.4

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

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Step 2.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{\partial N}{\partial x} = 12 y^{2} x^{3} - \frac{d}{d x} [x^{2}]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 12 y^{2} x^{3} - (2 x)$

Step 2.4.3

Multiply $2$ by $- 1$ .

$\frac{\partial N}{\partial x} = 12 y^{2} x^{3} - 2 x$

Step 2.5

Reorder terms.

$\frac{\partial N}{\partial x} = 12 x^{3} y^{2} - 2 x$

Step 3

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

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Step 3.1

Substitute $12 x^{3} y^{2} - 2 x$ for $\frac{\partial M}{\partial y}$ and $12 x^{3} y^{2} - 2 x$ for $\frac{\partial N}{\partial x}$ .

$12 x^{3} y^{2} - 2 x = 12 x^{3} y^{2} - 2 x$

Step 3.2

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

$12 x^{3} y^{2} - 2 x = 12 x^{3} y^{2} - 2 x$ is an identity.

Step 4

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

$f (x, y) = \int 4 x^{3} y^{3} - 2 x y d x$

Step 5

Integrate $M (x, y) = 4 x^{3} y^{3} - 2 x y$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 4 x^{3} y^{3} d x + \int - 2 x y d x$

Step 5.2

Since $4 y^{3}$ is constant with respect to $x$ , move $4 y^{3}$ out of the integral.

$f (x, y) = 4 y^{3} \int x^{3} d x + \int - 2 x y d x$

Step 5.3

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

$f (x, y) = 4 y^{3} (\frac{1}{4} x^{4} + C) + \int - 2 x y d x$

Step 5.4

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

$f (x, y) = 4 y^{3} (\frac{1}{4} x^{4} + C) - 2 y \int x d x$

Step 5.5

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

$f (x, y) = 4 y^{3} (\frac{1}{4} x^{4} + C) - 2 y (\frac{1}{2} x^{2} + C)$

Step 5.6

Simplify.

$f (x, y) = 4 \cdot \frac{1}{4} y^{3} x^{4} - 2 y (\frac{1}{2} x^{2}) + C$

Step 5.7

Simplify.

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Step 5.7.1

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

$f (x, y) = \frac{4}{4} y^{3} x^{4} - 2 y (\frac{1}{2} x^{2}) + C$

Step 5.7.2

Cancel the common factor of $4$ .

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Step 5.7.2.1

Cancel the common factor.

$f (x, y) = \frac{4}{4} y^{3} x^{4} - 2 y (\frac{1}{2} x^{2}) + C$

Step 5.7.2.2

Rewrite the expression.

$f (x, y) = 1 y^{3} x^{4} - 2 y (\frac{1}{2} x^{2}) + C$

Step 5.7.3

Multiply $y^{3}$ by $1$ .

$f (x, y) = y^{3} x^{4} - 2 y (\frac{1}{2} x^{2}) + C$

Step 5.7.4

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

$f (x, y) = y^{3} x^{4} - 2 y \frac{x^{2}}{2} + C$

Step 5.7.5

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

$f (x, y) = y^{3} x^{4} + y \frac{- 2 x^{2}}{2} + C$

Step 5.7.6

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

$f (x, y) = y^{3} x^{4} + \frac{y (- 2 x^{2})}{2} + C$

Step 5.7.7

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

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Step 5.7.7.1

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

$f (x, y) = y^{3} x^{4} + \frac{2 (y (- x^{2}))}{2} + C$

Step 5.7.7.2

Cancel the common factors.

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Step 5.7.7.2.1

Factor $2$ out of $2$ .

$f (x, y) = y^{3} x^{4} + \frac{2 (y (- x^{2}))}{2 (1)} + C$

Step 5.7.7.2.2

Cancel the common factor.

$f (x, y) = y^{3} x^{4} + \frac{2 (y (- x^{2}))}{2 \cdot 1} + C$

Step 5.7.7.2.3

Rewrite the expression.

$f (x, y) = y^{3} x^{4} + \frac{y (- x^{2})}{1} + C$

Step 5.7.7.2.4

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

$f (x, y) = y^{3} x^{4} + y (- x^{2}) + C$

Step 6

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

$f (x, y) = y^{3} x^{4} + y (- x^{2}) + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = 3 x^{4} y^{2} - x^{2}$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

By the Sum Rule, the derivative of $y^{3} x^{4} + y (- x^{2}) + g (y)$ with respect to $y$ is $\frac{d}{d y} [y^{3} x^{4}] + \frac{d}{d y} [y (- x^{2})] + \frac{d}{d y} [g (y)]$ .

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

Step 8.3

Evaluate $\frac{d}{d y} [y^{3} x^{4}]$ .

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Step 8.3.1

Since $x^{4}$ is constant with respect to $y$ , the derivative of $y^{3} x^{4}$ with respect to $y$ is $x^{4} \frac{d}{d y} [y^{3}]$ .

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

Step 8.3.2

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

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

Step 8.3.3

Move $3$ to the left of $x^{4}$ .

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $- 1$ by $1$ .

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

Step 8.5

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

$3 x^{4} y^{2} - x^{2} + \frac{d g}{d y} = 3 x^{4} y^{2} - x^{2}$

Step 8.6

Reorder terms.

$\frac{d g}{d y} - x^{2} + 3 x^{4} y^{2} = 3 x^{4} y^{2} - x^{2}$

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

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

$\frac{d g}{d y} + 3 x^{4} y^{2} = 3 x^{4} y^{2} - x^{2} + x^{2}$

Step 9.1.2

Subtract $3 x^{4} y^{2}$ from both sides of the equation.

$\frac{d g}{d y} = 3 x^{4} y^{2} - x^{2} + x^{2} - 3 x^{4} y^{2}$

Step 9.1.3

Combine the opposite terms in $3 x^{4} y^{2} - x^{2} + x^{2} - 3 x^{4} y^{2}$ .

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Step 9.1.3.1

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

$\frac{d g}{d y} = 3 x^{4} y^{2} + 0 - 3 x^{4} y^{2}$

Step 9.1.3.2

Add $3 x^{4} y^{2}$ and $0$ .

$\frac{d g}{d y} = 3 x^{4} y^{2} - 3 x^{4} y^{2}$

Step 9.1.3.3

Subtract $3 x^{4} y^{2}$ from $3 x^{4} y^{2}$ .

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

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 $f (x, y) = y^{3} x^{4} + y (- x^{2}) + g (y)$ .

$f (x, y) = y^{3} x^{4} + y (- x^{2}) + C$

Step 12

Rewrite using the commutative property of multiplication.

$f (x, y) = y^{3} x^{4} - y x^{2} + C$