Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $4 x^{2}$ is constant with respect to $y$ , the derivative of $4 x^{2}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [2 y^{2}]$

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

$\frac{\partial M}{\partial y} = 0 + 2 \frac{d}{d y} [y^{2}]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 + 2 (2 y)$

Step 1.3.3

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 0 + 4 y$

Step 1.4

Add $0$ and $4 y$ .

$\frac{\partial M}{\partial y} = 4 y$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = - y$

Step 3

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

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Step 3.1

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

$4 y = - y$

Step 3.2

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

$4 y = - y$ 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 $4 y$ for $\frac{\partial M}{\partial y}$ .

$\frac{4 y - \frac{\partial N}{\partial x}}{N}$

Step 4.2

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

$\frac{4 y - - y}{N}$

Step 4.3

Substitute $- (x y)$ for $N$ .

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Step 4.3.1

Substitute $- (x y)$ for $N$ .

$\frac{4 y - - y}{- (x y)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Factor $y$ out of $4 y - - y$ .

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Step 4.3.2.1.1

Factor $y$ out of $4 y$ .

$\frac{y \cdot 4 - - y}{- x y}$

Step 4.3.2.1.2

Factor $y$ out of $- - y$ .

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

Step 4.3.2.1.3

Factor $y$ out of $y \cdot 4 + y (- - 1)$ .

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

Step 4.3.2.2

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.3

Add $4$ and $1$ .

$\frac{y \cdot 5}{- x y}$

Step 4.3.3

Cancel the common factor of $y$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{y \cdot 5}{- x y}$

Step 4.3.3.2

Rewrite the expression.

$\frac{5}{- x}$

Step 4.3.4

Move the negative in front of the fraction.

$- \frac{5}{x}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{5}{x} d x}$

Step 5

Evaluate the integral $e^{\int - \frac{5}{x} d x}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{5}{x} d x}$

Step 5.2

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$μ (x, y) = e^{- (5 \int \frac{1}{x} d x)}$

Step 5.3

Multiply $5$ by $- 1$ .

$μ (x, y) = e^{- 5 \int \frac{1}{x} d x}$

Step 5.4

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

$μ (x, y) = e^{- 5 (\ln (| x |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 5 \ln (| x |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 5 \ln (| x |)$ by moving $- 5$ inside the logarithm.

$μ (x, y) = e^{\ln ({| x |}^{- 5})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 5} + C$

Step 5.6.3

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

$μ (x, y) = \frac{1}{{| x |}^{5}} + C$

Step 6

Multiply both sides of $(4 x^{2} + 2 y^{2}) d x - (x y) d y = 0$ by the integration factor $\frac{1}{x^{5}}$ .

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Step 6.1

Multiply $4 x^{2} + 2 y^{2}$ by $\frac{1}{x^{5}}$ .

$(4 x^{2} + 2 y^{2}) \frac{1}{x^{5}} d x - (x y) d y = 0$

Step 6.2

Multiply $4 x^{2} + 2 y^{2}$ by $\frac{1}{x^{5}}$ .

$\frac{4 x^{2} + 2 y^{2}}{x^{5}} d x - (x y) d y = 0$

Step 6.3

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

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Step 6.3.1

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

$\frac{2 (2 x^{2}) + 2 y^{2}}{x^{5}} d x - (x y) d y = 0$

Step 6.3.2

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

$\frac{2 (2 x^{2}) + 2 (y^{2})}{x^{5}} d x - (x y) d y = 0$

Step 6.3.3

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

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x - (x y) d y = 0$

Step 6.4

Multiply $- (x y)$ by $\frac{1}{x^{5}}$ .

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x - (x y) \frac{1}{x^{5}} d y = 0$

Step 6.5

Cancel the common factor of $x$ .

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Step 6.5.1

Factor $x$ out of $- (x y)$ .

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x + x (- (y)) \frac{1}{x^{5}} d y = 0$

Step 6.5.2

Factor $x$ out of $x^{5}$ .

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x + x (- (y)) \frac{1}{x \cdot x^{4}} d y = 0$

Step 6.5.3

Cancel the common factor.

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x + x (- (y)) \frac{1}{x \cdot x^{4}} d y = 0$

Step 6.5.4

Rewrite the expression.

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x - (y) \frac{1}{x^{4}} d y = 0$

Step 6.6

Combine $\frac{1}{x^{4}}$ and $y$ .

$\frac{2 (2 x^{2} + y^{2})}{x^{5}} d x - \frac{y}{x^{4}} d y = 0$

Step 7

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

$f (x, y) = \int - \frac{y}{x^{4}} d y$

Step 8

Integrate $N (x, y) = - \frac{y}{x^{4}}$ to find $f (x, y)$ .

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Step 8.1

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

$f (x, y) = - \int \frac{y}{x^{4}} d y$

Step 8.2

Since $\frac{1}{x^{4}}$ is constant with respect to $y$ , move $\frac{1}{x^{4}}$ out of the integral.

$f (x, y) = - (\frac{1}{x^{4}} \int y d y)$

Step 8.3

Remove parentheses.

$f (x, y) = - \frac{1}{x^{4}} \int y d y$

Step 8.4

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

$f (x, y) = - \frac{1}{x^{4}} (\frac{1}{2} y^{2} + C)$

Step 8.5

Simplify the answer.

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Step 8.5.1

Rewrite $- \frac{1}{x^{4}} (\frac{1}{2} y^{2} + C)$ as $- \frac{1}{x^{4}} \cdot \frac{1}{2} y^{2} + C$ .

$f (x, y) = - \frac{1}{x^{4}} \cdot \frac{1}{2} y^{2} + C$

Step 8.5.2

Simplify.

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Step 8.5.2.1

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

$f (x, y) = - \frac{1}{2 x^{4}} y^{2} + C$

Step 8.5.2.2

Combine $y^{2}$ and $\frac{1}{2 x^{4}}$ .

$f (x, y) = - \frac{y^{2}}{2 x^{4}} + C$

Step 9

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

$f (x, y) = - \frac{y^{2}}{2 x^{4}} + g (x)$

Step 10

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

$\frac{\partial f}{\partial x} = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [- \frac{y^{2}}{2 x^{4}} + g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.2

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

$\frac{d}{d x} [- \frac{y^{2}}{2 x^{4}}] + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3

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

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Step 11.3.1

Since $- \frac{y^{2}}{2}$ is constant with respect to $x$ , the derivative of $- \frac{y^{2}}{2 x^{4}}$ with respect to $x$ is $- \frac{y^{2}}{2} \frac{d}{d x} [\frac{1}{x^{4}}]$ .

$- \frac{y^{2}}{2} \frac{d}{d x} [\frac{1}{x^{4}}] + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.2

Rewrite $\frac{1}{x^{4}}$ as ${(x^{4})}^{- 1}$ .

$- \frac{y^{2}}{2} \frac{d}{d x} [{(x^{4})}^{- 1}] + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.3

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 $g (x) = x^{4}$ .

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Step 11.3.3.1

To apply the Chain Rule, set $u$ as $x^{4}$ .

$- \frac{y^{2}}{2} (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.3.2

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

$- \frac{y^{2}}{2} (- u^{- 2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.3.3

Replace all occurrences of $u$ with $x^{4}$ .

$- \frac{y^{2}}{2} (- {(x^{4})}^{- 2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.4

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

$- \frac{y^{2}}{2} (- {(x^{4})}^{- 2} (4 x^{3})) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.5

Multiply the exponents in ${(x^{4})}^{- 2}$ .

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Step 11.3.5.1

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

$- \frac{y^{2}}{2} (- x^{4 \cdot - 2} (4 x^{3})) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.5.2

Multiply $4$ by $- 2$ .

$- \frac{y^{2}}{2} (- x^{- 8} (4 x^{3})) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.6

Multiply $4$ by $- 1$ .

$- \frac{y^{2}}{2} (- 4 x^{- 8} x^{3}) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.7

Multiply $x^{- 8}$ by $x^{3}$ by adding the exponents.

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Step 11.3.7.1

Move $x^{3}$ .

$- \frac{y^{2}}{2} (- 4 (x^{3} x^{- 8})) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.7.2

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

$- \frac{y^{2}}{2} (- 4 x^{3 - 8}) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.7.3

Subtract $8$ from $3$ .

$- \frac{y^{2}}{2} (- 4 x^{- 5}) + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.8

Multiply $- 4$ by $- 1$ .

$4 \frac{y^{2}}{2} x^{- 5} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.9

Combine $4$ and $\frac{y^{2}}{2}$ .

$\frac{4 y^{2}}{2} x^{- 5} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.10

Combine $\frac{4 y^{2}}{2}$ and $x^{- 5}$ .

$\frac{4 y^{2} x^{- 5}}{2} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.11

Move $x^{- 5}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{4 y^{2}}{2 x^{5}} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.12

Cancel the common factor of $4$ and $2$ .

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Step 11.3.12.1

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

$\frac{2 (2 y^{2})}{2 x^{5}} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.12.2

Cancel the common factors.

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Step 11.3.12.2.1

Factor $2$ out of $2 x^{5}$ .

$\frac{2 (2 y^{2})}{2 (x^{5})} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.12.2.2

Cancel the common factor.

$\frac{2 (2 y^{2})}{2 x^{5}} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.3.12.2.3

Rewrite the expression.

$\frac{2 y^{2}}{x^{5}} + \frac{d}{d x} [g (x)] = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.4

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

$\frac{2 y^{2}}{x^{5}} + \frac{d g}{d x} = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

Step 11.5

Reorder terms.

$\frac{d g}{d x} + \frac{2 y^{2}}{x^{5}} = \frac{2 (2 x^{2} + y^{2})}{x^{5}}$

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

Subtract $\frac{2 (2 x^{2} + y^{2})}{x^{5}}$ from both sides of the equation.

$\frac{d g}{d x} + \frac{2 y^{2}}{x^{5}} - \frac{2 (2 x^{2} + y^{2})}{x^{5}} = 0$

Step 12.1.1.2

Combine the numerators over the common denominator.

$\frac{d g}{d x} + \frac{2 y^{2} - 2 (2 x^{2} + y^{2})}{x^{5}} = 0$

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

$\frac{d g}{d x} + \frac{2 y^{2} - 2 (2 x^{2}) - 2 y^{2}}{x^{5}} = 0$

Step 12.1.1.3.2

Multiply $2$ by $- 2$ .

$\frac{d g}{d x} + \frac{2 y^{2} - 4 x^{2} - 2 y^{2}}{x^{5}} = 0$

Step 12.1.1.4

Combine the opposite terms in $2 y^{2} - 4 x^{2} - 2 y^{2}$ .

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Step 12.1.1.4.1

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

$\frac{d g}{d x} + \frac{- 4 x^{2} + 0}{x^{5}} = 0$

Step 12.1.1.4.2

Add $- 4 x^{2}$ and $0$ .

$\frac{d g}{d x} + \frac{- 4 x^{2}}{x^{5}} = 0$

Step 12.1.1.5

Simplify each term.

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Step 12.1.1.5.1

Cancel the common factor of $x^{2}$ and $x^{5}$ .

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Step 12.1.1.5.1.1

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

$\frac{d g}{d x} + \frac{x^{2} \cdot - 4}{x^{5}} = 0$

Step 12.1.1.5.1.2

Cancel the common factors.

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Step 12.1.1.5.1.2.1

Factor $x^{2}$ out of $x^{5}$ .

$\frac{d g}{d x} + \frac{x^{2} \cdot - 4}{x^{2} x^{3}} = 0$

Step 12.1.1.5.1.2.2

Cancel the common factor.

$\frac{d g}{d x} + \frac{x^{2} \cdot - 4}{x^{2} x^{3}} = 0$

Step 12.1.1.5.1.2.3

Rewrite the expression.

$\frac{d g}{d x} + \frac{- 4}{x^{3}} = 0$

Step 12.1.1.5.2

Move the negative in front of the fraction.

$\frac{d g}{d x} - \frac{4}{x^{3}} = 0$

Step 12.1.2

Add $\frac{4}{x^{3}}$ to both sides of the equation.

$\frac{d g}{d x} = \frac{4}{x^{3}}$

Step 13

Find the antiderivative of $\frac{4}{x^{3}}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = \frac{4}{x^{3}}$ .

$\int \frac{d g}{d x} d x = \int \frac{4}{x^{3}} d x$

Step 13.2

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

$g (x) = \int \frac{4}{x^{3}} d x$

Step 13.3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$g (x) = 4 \int \frac{1}{x^{3}} d x$

Step 13.4

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

$g (x) = 4 \int {(x^{3})}^{- 1} d x$

Step 13.5

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

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Step 13.5.1

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

$g (x) = 4 \int x^{3 \cdot - 1} d x$

Step 13.5.2

Multiply $3$ by $- 1$ .

$g (x) = 4 \int x^{- 3} d x$

Step 13.6

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

$g (x) = 4 (- \frac{1}{2} x^{- 2} + C)$

Step 13.7

Simplify the answer.

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Step 13.7.1

Simplify.

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Step 13.7.1.1

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

$g (x) = 4 (- \frac{x^{- 2}}{2} + C)$

Step 13.7.1.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$g (x) = 4 (- \frac{1}{2 x^{2}} + C)$

Step 13.7.2

Simplify.

$g (x) = 4 (- \frac{1}{2 x^{2}}) + C$

Step 13.7.3

Simplify.

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Step 13.7.3.1

Multiply $- 1$ by $4$ .

$g (x) = - 4 \frac{1}{2 x^{2}} + C$

Step 13.7.3.2

Combine $- 4$ and $\frac{1}{2 x^{2}}$ .

$g (x) = \frac{- 4}{2 x^{2}} + C$

Step 13.7.3.3

Cancel the common factor of $- 4$ and $2$ .

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Step 13.7.3.3.1

Factor $2$ out of $- 4$ .

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

Step 13.7.3.3.2

Cancel the common factors.

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Step 13.7.3.3.2.1

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

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

Step 13.7.3.3.2.2

Cancel the common factor.

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

Step 13.7.3.3.2.3

Rewrite the expression.

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

Step 13.7.3.4

Move the negative in front of the fraction.

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

Step 14

Substitute for $g (x)$ in $f (x, y) = - \frac{y^{2}}{2 x^{4}} + g (x)$ .

$f (x, y) = - \frac{y^{2}}{2 x^{4}} - \frac{2}{x^{2}} + C$