Calculus Examples

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

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 $N (x, y) = y (2 x - 3)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Multiply $2$ by $1$ .

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

Step 2.7

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ 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 $N (x, y)$ .

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

Step 5

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

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Step 5.1

Since $2 x - 3$ is constant with respect to $y$ , move $2 x - 3$ out of the integral.

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

Step 5.2

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

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

Step 5.3

Rewrite $(2 x - 3) (\frac{1}{2} y^{2} + C)$ as $(2 x - 3) \frac{1}{2} y^{2} + C$ .

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

Step 6

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

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

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 $x$ .

$\frac{d}{d x} [(2 x - 3) \frac{1}{2} y^{2} + g (x)] = 6 x + y^{2}$

Step 8.2

By the Sum Rule, the derivative of $(2 x - 3) \frac{1}{2} y^{2} + g (x)$ with respect to $x$ is $\frac{d}{d x} [(2 x - 3) \frac{1}{2} y^{2}] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [(2 x - 3) \frac{1}{2} y^{2}] + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.3

Evaluate $\frac{d}{d x} [(2 x - 3) \frac{1}{2} y^{2}]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

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

$\frac{y^{2}}{2} (2 \cdot 1 + \frac{d}{d x} [- 3]) + \frac{d}{d x} [g (x)] = 6 x + y^{2}$

Step 8.3.6

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

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

Step 8.3.7

Multiply $2$ by $1$ .

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

Step 8.3.8

Add $2$ and $0$ .

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

Step 8.3.9

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

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

Step 8.3.10

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

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

Step 8.3.11

Cancel the common factor of $2$ .

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Step 8.3.11.1

Cancel the common factor.

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

Step 8.3.11.2

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

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

Step 8.4

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

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

Step 8.5

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

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

Step 12

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

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Step 12.1

Simplify each term.

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Step 12.1.1

Apply the distributive property.

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

Step 12.1.2

Cancel the common factor of $2$ .

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Step 12.1.2.1

Factor $2$ out of $2 x$ .

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

Step 12.1.2.2

Cancel the common factor.

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

Step 12.1.2.3

Rewrite the expression.

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

Step 12.1.3

Combine $- 3$ and $\frac{1}{2}$ .

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

Step 12.1.4

Move the negative in front of the fraction.

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

Step 12.1.5

Apply the distributive property.

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

Step 12.1.6

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

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

Step 12.1.7

Move $3$ to the left of $y^{2}$ .

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

Step 12.1.8

Reorder the factors of $x y^{2}$ .

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

Step 12.1.9

Combine $y^{2} x$ and $- \frac{3 y^{2}}{2}$ using a common denominator.

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Step 12.1.9.1

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

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

Step 12.1.9.2

To write $x y^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 12.1.9.3

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

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

Step 12.1.9.4

Combine the numerators over the common denominator.

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

Step 12.1.10

Simplify the numerator.

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Step 12.1.10.1

Factor $y^{2}$ out of $x y^{2} \cdot 2 - 3 y^{2}$ .

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Step 12.1.10.1.1

Factor $y^{2}$ out of $x y^{2} \cdot 2$ .

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

Step 12.1.10.1.2

Factor $y^{2}$ out of $- 3 y^{2}$ .

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

Step 12.1.10.1.3

Factor $y^{2}$ out of $y^{2} (x \cdot 2) + y^{2} \cdot - 3$ .

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

Step 12.1.10.2

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

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

Step 12.2

To write $3 x^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 12.3

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

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

Step 12.4

Combine the numerators over the common denominator.

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

Step 12.5

Simplify the numerator.

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Step 12.5.1

Apply the distributive property.

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

Step 12.5.2

Rewrite using the commutative property of multiplication.

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

Step 12.5.3

Move $- 3$ to the left of $y^{2}$ .

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

Step 12.5.4

Multiply $2$ by $3$ .

$f (x, y) = \frac{2 y^{2} x - 3 y^{2} + 6 x^{2}}{2} + C$