Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [3 x^{2} y^{4}]$ .

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Step 1.3.1

Since $3 x^{2}$ is constant with respect to $y$ , the derivative of $3 x^{2} y^{4}$ with respect to $y$ is $3 x^{2} \frac{d}{d y} [y^{4}]$ .

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $4$ by $3$ .

$\frac{\partial M}{\partial y} = 12 x^{2} y^{3} + \frac{d}{d y} [2 x y]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 12 x^{2} y^{3} + 2 x \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 12 x^{2} y^{3} + 2 x \cdot 1$

Step 1.4.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 12 x^{2} y^{3} + 2 x$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = 12 y^{3} x^{2} + 2 x$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $3$ by $2$ .

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

Step 2.4

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

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Step 2.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 6 y^{3} x^{2} - (2 x)$

Step 2.4.3

Multiply $2$ by $- 1$ .

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

Step 3

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

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Step 3.1

Substitute $12 y^{3} x^{2} + 2 x$ for $\frac{\partial M}{\partial y}$ and $6 y^{3} x^{2} - 2 x$ for $\frac{\partial N}{\partial x}$ .

$12 y^{3} x^{2} + 2 x = 6 y^{3} x^{2} - 2 x$

Step 3.2

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

$12 y^{3} x^{2} + 2 x = 6 y^{3} x^{2} - 2 x$ is not an identity.

Step 4

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

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Step 4.1

Substitute $6 y^{3} x^{2} - 2 x$ for $\frac{\partial N}{\partial x}$ .

$\frac{6 y^{3} x^{2} - 2 x - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $12 y^{3} x^{2} + 2 x$ for $\frac{\partial M}{\partial y}$ .

$\frac{6 y^{3} x^{2} - 2 x - (12 y^{3} x^{2} + 2 x)}{M}$

Step 4.3

Substitute $3 x^{2} y^{4} + 2 x y$ for $M$ .

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Step 4.3.1

Substitute $3 x^{2} y^{4} + 2 x y$ for $M$ .

$\frac{6 y^{3} x^{2} - 2 x - (12 y^{3} x^{2} + 2 x)}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{6 y^{3} x^{2} - 2 x - (12 y^{3} x^{2}) - (2 x)}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2.2

Multiply $12$ by $- 1$ .

$\frac{6 y^{3} x^{2} - 2 x - 12 (y^{3} x^{2}) - (2 x)}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2.3

Multiply $2$ by $- 1$ .

$\frac{6 y^{3} x^{2} - 2 x - 12 (y^{3} x^{2}) - 2 x}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2.4

Subtract $12 y^{3} x^{2}$ from $6 y^{3} x^{2}$ .

$\frac{- 6 y^{3} x^{2} - 2 x - 2 x}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2.5

Subtract $2 x$ from $- 2 x$ .

$\frac{- 6 y^{3} x^{2} - 4 x}{3 x^{2} y^{4} + 2 x y}$

Step 4.3.2.6

Factor $2 x$ out of $- 6 y^{3} x^{2} - 4 x$ .

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Step 4.3.2.6.1

Factor $2 x$ out of $- 6 y^{3} x^{2}$ .

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

Step 4.3.2.6.2

Factor $2 x$ out of $- 4 x$ .

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

Step 4.3.2.6.3

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

Factor $x y$ out of $2 x y$ .

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

Step 4.3.3.3

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

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

Step 4.3.4

Cancel the common factor of $x$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Rewrite the expression.

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

Step 4.3.5

Cancel the common factor of $- 3 y^{3} x - 2$ and $3 x y^{3} + 2$ .

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Step 4.3.5.1

Factor $- 1$ out of $- 3 y^{3} x$ .

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

Step 4.3.5.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 4.3.5.3

Factor $- 1$ out of $- (3 y^{3} x) - 1 (2)$ .

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

Step 4.3.5.4

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

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

Step 4.3.5.5

Reorder terms.

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

Step 4.3.5.6

Cancel the common factor.

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

Step 4.3.5.7

Rewrite the expression.

$\frac{2 \cdot (- 1)}{y}$

Step 4.3.6

Multiply $2$ by $- 1$ .

$\frac{- 2}{y}$

Step 4.3.7

Substitute $3 x^{2} y^{4} + 2 x y$ for $M$ .

$- \frac{2}{y}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{2}{y} d y}$

Step 5

Evaluate the integral $e^{\int - \frac{2}{y} d y}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{2}{y} d y}$

Step 5.2

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

$μ (x, y) = e^{- (2 \int \frac{1}{y} d y)}$

Step 5.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{1}{y} d y}$

Step 5.4

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

$μ (x, y) = e^{- 2 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 2 \ln (| y |)$ by moving $- 2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 2})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

Remove the absolute value in ${| y |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

Multiply both sides of $(3 x^{2} y^{4} + 2 x y) d x + (2 x^{3} y^{3} - x^{2}) d y = 0$ by the integration factor $\frac{1}{y^{2}}$ .

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Step 6.1

Multiply $3 x^{2} y^{4} + 2 x y$ by $\frac{1}{y^{2}}$ .

$(3 x^{2} y^{4} + 2 x y) \frac{1}{y^{2}} d x + (2 x^{3} y^{3} - x^{2}) d y = 0$

Step 6.2

Multiply $3 x^{2} y^{4} + 2 x y$ by $\frac{1}{y^{2}}$ .

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Factor $x y$ out of $2 x y$ .

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

Step 6.3.3

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

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

Step 6.4

Cancel the common factor of $y$ and $y^{2}$ .

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Step 6.4.1

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

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

Step 6.4.2

Cancel the common factors.

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Step 6.4.2.1

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

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

Step 6.4.2.2

Cancel the common factor.

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

Step 6.4.2.3

Rewrite the expression.

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

Step 6.5

Multiply $2 x^{3} y^{3} - x^{2}$ by $\frac{1}{y^{2}}$ .

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

Step 6.6

Multiply $2 x^{3} y^{3} - x^{2}$ by $\frac{1}{y^{2}}$ .

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

Step 6.7

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

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Step 6.7.1

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

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

Step 6.7.2

Factor $x^{2}$ out of $- x^{2}$ .

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

Step 6.7.3

Factor $x^{2}$ out of $x^{2} (2 x y^{3}) + x^{2} \cdot - 1$ .

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{x (3 x y^{3} + 2)}{y}$ to find $f (x, y)$ .

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Step 8.1

Since $\frac{1}{y}$ is constant with respect to $x$ , move $\frac{1}{y}$ out of the integral.

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

Step 8.2

Expand $x (3 x y^{3} + 2)$ .

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Step 8.2.1

Apply the distributive property.

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

Step 8.2.2

Move parentheses.

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

Step 8.2.3

Remove parentheses.

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

Step 8.2.4

Reorder $x$ and $3$ .

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

Step 8.2.5

Reorder $x$ and $2$ .

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

Step 8.2.6

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

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

Step 8.2.7

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

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

Step 8.2.8

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

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

Step 8.2.9

Add $1$ and $1$ .

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

Step 8.3

Split the single integral into multiple integrals.

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

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

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

Step 8.7

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

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

Step 8.8

Simplify.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = \frac{1}{y}$ and $g (y) = y^{3} x^{3} + x^{2}$ .

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

$\frac{1}{y} (x^{3} (3 y^{2}) + 0) + (y^{3} x^{3} + x^{2}) \frac{d}{d y} [\frac{1}{y}] + \frac{d}{d y} [g (y)] = \frac{x^{2} (2 x y^{3} - 1)}{y^{2}}$

Step 11.3.6

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

$\frac{1}{y} (x^{3} (3 y^{2}) + 0) + (y^{3} x^{3} + x^{2}) \frac{d}{d y} [y^{- 1}] + \frac{d}{d y} [g (y)] = \frac{x^{2} (2 x y^{3} - 1)}{y^{2}}$

Step 11.3.7

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

$\frac{1}{y} (x^{3} (3 y^{2}) + 0) + (y^{3} x^{3} + x^{2}) (- y^{- 2}) + \frac{d}{d y} [g (y)] = \frac{x^{2} (2 x y^{3} - 1)}{y^{2}}$

Step 11.3.8

Move $3$ to the left of $x^{3}$ .

$\frac{1}{y} (3 \cdot x^{3} y^{2} + 0) + (y^{3} x^{3} + x^{2}) (- y^{- 2}) + \frac{d}{d y} [g (y)] = \frac{x^{2} (2 x y^{3} - 1)}{y^{2}}$

Step 11.3.9

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

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

Step 11.3.10

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

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

Step 11.3.11

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

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

Step 11.3.12

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

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

Step 11.3.13

Move $3$ to the left of $x^{3}$ .

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

Step 11.3.14

Cancel the common factor of $y^{2}$ and $y$ .

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Step 11.3.14.1

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

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

Step 11.3.14.2

Cancel the common factors.

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Step 11.3.14.2.1

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

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

Step 11.3.14.2.2

Factor $y$ out of $y^{1}$ .

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

Step 11.3.14.2.3

Cancel the common factor.

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

Step 11.3.14.2.4

Rewrite the expression.

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

Step 11.3.14.2.5

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 11.5.2

Apply the distributive property.

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

Step 11.5.3

Combine terms.

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Step 11.5.3.1

Combine $y^{3}$ and $\frac{1}{y^{2}}$ .

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

Step 11.5.3.2

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

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

Step 11.5.3.3

Cancel the common factor of $y^{3}$ and $y^{2}$ .

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Step 11.5.3.3.1

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

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

Step 11.5.3.3.2

Cancel the common factors.

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Step 11.5.3.3.2.1

Multiply by $1$ .

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

Step 11.5.3.3.2.2

Cancel the common factor.

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

Step 11.5.3.3.2.3

Rewrite the expression.

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

Step 11.5.3.3.2.4

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

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

Step 11.5.3.4

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

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

Step 11.5.3.5

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

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

Step 11.5.4

Reorder terms.

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

Step 12

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

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Step 12.1

Move all terms containing variables to the left side of the equation.

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Step 12.1.1

Subtract $\frac{x^{2} (2 x y^{3} - 1)}{y^{2}}$ from both sides of the equation.

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

Step 12.1.2

Combine the numerators over the common denominator.

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

Step 12.1.3

Simplify each term.

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Step 12.1.3.1

Apply the distributive property.

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

Step 12.1.3.2

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

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Step 12.1.3.2.1

Move $x$ .

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

Step 12.1.3.2.2

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

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Step 12.1.3.2.2.1

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

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

Step 12.1.3.2.2.2

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

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

Step 12.1.3.2.3

Add $1$ and $2$ .

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

Step 12.1.3.3

Multiply $- x^{2} \cdot - 1$ .

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Step 12.1.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.1.3.3.2

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

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

Step 12.1.3.4

Simplify each term.

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Step 12.1.3.4.1

Rewrite using the commutative property of multiplication.

$\frac{d g}{d y} + 2 x^{3} y + \frac{- x^{2} - 1 \cdot 2 x^{3} y^{3} + x^{2}}{y^{2}} = 0$

Step 12.1.3.4.2

Multiply $- 1$ by $2$ .

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

Step 12.1.4

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

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Step 12.1.4.1

Add $- x^{2}$ and $x^{2}$ .

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

Step 12.1.4.2

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

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

Step 12.1.5

Cancel the common factor of $y^{3}$ and $y^{2}$ .

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Step 12.1.5.1

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

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

Step 12.1.5.2

Cancel the common factors.

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Step 12.1.5.2.1

Multiply by $1$ .

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

Step 12.1.5.2.2

Cancel the common factor.

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

Step 12.1.5.2.3

Rewrite the expression.

$\frac{d g}{d y} + 2 x^{3} y + \frac{- 2 x^{3} y}{1} = 0$

Step 12.1.5.2.4

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

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

Step 12.1.6

Combine the opposite terms in $\frac{d g}{d y} + 2 x^{3} y - 2 x^{3} y$ .

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Step 12.1.6.1

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

$\frac{d g}{d y} + 0 = 0$

Step 12.1.6.2

Add $\frac{d g}{d y}$ and $0$ .

$\frac{d g}{d y} = 0$

Step 13

Find the antiderivative of $0$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 13.2

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

$g (y) = \int 0 d y$

Step 13.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 13.4

Add $0$ and $C$ .

$g (y) = C$

Step 14

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

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

Step 15

Simplify each term.

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Step 15.1

Multiply $\frac{1}{y}$ by $y^{3} x^{3} + x^{2}$ .

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

Step 15.2

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

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Step 15.2.1

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

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

Step 15.2.2

Multiply by $1$ .

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

Step 15.2.3

Factor $x^{2}$ out of $x^{2} (y^{3} x) + x^{2} \cdot 1$ .

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