Calculus Examples

$(6 x + y^{2}) d x + y (2 x + 3 y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 6 x + y^{2}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [6 x + y^{2}]$

Step 1.2

By the Sum Rule, the derivative of $6 x + y^{2}$ with respect to $y$ is $\frac{d}{d y} [6 x] + \frac{d}{d y} [y^{2}]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [6 x] + \frac{d}{d y} [y^{2}]$

Step 1.3

Since $6 x$ is constant with respect to $y$ , the derivative of $6 x$ with respect to $y$ is $0$ .

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

Step 1.4

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

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

Step 1.5

Add $0$ and $2 y$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Multiply $2$ by $1$ .

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

Step 2.7

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

$\frac{\partial N}{\partial x} = y (2 + 0)$

Step 2.8

Simplify the expression.

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Step 2.8.1

Add $2$ and $0$ .

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

Step 2.8.2

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

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

Step 3

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

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Step 3.1

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

$2 y = 2 y$

Step 3.2

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

$2 y = 2 y$ is an identity.

Step 4

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

$f (x, y) = \int y (2 x + 3 y) d y$

Step 5

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

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Step 5.1

Expand $y (2 x + 3 y)$ .

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Step 5.1.1

Apply the distributive property.

$f (x, y) = \int y (2 x) + y (3 y) d y$

Step 5.1.2

Remove parentheses.

$f (x, y) = \int y \cdot 2 x + y (3 y) d y$

Step 5.1.3

Reorder $y$ and $2$ .

$f (x, y) = \int 2 \cdot y x + y (3 y) d y$

Step 5.1.4

Remove parentheses.

$f (x, y) = \int 2 \cdot y x + y \cdot 3 y d y$

Step 5.1.5

Reorder $y$ and $3$ .

$f (x, y) = \int 2 y x + 3 y \cdot y d y$

Step 5.1.6

Raise $y$ to the power of $1$ .

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

Step 5.1.7

Raise $y$ to the power of $1$ .

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

Step 5.1.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.1.9

Add $1$ and $1$ .

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

Step 5.1.10

Reorder $2 y x$ and $3 y^{2}$ .

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

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 5.8

Simplify.

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Step 5.8.1

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

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

Step 5.8.2

Cancel the common factor of $3$ .

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Step 5.8.2.1

Cancel the common factor.

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

Step 5.8.2.2

Rewrite the expression.

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

Step 5.8.3

Multiply $y^{3}$ by $1$ .

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

Step 5.8.4

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

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

Step 5.8.5

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

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

Step 5.8.6

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

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

Step 5.8.7

Multiply $2$ by $x$ .

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

Step 5.8.8

Cancel the common factor of $2$ .

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Step 5.8.8.1

Cancel the common factor.

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

Step 5.8.8.2

Divide $x$ by $1$ .

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 6 x + y^{2}$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [y^{3} + x y^{2} + g (x)] = 6 x + y^{2}$

Step 8.2

Differentiate.

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Step 8.2.1

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

$\frac{d}{d x} [y^{3}] + \frac{d}{d x} [x y^{2}] + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.2.2

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

$0 + \frac{d}{d x} [x y^{2}] + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.3

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

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Step 8.3.1

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

$0 + y^{2} \frac{d}{d x} [x] + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.3.2

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

$0 + y^{2} \cdot 1 + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.3.3

Multiply $y^{2}$ by $1$ .

$0 + y^{2} + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.4

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

$0 + y^{2} + \frac{d g}{d x} = 6 x + y^{2}$

Step 8.5

Simplify.

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Step 8.5.1

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

$y^{2} + \frac{d g}{d x} = 6 x + y^{2}$

Step 8.5.2

Reorder terms.

$\frac{d g}{d x} + y^{2} = 6 x + y^{2}$

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

$\frac{d g}{d x} = 6 x + y^{2} - y^{2}$

Step 9.1.2

Combine the opposite terms in $6 x + y^{2} - y^{2}$ .

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Step 9.1.2.1

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

$\frac{d g}{d x} = 6 x + 0$

Step 9.1.2.2

Add $6 x$ and $0$ .

$\frac{d g}{d x} = 6 x$

Step 10

Find the antiderivative of $6 x$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 6 x$ .

$\int \frac{d g}{d x} d x = \int 6 x d x$

Step 10.2

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

$g (x) = \int 6 x d x$

Step 10.3

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$g (x) = 6 \int x d x$

Step 10.4

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

$g (x) = 6 (\frac{1}{2} x^{2} + C)$

Step 10.5

Simplify the answer.

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Step 10.5.1

Rewrite $6 (\frac{1}{2} x^{2} + C)$ as $6 (\frac{1}{2}) x^{2} + C$ .

$g (x) = 6 (\frac{1}{2}) x^{2} + C$

Step 10.5.2

Simplify.

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Step 10.5.2.1

Combine $6$ and $\frac{1}{2}$ .

$g (x) = \frac{6}{2} x^{2} + C$

Step 10.5.2.2

Cancel the common factor of $6$ and $2$ .

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Step 10.5.2.2.1

Factor $2$ out of $6$ .

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

Step 10.5.2.2.2

Cancel the common factors.

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Step 10.5.2.2.2.1

Factor $2$ out of $2$ .

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

Step 10.5.2.2.2.2

Cancel the common factor.

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

Step 10.5.2.2.2.3

Rewrite the expression.

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

Step 10.5.2.2.2.4

Divide $3$ by $1$ .

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

Step 11

Substitute for $g (x)$ in $f (x, y) = y^{3} + x y^{2} + g (x)$ .

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