Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - 2 y^{3} + 1$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $- 2$ .

$\frac{\partial M}{\partial y} = - 6 y^{2} + \frac{d}{d y} [1]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = - 6 y^{2} + 0$

Step 1.4.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $3$ by $1$ .

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

Step 2.4

Differentiate using the Power Rule.

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Step 2.4.1

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

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

Step 2.4.2

Reorder terms.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$- 6 y^{2} = 3 x^{2} + 3 y^{2}$ 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 $- 6 y^{2}$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.2

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

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $3$ by $- 1$ .

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

Step 4.3.2.3

Multiply $3$ by $- 1$ .

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

Step 4.3.2.4

Subtract $3 y^{2}$ from $- 6 y^{2}$ .

$\frac{- 3 x^{2} - 9 y^{2}}{3 x y^{2} + x^{3}}$

Step 4.3.2.5

Factor $3$ out of $- 3 x^{2} - 9 y^{2}$ .

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Step 4.3.2.5.1

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

$\frac{3 (- x^{2}) - 9 y^{2}}{3 x y^{2} + x^{3}}$

Step 4.3.2.5.2

Factor $3$ out of $- 9 y^{2}$ .

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

Step 4.3.2.5.3

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.3.3

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

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

Step 4.3.4

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

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Step 4.3.4.1

Factor $- 1$ out of $- x^{2}$ .

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

Step 4.3.4.2

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

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

Step 4.3.4.3

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

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

Step 4.3.4.4

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

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

Step 4.3.4.5

Reorder terms.

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

Step 4.3.4.6

Cancel the common factor.

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

Step 4.3.4.7

Rewrite the expression.

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

Step 4.3.5

Multiply $3$ by $- 1$ .

$\frac{- 3}{x}$

Step 4.3.6

Move the negative in front of the fraction.

$- \frac{3}{x}$

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Multiply $3$ by $- 1$ .

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

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| x |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 3} + C$

Step 5.6.3

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

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

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

Step 6.6

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

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

Step 6.7

Move the negative in front of the fraction.

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

Step 6.8

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

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

Step 6.9

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

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

Step 6.10

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

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Step 6.10.1

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

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

Step 6.10.2

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

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

Step 6.10.3

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

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

Step 6.11

Cancel the common factors.

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Step 6.11.1

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

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

Step 6.11.2

Cancel the common factor.

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

Step 6.11.3

Rewrite the expression.

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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

Step 8.2

Split the single integral into multiple integrals.

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

Apply the constant rule.

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

Step 8.6

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

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

Step 8.7

Simplify.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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Step 11.3.7.1

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

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

Step 11.3.7.2

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

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

Step 11.3.7.3

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

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

Step 11.3.8

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

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

Step 11.3.9

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

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

Step 11.3.10

Add $0$ and $2 y x$ .

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

Step 11.3.11

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

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

Step 11.3.12

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

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

Step 11.3.13

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

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

Step 11.3.14

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

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

Step 11.3.15

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

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Step 11.3.15.1

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

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

Step 11.3.15.2

Cancel the common factors.

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Step 11.3.15.2.1

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

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

Step 11.3.15.2.2

Cancel the common factor.

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

Step 11.3.15.2.3

Rewrite the expression.

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

Step 11.3.16

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

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Step 11.3.16.1

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

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

Step 11.3.16.2

Multiply $2$ by $- 2$ .

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

Step 11.3.17

Multiply $2$ by $- 1$ .

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

Step 11.3.18

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

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

Step 11.3.19

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

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

Step 11.3.20

Subtract $4$ from $1$ .

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

Step 11.3.21

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

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

Step 11.3.22

Combine the numerators over the common denominator.

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

Step 11.3.23

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

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

Step 11.3.24

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

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

Step 11.3.25

Subtract $3$ from $1$ .

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

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

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

Step 11.5.2

Apply the distributive property.

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

Step 11.5.3

Combine terms.

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Step 11.5.3.1

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

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

Step 11.5.3.2

Move the negative in front of the fraction.

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

Step 11.5.3.3

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

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

Step 11.5.3.4

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

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

Step 11.5.3.5

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

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

Step 11.5.3.6

Move the negative in front of the fraction.

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

Step 11.5.3.7

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

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

Step 11.5.3.8

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

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

Step 11.5.3.9

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

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

Step 11.5.3.10

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

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

Step 11.5.3.11

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

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Step 11.5.3.11.1

Cancel the common factor.

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

Step 11.5.3.11.2

Divide $2 y$ by $1$ .

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

Step 11.5.3.12

Multiply $2$ by $- 1$ .

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

Step 11.5.3.13

Subtract $2 y$ from $2 y$ .

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

Step 11.5.3.14

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

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

Step 11.5.3.15

Rewrite $\frac{- \frac{2 y^{3}}{x^{2}}}{x}$ as a product.

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

Step 11.5.3.16

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

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

Step 11.5.3.17

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

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Step 11.5.3.17.1

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

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Step 11.5.3.17.1.1

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

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

Step 11.5.3.17.1.2

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

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

Step 11.5.3.17.2

Add $1$ and $2$ .

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

Step 11.5.4

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

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

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Step 12.1.1.1

Add $\frac{2 y^{3} - 1}{x^{3}}$ to both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

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

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

Step 12.1.1.4

Subtract $1$ from $0$ .

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

Step 12.1.1.5

Move the negative in front of the fraction.

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

Step 12.1.2

Add $\frac{1}{x^{3}}$ to both sides of the equation.

$\frac{d g}{d x} = \frac{1}{x^{3}}$

Step 13

Find the antiderivative of $\frac{1}{x^{3}}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = \frac{1}{x^{3}}$ .

$\int \frac{d g}{d x} d x = \int \frac{1}{x^{3}} d x$

Step 13.2

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

$g (x) = \int \frac{1}{x^{3}} d x$

Step 13.3

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

$g (x) = \int {(x^{3})}^{- 1} d x$

Step 13.4

Multiply the exponents in ${(x^{3})}^{- 1}$ .

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Step 13.4.1

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

$g (x) = \int x^{3 \cdot - 1} d x$

Step 13.4.2

Multiply $3$ by $- 1$ .

$g (x) = \int x^{- 3} d x$

Step 13.5

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

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

Step 13.6

Simplify the answer.

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Step 13.6.1

Rewrite $- \frac{1}{2} x^{- 2} + C$ as $- \frac{1}{2} \cdot \frac{1}{x^{2}} + C$ .

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

Step 13.6.2

Simplify.

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Step 13.6.2.1

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

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

Step 13.6.2.2

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

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

Step 13.6.2.3

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

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

Step 14

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

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

Step 15

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

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Step 15.1

Simplify each term.

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Step 15.1.1

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

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

Step 15.1.2

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

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Step 15.1.2.1

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

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

Step 15.1.2.2

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

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

Step 15.1.2.3

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

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

Step 15.2

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

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

Step 15.3

Write each expression with a common denominator of $2 x^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 15.3.1

Multiply $\frac{y (y^{2} + x^{2})}{x^{2}}$ by $\frac{2}{2}$ .

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

Step 15.3.2

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

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

Step 15.4

Combine the numerators over the common denominator.

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

Step 15.5

Simplify the numerator.

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Step 15.5.1

Apply the distributive property.

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

Step 15.5.2

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

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Step 15.5.2.1

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

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Step 15.5.2.1.1

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

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

Step 15.5.2.1.2

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

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

Step 15.5.2.2

Add $1$ and $2$ .

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

Step 15.5.3

Apply the distributive property.

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

Step 15.5.4

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

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

Step 15.5.5

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

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

Step 15.5.6

Remove parentheses.

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