Calculus Examples

$(2 y^{2} + 3 x y - 2 y + 6 x) d x + x (x + 2 y - 1) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 y^{2} + 3 x y - 2 y + 6 x$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y^{2} + 3 x y - 2 y + 6 x]$

Step 1.2

By the Sum Rule, the derivative of $2 y^{2} + 3 x y - 2 y + 6 x$ with respect to $y$ is $\frac{d}{d y} [2 y^{2}] + \frac{d}{d y} [3 x y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y^{2}] + \frac{d}{d y} [3 x y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 2 \frac{d}{d y} [y^{2}] + \frac{d}{d y} [3 x y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 4 y + \frac{d}{d y} [3 x y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 4 y + 3 x \frac{d}{d y} [y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 4 y + 3 x \cdot 1 + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.4.3

Multiply $3$ by $1$ .

$\frac{\partial M}{\partial y} = 4 y + 3 x + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [6 x]$

Step 1.5

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

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Step 1.5.1

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

$\frac{\partial M}{\partial y} = 4 y + 3 x - 2 \frac{d}{d y} [y] + \frac{d}{d y} [6 x]$

Step 1.5.2

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

$\frac{\partial M}{\partial y} = 4 y + 3 x - 2 \cdot 1 + \frac{d}{d y} [6 x]$

Step 1.5.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 4 y + 3 x - 2 + \frac{d}{d y} [6 x]$

Step 1.6

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

$\frac{\partial M}{\partial y} = 4 y + 3 x - 2 + 0$

Step 1.7

Simplify.

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Step 1.7.1

Add $4 y + 3 x - 2$ and $0$ .

$\frac{\partial M}{\partial y} = 4 y + 3 x - 2$

Step 1.7.2

Reorder terms.

$\frac{\partial M}{\partial y} = 3 x + 4 y - 2$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = x + 2 y - 1$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

$\frac{\partial N}{\partial x} = x (1 + 0 + \frac{d}{d x} [- 1]) + (x + 2 y - 1) \frac{d}{d x} [x]$

Step 2.3.4

Add $1$ and $0$ .

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

Step 2.3.5

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

$\frac{\partial N}{\partial x} = x (1 + 0) + (x + 2 y - 1) \frac{d}{d x} [x]$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $1$ and $0$ .

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

Step 2.3.6.2

Multiply $x$ by $1$ .

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

Step 2.3.7

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

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

Step 2.3.8

Simplify by adding terms.

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Step 2.3.8.1

Multiply $x + 2 y - 1$ by $1$ .

$\frac{\partial N}{\partial x} = x + x + 2 y - 1$

Step 2.3.8.2

Add $x$ and $x$ .

$\frac{\partial N}{\partial x} = 2 x + 2 y - 1$

Step 3

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

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Step 3.1

Substitute $3 x + 4 y - 2$ for $\frac{\partial M}{\partial y}$ and $2 x + 2 y - 1$ for $\frac{\partial N}{\partial x}$ .

$3 x + 4 y - 2 = 2 x + 2 y - 1$

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$3 x + 4 y - 2 = 2 x + 2 y - 1$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

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Step 4.1

Substitute $3 x + 4 y - 2$ for $\frac{\partial M}{\partial y}$ .

$\frac{3 x + 4 y - 2 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $2 x + 2 y - 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{3 x + 4 y - 2 - (2 x + 2 y - 1)}{N}$

Step 4.3

Substitute $x (x + 2 y - 1)$ for $N$ .

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Step 4.3.1

Substitute $x (x + 2 y - 1)$ for $N$ .

$\frac{3 x + 4 y - 2 - (2 x + 2 y - 1)}{x (x + 2 y - 1)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{3 x + 4 y - 2 - (2 x) - (2 y) - - 1}{x (x + 2 y - 1)}$

Step 4.3.2.2

Simplify.

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Step 4.3.2.2.1

Multiply $2$ by $- 1$ .

$\frac{3 x + 4 y - 2 - 2 x - (2 y) - - 1}{x (x + 2 y - 1)}$

Step 4.3.2.2.2

Multiply $2$ by $- 1$ .

$\frac{3 x + 4 y - 2 - 2 x - 2 y - - 1}{x (x + 2 y - 1)}$

Step 4.3.2.2.3

Multiply $- 1$ by $- 1$ .

$\frac{3 x + 4 y - 2 - 2 x - 2 y + 1}{x (x + 2 y - 1)}$

Step 4.3.2.3

Subtract $2 x$ from $3 x$ .

$\frac{x + 4 y - 2 - 2 y + 1}{x (x + 2 y - 1)}$

Step 4.3.2.4

Subtract $2 y$ from $4 y$ .

$\frac{x + 2 y - 2 + 1}{x (x + 2 y - 1)}$

Step 4.3.2.5

Add $- 2$ and $1$ .

$\frac{x + 2 y - 1}{x (x + 2 y - 1)}$

Step 4.3.3

Cancel the common factor of $x + 2 y - 1$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{x + 2 y - 1}{x (x + 2 y - 1)}$

Step 4.3.3.2

Rewrite the expression.

$\frac{1}{x}$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

$μ (x, y) = e^{\int \frac{1}{x} d x}$

Step 5

Evaluate the integral $e^{\int \frac{1}{x} d x}$ .

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Step 5.1

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$μ (x, y) = e^{\ln (| x |) + C}$

Step 5.2

Simplify the answer.

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Step 5.2.1

Simplify.

$μ (x, y) = e^{\ln (| x |)} + C$

Step 5.2.2

Exponentiation and log are inverse functions.

$μ (x, y) = | x | + C$

Step 6

Multiply both sides of $(2 y^{2} + 3 x y - 2 y + 6 x) d x + x (x + 2 y - 1) d y = 0$ by the integration factor $x$ .

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Step 6.1

Multiply $2 y^{2} + 3 x y - 2 y + 6 x$ by $x$ .

$(2 y^{2} + 3 x y - 2 y + 6 x) x d x + x (x + 2 y - 1) d y = 0$

Step 6.2

Apply the distributive property.

$(2 y^{2} x + 3 x y x - 2 y x + 6 x \cdot x) d x + x (x + 2 y - 1) d y = 0$

Step 6.3

Simplify.

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Step 6.3.1

Multiply $x$ by $x$ by adding the exponents.

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Step 6.3.1.1

Move $x$ .

$(2 y^{2} x + 3 (x \cdot x) y - 2 y x + 6 x \cdot x) d x + x (x + 2 y - 1) d y = 0$

Step 6.3.1.2

Multiply $x$ by $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x \cdot x) d x + x (x + 2 y - 1) d y = 0$

Step 6.3.2

Multiply $x$ by $x$ by adding the exponents.

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Step 6.3.2.1

Move $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 (x \cdot x)) d x + x (x + 2 y - 1) d y = 0$

Step 6.3.2.2

Multiply $x$ by $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + x (x + 2 y - 1) d y = 0$

Step 6.4

Multiply $x (x + 2 y - 1)$ by $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + x (x + 2 y - 1) x d y = 0$

Step 6.5

Multiply $x$ by $x$ by adding the exponents.

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Step 6.5.1

Move $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + x \cdot x (x + 2 y - 1) d y = 0$

Step 6.5.2

Multiply $x$ by $x$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + x^{2} (x + 2 y - 1) d y = 0$

Step 6.6

Apply the distributive property.

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{2} x + x^{2} (2 y) + x^{2} \cdot - 1) d y = 0$

Step 6.7

Simplify.

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Step 6.7.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 6.7.1.1

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

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Step 6.7.1.1.1

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

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{2} x^{1} + x^{2} (2 y) + x^{2} \cdot - 1) d y = 0$

Step 6.7.1.1.2

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

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{2 + 1} + x^{2} (2 y) + x^{2} \cdot - 1) d y = 0$

Step 6.7.1.2

Add $2$ and $1$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{3} + x^{2} (2 y) + x^{2} \cdot - 1) d y = 0$

Step 6.7.2

Rewrite using the commutative property of multiplication.

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{3} + 2 x^{2} y + x^{2} \cdot - 1) d y = 0$

Step 6.7.3

Move $- 1$ to the left of $x^{2}$ .

$(2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}) d x + (x^{3} + 2 x^{2} y - 1 \cdot x^{2}) d y = 0$

Step 6.8

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

Split the single integral into multiple integrals.

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

Step 8.2

Apply the constant rule.

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

Apply the constant rule.

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

Step 8.6

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

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

Step 8.7

Simplify.

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial x} = 2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 11.3.3

Move $3$ to the left of $y$ .

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

Step 11.4

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

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Step 11.4.1

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

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

Step 11.4.2

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

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

Step 11.4.3

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

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

Step 11.5

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

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Step 11.5.1

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

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

Step 11.5.2

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

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

Step 11.5.3

Multiply $2$ by $- 1$ .

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

Step 11.6

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

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

Step 11.7

Reorder terms.

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

Step 12

Solve for $\frac{d g}{d x}$ .

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Step 12.1

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 12.1.1

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

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

Step 12.1.2

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

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

Step 12.1.3

Add $2 x y$ to both sides of the equation.

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

Step 12.1.4

Combine the opposite terms in $2 y^{2} x + 3 x^{2} y - 2 y x + 6 x^{2} - 3 x^{2} y - 2 y^{2} x + 2 x y$ .

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Step 12.1.4.1

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

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

Step 12.1.4.2

Add $2 y^{2} x - 2 y x + 6 x^{2}$ and $0$ .

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

Step 12.1.4.3

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

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

Step 12.1.4.4

Add $- 2 y x + 6 x^{2}$ and $0$ .

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

Step 12.1.4.5

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

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

Step 12.1.4.6

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

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

Step 12.1.4.7

Add $6 x^{2}$ and $0$ .

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

Step 13

Find the antiderivative of $6 x^{2}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = 6 x^{2}$ .

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

Step 13.2

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

$g (x) = \int 6 x^{2} d x$

Step 13.3

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

$g (x) = 6 \int x^{2} d x$

Step 13.4

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

$g (x) = 6 (\frac{1}{3} x^{3} + C)$

Step 13.5

Simplify the answer.

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Step 13.5.1

Rewrite $6 (\frac{1}{3} x^{3} + C)$ as $6 (\frac{1}{3}) x^{3} + C$ .

$g (x) = 6 (\frac{1}{3}) x^{3} + C$

Step 13.5.2

Simplify.

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Step 13.5.2.1

Combine $6$ and $\frac{1}{3}$ .

$g (x) = \frac{6}{3} x^{3} + C$

Step 13.5.2.2

Cancel the common factor of $6$ and $3$ .

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Step 13.5.2.2.1

Factor $3$ out of $6$ .

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

Step 13.5.2.2.2

Cancel the common factors.

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Step 13.5.2.2.2.1

Factor $3$ out of $3$ .

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

Step 13.5.2.2.2.2

Cancel the common factor.

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

Step 13.5.2.2.2.3

Rewrite the expression.

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

Step 13.5.2.2.2.4

Divide $2$ by $1$ .

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

Step 14

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

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