Calculus Examples

$(2 x y^{2} + 1) d x + (2 x^{2} y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y^{2} + 1$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 2 x (2 y) + \frac{d}{d y} [1]$

Step 1.3.3

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 4 x y + \frac{d}{d y} [1]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

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

Step 1.4.2

Add $4 x y$ and $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Simplify the expression.

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Step 2.4.1

Multiply $2$ by $2$ .

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

Step 2.4.2

Reorder the factors of $4 y x$ .

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

Step 3

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

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Step 3.1

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

$4 x y = 4 x y$

Step 3.2

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

$4 x y = 4 x y$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the answer.

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Step 5.3.1

Rewrite $2 x^{2} (\frac{1}{2} y^{2} + C)$ as $2 x^{2} \frac{1}{2} y^{2} + C$ .

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

Step 5.3.2

Simplify.

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

Move $2$ to the left of $x^{2}$ .

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

Step 5.3.2.4

Multiply $2$ by $x^{2}$ .

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

Step 5.3.2.5

Cancel the common factor of $2$ .

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Step 5.3.2.5.1

Cancel the common factor.

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

Step 5.3.2.5.2

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

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 2 x y^{2} + 1$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Move $2$ to the left of $y^{2}$ .

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

Step 8.4

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

$2 y^{2} x + \frac{d g}{d x} = 2 x y^{2} + 1$

Step 8.5

Reorder terms.

$\frac{d g}{d x} + 2 y^{2} x = 2 x y^{2} + 1$

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

$\frac{d g}{d x} = 2 x y^{2} + 1 - 2 y^{2} x$

Step 9.1.2

Combine the opposite terms in $2 x y^{2} + 1 - 2 y^{2} x$ .

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Step 9.1.2.1

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

$\frac{d g}{d x} = 2 y^{2} x + 1 - 2 y^{2} x$

Step 9.1.2.2

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

$\frac{d g}{d x} = 0 + 1$

Step 9.1.2.3

Add $0$ and $1$ .

$\frac{d g}{d x} = 1$

Step 10

Find the antiderivative of $1$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 1$ .

$\int \frac{d g}{d x} d x = \int d x$

Step 10.2

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

$g (x) = \int d x$

Step 10.3

Apply the constant rule.

$g (x) = x + C$

Step 11

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

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