Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $2$ .

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 4 x^{3} y + 0$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 4 x^{3} y$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Move $4$ to the left of $y$ .

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

Step 2.4

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

$\frac{\partial N}{\partial x} = 4 y x^{3} + 0$

Step 2.5

Simplify.

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Step 2.5.1

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

$\frac{\partial N}{\partial x} = 4 y x^{3}$

Step 2.5.2

Reorder the factors of $4 y x^{3}$ .

$\frac{\partial N}{\partial x} = 4 x^{3} y$

Step 3

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

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Step 3.1

Substitute $4 x^{3} y$ for $\frac{\partial M}{\partial y}$ and $4 x^{3} y$ for $\frac{\partial N}{\partial x}$ .

$4 x^{3} y = 4 x^{3} y$

Step 3.2

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

$4 x^{3} y = 4 x^{3} y$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 5.7

Simplify.

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Step 5.7.1

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

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

Step 5.7.2

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

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

Step 5.7.3

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

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

Step 5.7.4

Combine $2$ and $\frac{y^{4}}{4}$ .

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

Step 5.7.5

Cancel the common factor of $2$ and $4$ .

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Step 5.7.5.1

Factor $2$ out of $2 y^{4}$ .

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

Step 5.7.5.2

Cancel the common factors.

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Step 5.7.5.2.1

Factor $2$ out of $4$ .

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

Step 5.7.5.2.2

Cancel the common factor.

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

Step 5.7.5.2.3

Rewrite the expression.

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

Step 5.8

Reorder terms.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d x} [\frac{1}{2} x^{4} y^{2}]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

Combine $4$ and $\frac{y^{2}}{2}$ .

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

Step 8.3.6

Combine $\frac{4 y^{2}}{2}$ and $x^{3}$ .

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

Step 8.3.7

Cancel the common factor of $4$ and $2$ .

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Step 8.3.7.1

Factor $2$ out of $4 y^{2} x^{3}$ .

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

Step 8.3.7.2

Cancel the common factors.

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Step 8.3.7.2.1

Factor $2$ out of $2$ .

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

Step 8.3.7.2.2

Cancel the common factor.

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

Step 8.3.7.2.3

Rewrite the expression.

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

Step 8.3.7.2.4

Divide $2 y^{2} x^{3}$ by $1$ .

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

Step 8.4

Since $\frac{1}{2} y^{4}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} y^{4}$ with respect to $x$ is $0$ .

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

Step 8.5

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

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

Step 8.6

Simplify.

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Step 8.6.1

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

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

Step 8.6.2

Reorder terms.

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

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

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

Step 9.1.2

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

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Step 9.1.2.1

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

$\frac{d g}{d x} = 0 - e^{x}$

Step 9.1.2.2

Subtract $e^{x}$ from $0$ .

$\frac{d g}{d x} = - e^{x}$

Step 10

Find the antiderivative of $- e^{x}$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = - e^{x}$ .

$\int \frac{d g}{d x} d x = \int - e^{x} d x$

Step 10.2

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

$g (x) = \int - e^{x} d x$

Step 10.3

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

$g (x) = - \int e^{x} d x$

Step 10.4

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

$g (x) = - (e^{x} + C)$

Step 10.5

Simplify.

$g (x) = - e^{x} + C$

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

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

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

Step 12.2

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

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

Step 12.3

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

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