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

Multiply $2$ by $- 1$ .

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

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

Subtract $2 x y$ from $0$ .

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

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

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

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

Step 2.3

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 2.4

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 2.5

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 2.6

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

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

Step 2.7

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 2.8

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 2.9

Multiply $2$ by $1$ .

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

Step 2.10

Simplify.

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Step 2.10.1

Apply the distributive property.

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

Step 2.10.2

Combine terms.

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Step 2.10.2.1

Multiply $2$ by $- 1$ .

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

Step 2.10.2.2

Multiply $- 1$ by $2$ .

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

Step 2.10.3

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

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

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

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

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 5.4

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 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 5.8

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

Differentiate using the Sum Rule.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

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

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

Step 8.2.1.2

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

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

Step 8.2.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

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

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

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

Step 8.3.6

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

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

Step 8.3.7

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

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

Step 8.3.8

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

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

Step 8.3.9

Cancel the common factor of $2$ .

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Step 8.3.9.1

Cancel the common factor.

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

Step 8.3.9.2

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

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

Step 8.3.10

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 8.4

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 8.5

Simplify.

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Step 8.5.1

Apply the distributive property.

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

Step 8.5.2

Multiply $2$ by $- 1$ .

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

Step 8.5.3

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

Add $y^{2} x$ to 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

Add $2 y$ to 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

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

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

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

Add $- 2 y$ and $2 y$ .

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

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

Simplify each term.

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Step 12.1.1

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 12.1.2

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 12.2

Apply the distributive property.

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

Step 12.3

Multiply $2$ by $- 1$ .

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

Step 12.4

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

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