Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $2 x^{3}$ is constant with respect to $y$ , the derivative of $2 x^{3}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [x y^{2}] + \frac{d}{d y} [2 y] + \frac{d}{d y} [3]$

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

$\frac{\partial M}{\partial y} = 0 + x \frac{d}{d y} [y^{2}] + \frac{d}{d y} [2 y] + \frac{d}{d y} [3]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 + x (2 y) + \frac{d}{d y} [2 y] + \frac{d}{d y} [3]$

Step 1.3.3

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

$\frac{\partial M}{\partial y} = 0 + 2 x y + \frac{d}{d y} [2 y] + \frac{d}{d y} [3]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 0 + 2 x y + 2 \frac{d}{d y} [y] + \frac{d}{d y} [3]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 0 + 2 x y + 2 \cdot 1 + \frac{d}{d y} [3]$

Step 1.4.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 + 2 x y + 2 + \frac{d}{d y} [3]$

Step 1.5

Since $3$ is constant with respect to $y$ , the derivative of $3$ with respect to $y$ is $0$ .

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

Step 1.6

Combine terms.

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Step 1.6.1

Add $0$ and $2 x y$ .

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

Step 1.6.2

Add $2 x y + 2$ and $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $2$ by $1$ .

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

Step 2.5

Reorder terms.

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

Step 3

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

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Step 3.1

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

$2 x y + 2 = 2 x y + 2$

Step 3.2

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

$2 x y + 2 = 2 x y + 2$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 5.7

Reorder terms.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

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

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

Step 8.3.6

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

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

Step 8.3.7

Cancel the common factor of $2$ .

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Step 8.3.7.1

Cancel the common factor.

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

Step 8.3.7.2

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

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $2$ by $1$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

Step 9

Solve for $\frac{d g}{d x}$ .

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Step 9.1

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 9.1.1

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

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

Step 9.1.2

Subtract $2 y$ from both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

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

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

Step 9.1.3.4

Subtract $2 y$ from $2 y$ .

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

Step 9.1.3.5

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

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

Step 10

Find the antiderivative of $2 x^{3} + 3$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 2 x^{3} + 3$ .

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

Step 10.2

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

$g (x) = \int 2 x^{3} + 3 d x$

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int 2 x^{3} d x + \int 3 d x$

Step 10.4

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

$g (x) = 2 \int x^{3} d x + \int 3 d x$

Step 10.5

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

$g (x) = 2 (\frac{1}{4} x^{4} + C) + \int 3 d x$

Step 10.6

Apply the constant rule.

$g (x) = 2 (\frac{1}{4} x^{4} + C) + 3 x + C$

Step 10.7

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

$g (x) = 2 (\frac{x^{4}}{4} + C) + 3 x + C$

Step 10.8

Simplify.

$g (x) = \frac{x^{4}}{2} + 3 x + C$

Step 10.9

Reorder terms.

$g (x) = \frac{1}{2} x^{4} + 3 x + C$

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

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

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

Step 12.2

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

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

Step 12.3

Combine $\frac{1}{2}$ and $x^{4}$ .

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