Calculus Examples

$(21 x^{2} y^{4} - 8 x^{3} y) d x + (28 x^{3} y^{3} - 2 x^{4}) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 21 x^{2} y^{4} - 8 x^{3} y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [21 x^{2} y^{4} - 8 x^{3} y]$

Step 1.2

By the Sum Rule, the derivative of $21 x^{2} y^{4} - 8 x^{3} y$ with respect to $y$ is $\frac{d}{d y} [21 x^{2} y^{4}] + \frac{d}{d y} [- 8 x^{3} y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [21 x^{2} y^{4}] + \frac{d}{d y} [- 8 x^{3} y]$

Step 1.3

Evaluate $\frac{d}{d y} [21 x^{2} y^{4}]$ .

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Step 1.3.1

Since $21 x^{2}$ is constant with respect to $y$ , the derivative of $21 x^{2} y^{4}$ with respect to $y$ is $21 x^{2} \frac{d}{d y} [y^{4}]$ .

$\frac{\partial M}{\partial y} = 21 x^{2} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [- 8 x^{3} y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 21 x^{2} (4 y^{3}) + \frac{d}{d y} [- 8 x^{3} y]$

Step 1.3.3

Multiply $4$ by $21$ .

$\frac{\partial M}{\partial y} = 84 x^{2} y^{3} + \frac{d}{d y} [- 8 x^{3} y]$

Step 1.4

Evaluate $\frac{d}{d y} [- 8 x^{3} y]$ .

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Step 1.4.1

Since $- 8 x^{3}$ is constant with respect to $y$ , the derivative of $- 8 x^{3} y$ with respect to $y$ is $- 8 x^{3} \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = 84 x^{2} y^{3} - 8 x^{3} \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 84 x^{2} y^{3} - 8 x^{3} \cdot 1$

Step 1.4.3

Multiply $- 8$ by $1$ .

$\frac{\partial M}{\partial y} = 84 x^{2} y^{3} - 8 x^{3}$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = - 8 x^{3} + 84 y^{3} x^{2}$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 28 x^{3} y^{3} - 2 x^{4}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [28 x^{3} y^{3} - 2 x^{4}]$

Step 2.2

By the Sum Rule, the derivative of $28 x^{3} y^{3} - 2 x^{4}$ with respect to $x$ is $\frac{d}{d x} [28 x^{3} y^{3}] + \frac{d}{d x} [- 2 x^{4}]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [28 x^{3} y^{3}] + \frac{d}{d x} [- 2 x^{4}]$

Step 2.3

Evaluate $\frac{d}{d x} [28 x^{3} y^{3}]$ .

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Step 2.3.1

Since $28 y^{3}$ is constant with respect to $x$ , the derivative of $28 x^{3} y^{3}$ with respect to $x$ is $28 y^{3} \frac{d}{d x} [x^{3}]$ .

$\frac{\partial N}{\partial x} = 28 y^{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2 x^{4}]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 28 y^{3} (3 x^{2}) + \frac{d}{d x} [- 2 x^{4}]$

Step 2.3.3

Multiply $3$ by $28$ .

$\frac{\partial N}{\partial x} = 84 y^{3} x^{2} + \frac{d}{d x} [- 2 x^{4}]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 84 y^{3} x^{2} - 2 \frac{d}{d x} [x^{4}]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 84 y^{3} x^{2} - 2 (4 x^{3})$

Step 2.4.3

Multiply $4$ by $- 2$ .

$\frac{\partial N}{\partial x} = 84 y^{3} x^{2} - 8 x^{3}$

Step 2.5

Reorder terms.

$\frac{\partial N}{\partial x} = - 8 x^{3} + 84 y^{3} x^{2}$

Step 3

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

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Step 3.1

Substitute $- 8 x^{3} + 84 y^{3} x^{2}$ for $\frac{\partial M}{\partial y}$ and $- 8 x^{3} + 84 y^{3} x^{2}$ for $\frac{\partial N}{\partial x}$ .

$- 8 x^{3} + 84 y^{3} x^{2} = - 8 x^{3} + 84 y^{3} x^{2}$

Step 3.2

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

$- 8 x^{3} + 84 y^{3} x^{2} = - 8 x^{3} + 84 y^{3} x^{2}$ is an identity.

Step 4

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

$f (x, y) = \int 21 x^{2} y^{4} - 8 x^{3} y d x$

Step 5

Integrate $M (x, y) = 21 x^{2} y^{4} - 8 x^{3} y$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 21 x^{2} y^{4} d x + \int - 8 x^{3} y d x$

Step 5.2

Since $21 y^{4}$ is constant with respect to $x$ , move $21 y^{4}$ out of the integral.

$f (x, y) = 21 y^{4} \int x^{2} d x + \int - 8 x^{3} y d x$

Step 5.3

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

$f (x, y) = 21 y^{4} (\frac{1}{3} x^{3} + C) + \int - 8 x^{3} y d x$

Step 5.4

Since $- 8 y$ is constant with respect to $x$ , move $- 8 y$ out of the integral.

$f (x, y) = 21 y^{4} (\frac{1}{3} x^{3} + C) - 8 y \int x^{3} d x$

Step 5.5

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

$f (x, y) = 21 y^{4} (\frac{1}{3} x^{3} + C) - 8 y (\frac{1}{4} x^{4} + C)$

Step 5.6

Simplify.

$f (x, y) = 21 \cdot \frac{1}{3} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7

Simplify.

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Step 5.7.1

Combine $21$ and $\frac{1}{3}$ .

$f (x, y) = \frac{21}{3} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.2

Cancel the common factor of $21$ and $3$ .

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Step 5.7.2.1

Factor $3$ out of $21$ .

$f (x, y) = \frac{3 \cdot 7}{3} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.2.2

Cancel the common factors.

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Step 5.7.2.2.1

Factor $3$ out of $3$ .

$f (x, y) = \frac{3 \cdot 7}{3 (1)} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.2.2.2

Cancel the common factor.

$f (x, y) = \frac{3 \cdot 7}{3 \cdot 1} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.2.2.3

Rewrite the expression.

$f (x, y) = \frac{7}{1} y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.2.2.4

Divide $7$ by $1$ .

$f (x, y) = 7 y^{4} x^{3} - 8 y (\frac{1}{4} x^{4}) + C$

Step 5.7.3

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

$f (x, y) = 7 y^{4} x^{3} - 8 y \frac{x^{4}}{4} + C$

Step 5.7.4

Combine $- 8$ and $\frac{x^{4}}{4}$ .

$f (x, y) = 7 y^{4} x^{3} + y \frac{- 8 x^{4}}{4} + C$

Step 5.7.5

Combine $y$ and $\frac{- 8 x^{4}}{4}$ .

$f (x, y) = 7 y^{4} x^{3} + \frac{y (- 8 x^{4})}{4} + C$

Step 5.7.6

Cancel the common factor of $- 8$ and $4$ .

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Step 5.7.6.1

Factor $4$ out of $y (- 8 x^{4})$ .

$f (x, y) = 7 y^{4} x^{3} + \frac{4 (y (- 2 x^{4}))}{4} + C$

Step 5.7.6.2

Cancel the common factors.

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Step 5.7.6.2.1

Factor $4$ out of $4$ .

$f (x, y) = 7 y^{4} x^{3} + \frac{4 (y (- 2 x^{4}))}{4 (1)} + C$

Step 5.7.6.2.2

Cancel the common factor.

$f (x, y) = 7 y^{4} x^{3} + \frac{4 (y (- 2 x^{4}))}{4 \cdot 1} + C$

Step 5.7.6.2.3

Rewrite the expression.

$f (x, y) = 7 y^{4} x^{3} + \frac{y (- 2 x^{4})}{1} + C$

Step 5.7.6.2.4

Divide $y (- 2 x^{4})$ by $1$ .

$f (x, y) = 7 y^{4} x^{3} + y (- 2 x^{4}) + C$

Step 6

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

$f (x, y) = 7 y^{4} x^{3} + y (- 2 x^{4}) + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = 28 x^{3} y^{3} - 2 x^{4}$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [7 y^{4} x^{3} + y (- 2 x^{4}) + g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.2

By the Sum Rule, the derivative of $7 y^{4} x^{3} + y (- 2 x^{4}) + g (y)$ with respect to $y$ is $\frac{d}{d y} [7 y^{4} x^{3}] + \frac{d}{d y} [y (- 2 x^{4})] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [7 y^{4} x^{3}] + \frac{d}{d y} [y (- 2 x^{4})] + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.3

Evaluate $\frac{d}{d y} [7 y^{4} x^{3}]$ .

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Step 8.3.1

Since $7 x^{3}$ is constant with respect to $y$ , the derivative of $7 y^{4} x^{3}$ with respect to $y$ is $7 x^{3} \frac{d}{d y} [y^{4}]$ .

$7 x^{3} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [y (- 2 x^{4})] + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.3.2

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

$7 x^{3} (4 y^{3}) + \frac{d}{d y} [y (- 2 x^{4})] + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.3.3

Multiply $4$ by $7$ .

$28 x^{3} y^{3} + \frac{d}{d y} [y (- 2 x^{4})] + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.4

Evaluate $\frac{d}{d y} [y (- 2 x^{4})]$ .

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Step 8.4.1

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

$28 x^{3} y^{3} - 2 x^{4} \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.4.2

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

$28 x^{3} y^{3} - 2 x^{4} \cdot 1 + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.4.3

Multiply $- 2$ by $1$ .

$28 x^{3} y^{3} - 2 x^{4} + \frac{d}{d y} [g (y)] = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.5

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

$28 x^{3} y^{3} - 2 x^{4} + \frac{d g}{d y} = 28 x^{3} y^{3} - 2 x^{4}$

Step 8.6

Reorder terms.

$\frac{d g}{d y} - 2 x^{4} + 28 x^{3} y^{3} = 28 x^{3} y^{3} - 2 x^{4}$

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 $2 x^{4}$ to both sides of the equation.

$\frac{d g}{d y} + 28 x^{3} y^{3} = 28 x^{3} y^{3} - 2 x^{4} + 2 x^{4}$

Step 9.1.2

Subtract $28 x^{3} y^{3}$ from both sides of the equation.

$\frac{d g}{d y} = 28 x^{3} y^{3} - 2 x^{4} + 2 x^{4} - 28 x^{3} y^{3}$

Step 9.1.3

Combine the opposite terms in $28 x^{3} y^{3} - 2 x^{4} + 2 x^{4} - 28 x^{3} y^{3}$ .

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Step 9.1.3.1

Add $- 2 x^{4}$ and $2 x^{4}$ .

$\frac{d g}{d y} = 28 x^{3} y^{3} + 0 - 28 x^{3} y^{3}$

Step 9.1.3.2

Add $28 x^{3} y^{3}$ and $0$ .

$\frac{d g}{d y} = 28 x^{3} y^{3} - 28 x^{3} y^{3}$

Step 9.1.3.3

Subtract $28 x^{3} y^{3}$ from $28 x^{3} y^{3}$ .

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

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 $f (x, y) = 7 y^{4} x^{3} + y (- 2 x^{4}) + g (y)$ .

$f (x, y) = 7 y^{4} x^{3} + y (- 2 x^{4}) + C$

Step 12

Rewrite using the commutative property of multiplication.

$f (x, y) = 7 y^{4} x^{3} - 2 y x^{4} + C$