Calculus Examples

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

Step 1

Write the problem as a mathematical expression.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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 $f (y) = y$ and $g (y) = y^{3} - x$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Add $1$ and $2$ .

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

Step 2.7

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

$\frac{\partial M}{\partial y} = 3 y^{3} + (y^{3} - x) \cdot 1$

Step 2.8

Simplify by adding terms.

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Step 2.8.1

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

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

Step 2.8.2

Add $3 y^{3}$ and $y^{3}$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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 $g (x) = y^{3} + x$ .

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

Step 3.3

Differentiate.

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Step 3.3.1

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

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

Step 3.3.2

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

$\frac{\partial N}{\partial x} = x (0 + \frac{d}{d x} [x]) + (y^{3} + x) \frac{d}{d x} [x]$

Step 3.3.3

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 3.3.4

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

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

Step 3.3.5

Multiply $x$ by $1$ .

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

Step 3.3.6

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

$\frac{\partial N}{\partial x} = x + (y^{3} + x) \cdot 1$

Step 3.3.7

Simplify by adding terms.

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Step 3.3.7.1

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

$\frac{\partial N}{\partial x} = x + y^{3} + x$

Step 3.3.7.2

Add $x$ and $x$ .

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

Step 4

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

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Step 4.1

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

$4 y^{3} - x = y^{3} + 2 x$

Step 4.2

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

$4 y^{3} - x = y^{3} + 2 x$ is not an identity.

Step 5

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 5.1

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

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

Step 5.2

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

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

Step 5.3

Substitute $y (y^{3} - x)$ for $M$ .

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Step 5.3.1

Substitute $y (y^{3} - x)$ for $M$ .

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

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

Step 5.3.2.2

Multiply $4$ by $- 1$ .

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

Step 5.3.2.3

Multiply $- - x$ .

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Step 5.3.2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.2.3.2

Multiply $x$ by $1$ .

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

Step 5.3.2.4

Subtract $4 y^{3}$ from $y^{3}$ .

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

Step 5.3.2.5

Add $2 x$ and $x$ .

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

Step 5.3.2.6

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

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Step 5.3.2.6.1

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

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

Step 5.3.2.6.2

Factor $3$ out of $3 x$ .

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

Step 5.3.2.6.3

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

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

Factor $- 1$ out of $x$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

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

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

Step 5.3.3.5

Cancel the common factor.

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

Step 5.3.3.6

Rewrite the expression.

$\frac{3 \cdot (- 1)}{y}$

Step 5.3.4

Multiply $3$ by $- 1$ .

$\frac{- 3}{y}$

Step 5.3.5

Substitute $y (y^{3} - x)$ for $M$ .

$- \frac{3}{y}$

Step 5.4

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

$μ (x, y) = e^{\int - \frac{3}{y} d y}$

Step 6

Evaluate the integral $e^{\int - \frac{3}{y} d y}$ .

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Step 6.1

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

$μ (x, y) = e^{- \int \frac{3}{y} d y}$

Step 6.2

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

$μ (x, y) = e^{- (3 \int \frac{1}{y} d y)}$

Step 6.3

Multiply $3$ by $- 1$ .

$μ (x, y) = e^{- 3 \int \frac{1}{y} d y}$

Step 6.4

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

$μ (x, y) = e^{- 3 (\ln (| y |) + C)}$

Step 6.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| y |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 3 \ln (| y |)$ by moving $- 3$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 3})} + C$

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 3} + C$

Step 6.6.3

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

Cancel the common factor of $y$ .

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Step 7.2.1

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

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

Step 7.2.2

Cancel the common factor.

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

Step 7.2.3

Rewrite the expression.

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

Step 7.3

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

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

Step 7.4

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

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

Step 7.5

Apply the distributive property.

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

Step 7.6

Multiply $x$ by $x$ .

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

Step 7.7

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

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

Step 7.8

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

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Step 7.8.1

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

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

Step 7.8.2

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

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

Step 7.8.3

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Since $\frac{1}{y^{2}}$ is constant with respect to $x$ , move $\frac{1}{y^{2}}$ out of the integral.

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

Step 9.2

Split the single integral into multiple integrals.

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

Step 9.3

Apply the constant rule.

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

Step 9.4

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

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

Step 9.5

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

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

Step 9.6

Simplify.

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

Step 10

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

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

Step 11

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

$\frac{\partial f}{\partial y} = \frac{x (y^{3} + x)}{y^{3}}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

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

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

Step 12.3.6

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

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

Step 12.3.7

Rewrite $\frac{1}{y^{2}}$ as ${(y^{2})}^{- 1}$ .

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

Step 12.3.8

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{2}$ .

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Step 12.3.8.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

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

Step 12.3.8.2

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

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

Step 12.3.8.3

Replace all occurrences of $u$ with $y^{2}$ .

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

Step 12.3.9

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

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

Step 12.3.10

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

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

Step 12.3.11

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

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

Step 12.3.12

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

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

Step 12.3.13

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

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

Step 12.3.14

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

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

Step 12.3.15

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

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

Step 12.3.16

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

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Step 12.3.16.1

Cancel the common factor.

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

Step 12.3.16.2

Divide $3 x$ by $1$ .

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

Step 12.3.17

Multiply the exponents in ${(y^{2})}^{- 2}$ .

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Step 12.3.17.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 12.3.17.2

Multiply $2$ by $- 2$ .

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

Step 12.3.18

Multiply $2$ by $- 1$ .

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

Step 12.3.19

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

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

Step 12.3.20

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

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

Step 12.3.21

Subtract $4$ from $1$ .

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

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

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

Step 12.5.2

Apply the distributive property.

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

Step 12.5.3

Combine terms.

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Step 12.5.3.1

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

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

Step 12.5.3.2

Move the negative in front of the fraction.

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

Step 12.5.3.3

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

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

Step 12.5.3.4

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

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

Step 12.5.3.5

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

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

Step 12.5.3.6

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

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

Step 12.5.3.7

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

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Step 12.5.3.7.1

Cancel the common factor.

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

Step 12.5.3.7.2

Divide $2 x$ by $1$ .

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

Step 12.5.3.8

Multiply $2$ by $- 1$ .

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

Step 12.5.3.9

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

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

Step 12.5.3.10

Move the negative in front of the fraction.

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

Step 12.5.3.11

Multiply $- 1$ by $- 1$ .

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

Step 12.5.3.12

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

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

Step 12.5.3.13

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

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

Step 12.5.3.14

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

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

Step 12.5.3.15

Cancel the common factor of $2$ .

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Step 12.5.3.15.1

Cancel the common factor.

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

Step 12.5.3.15.2

Rewrite the expression.

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

Step 12.5.3.16

Subtract $2 x$ from $3 x$ .

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

Step 12.5.4

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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

Step 13.1.2

Combine the numerators over the common denominator.

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

Step 13.1.3

Simplify each term.

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Step 13.1.3.1

Apply the distributive property.

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

Step 13.1.3.2

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

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Step 13.1.3.2.1

Move $x$ .

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

Step 13.1.3.2.2

Multiply $x$ by $x$ .

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

Step 13.1.4

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

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Step 13.1.4.1

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

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

Step 13.1.4.2

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

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

Step 13.1.5

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

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Step 13.1.5.1

Cancel the common factor.

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

Step 13.1.5.2

Divide $- x$ by $1$ .

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

Step 13.1.6

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

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Step 13.1.6.1

Subtract $x$ from $x$ .

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

Step 13.1.6.2

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

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

$g (y) = 0 + C$

Step 14.4

Add $0$ and $C$ .

$g (y) = C$

Step 15

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

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

Step 16

Simplify each term.

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Step 16.1

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

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

Step 16.2

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

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

Step 16.3

Simplify the numerator.

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Step 16.3.1

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

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Step 16.3.1.1

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

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

Step 16.3.1.2

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

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

Step 16.3.1.3

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

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

Step 16.3.2

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

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

Step 16.3.3

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

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

Step 16.3.4

Combine the numerators over the common denominator.

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

Step 16.3.5

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

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

Step 16.4

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

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

Step 16.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 16.6

Combine.

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

Step 16.7

Multiply $x$ by $1$ .

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