Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 (2 y^{2} + 5 x y - 2 y + 4)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Since $2$ is constant with respect to $y$ , the derivative of $2 (2 y^{2} + 5 x y - 2 y + 4)$ with respect to $y$ is $2 \frac{d}{d y} [2 y^{2} + 5 x y - 2 y + 4]$ .

$\frac{\partial M}{\partial y} = 2 \frac{d}{d y} [2 y^{2} + 5 x y - 2 y + 4]$

Step 1.3

By the Sum Rule, the derivative of $2 y^{2} + 5 x y - 2 y + 4$ with respect to $y$ is $\frac{d}{d y} [2 y^{2}] + \frac{d}{d y} [5 x y] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [4]$ .

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

Step 1.4

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

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

Step 1.5

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

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

Step 1.6

Multiply $2$ by $2$ .

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

Step 1.7

Since $5 x$ is constant with respect to $y$ , the derivative of $5 x y$ with respect to $y$ is $5 x \frac{d}{d y} [y]$ .

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

Step 1.8

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

$\frac{\partial M}{\partial y} = 2 (4 y + 5 x \cdot 1 + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [4])$

Step 1.9

Multiply $5$ by $1$ .

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

Step 1.10

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

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

Step 1.11

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

$\frac{\partial M}{\partial y} = 2 (4 y + 5 x - 2 \cdot 1 + \frac{d}{d y} [4])$

Step 1.12

Multiply $- 2$ by $1$ .

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

Step 1.13

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

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

Step 1.14

Add $4 y + 5 x - 2$ and $0$ .

$\frac{\partial M}{\partial y} = 2 (4 y + 5 x - 2)$

Step 1.15

Simplify.

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Step 1.15.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = 2 (4 y) + 2 (5 x) + 2 \cdot - 2$

Step 1.15.2

Combine terms.

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Step 1.15.2.1

Multiply $4$ by $2$ .

$\frac{\partial M}{\partial y} = 8 y + 2 (5 x) + 2 \cdot - 2$

Step 1.15.2.2

Multiply $5$ by $2$ .

$\frac{\partial M}{\partial y} = 8 y + 10 x + 2 \cdot - 2$

Step 1.15.2.3

Multiply $2$ by $- 2$ .

$\frac{\partial M}{\partial y} = 8 y + 10 x - 4$

Step 1.15.3

Reorder terms.

$\frac{\partial M}{\partial y} = 10 x + 8 y - 4$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $1$ .

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

Step 2.3.5

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

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

Step 2.3.6

Add $2$ and $0$ .

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

Step 2.3.7

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

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

Step 2.3.8

Simplify the expression.

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Step 2.3.8.1

Add $2$ and $0$ .

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

Step 2.3.8.2

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

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

Step 2.3.9

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

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

Step 2.3.10

Simplify by adding terms.

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Step 2.3.10.1

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

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

Step 2.3.10.2

Add $2 x$ and $2 x$ .

$\frac{\partial N}{\partial x} = 4 x + 2 y - 1$

Step 3

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

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Step 3.1

Substitute $10 x + 8 y - 4$ for $\frac{\partial M}{\partial y}$ and $4 x + 2 y - 1$ for $\frac{\partial N}{\partial x}$ .

$10 x + 8 y - 4 = 4 x + 2 y - 1$

Step 3.2

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

$10 x + 8 y - 4 = 4 x + 2 y - 1$ is not an identity.

Step 4

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

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Step 4.1

Substitute $10 x + 8 y - 4$ for $\frac{\partial M}{\partial y}$ .

$\frac{10 x + 8 y - 4 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

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

$\frac{10 x + 8 y - 4 - (4 x + 2 y - 1)}{N}$

Step 4.3

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

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Step 4.3.1

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

$\frac{10 x + 8 y - 4 - (4 x + 2 y - 1)}{x (2 x + 2 y - 1)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{10 x + 8 y - 4 - (4 x) - (2 y) - - 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.2

Simplify.

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Step 4.3.2.2.1

Multiply $4$ by $- 1$ .

$\frac{10 x + 8 y - 4 - 4 x - (2 y) - - 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.2.2

Multiply $2$ by $- 1$ .

$\frac{10 x + 8 y - 4 - 4 x - 2 y - - 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.2.3

Multiply $- 1$ by $- 1$ .

$\frac{10 x + 8 y - 4 - 4 x - 2 y + 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.3

Subtract $4 x$ from $10 x$ .

$\frac{6 x + 8 y - 4 - 2 y + 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.4

Subtract $2 y$ from $8 y$ .

$\frac{6 x + 6 y - 4 + 1}{x (2 x + 2 y - 1)}$

Step 4.3.2.5

Add $- 4$ and $1$ .

$\frac{6 x + 6 y - 3}{x (2 x + 2 y - 1)}$

Step 4.3.2.6

Factor $3$ out of $6 x + 6 y - 3$ .

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Step 4.3.2.6.1

Factor $3$ out of $6 x$ .

$\frac{3 (2 x) + 6 y - 3}{x (2 x + 2 y - 1)}$

Step 4.3.2.6.2

Factor $3$ out of $6 y$ .

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

Step 4.3.2.6.3

Factor $3$ out of $- 3$ .

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

Step 4.3.2.6.4

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

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

Step 4.3.2.6.5

Factor $3$ out of $3 (2 x + 2 y) + 3 (- 1)$ .

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

Step 4.3.3

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

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Step 4.3.3.1

Cancel the common factor.

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

Step 4.3.3.2

Rewrite the expression.

$\frac{3}{x}$

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

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

Step 5.2

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

$μ (x, y) = e^{3 (\ln (| x |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{3 \ln (| x |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

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

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{3} + C$

Step 6

Multiply both sides of $2 (2 y^{2} + 5 x y - 2 y + 4) d x + x (2 x + 2 y - 1) d y = 0$ by the integration factor $x^{3}$ .

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Step 6.1

Multiply $2 (2 y^{2} + 5 x y - 2 y + 4)$ by $x^{3}$ .

$2 (2 y^{2} + 5 x y - 2 y + 4) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.2

Apply the distributive property.

$(2 (2 y^{2}) + 2 (5 x y) + 2 (- 2 y) + 2 \cdot 4) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.3

Simplify.

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Step 6.3.1

Multiply $2$ by $2$ .

$(4 y^{2} + 2 (5 x y) + 2 (- 2 y) + 2 \cdot 4) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.3.2

Multiply $5$ by $2$ .

$(4 y^{2} + 10 (x y) + 2 (- 2 y) + 2 \cdot 4) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.3.3

Multiply $- 2$ by $2$ .

$(4 y^{2} + 10 (x y) - 4 y + 2 \cdot 4) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.3.4

Multiply $2$ by $4$ .

$(4 y^{2} + 10 (x y) - 4 y + 8) x^{3} d x + x (2 x + 2 y - 1) d y = 0$

Step 6.4

Apply the distributive property.

$(4 y^{2} x^{3} + 10 x y x^{3} - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) d y = 0$

Step 6.5

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

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Step 6.5.1

Move $x^{3}$ .

$(4 y^{2} x^{3} + 10 (x^{3} x) y - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) d y = 0$

Step 6.5.2

Multiply $x^{3}$ by $x$ .

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Step 6.5.2.1

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

$(4 y^{2} x^{3} + 10 (x^{3} x^{1}) y - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) d y = 0$

Step 6.5.2.2

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

$(4 y^{2} x^{3} + 10 x^{3 + 1} y - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) d y = 0$

Step 6.5.3

Add $3$ and $1$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) d y = 0$

Step 6.6

Multiply $x (2 x + 2 y - 1)$ by $x^{3}$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x (2 x + 2 y - 1) x^{3} d y = 0$

Step 6.7

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

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Step 6.7.1

Move $x^{3}$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x^{3} x (2 x + 2 y - 1) d y = 0$

Step 6.7.2

Multiply $x^{3}$ by $x$ .

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Step 6.7.2.1

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

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x^{3} x^{1} (2 x + 2 y - 1) d y = 0$

Step 6.7.2.2

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

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x^{3 + 1} (2 x + 2 y - 1) d y = 0$

Step 6.7.3

Add $3$ and $1$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + x^{4} (2 x + 2 y - 1) d y = 0$

Step 6.8

Apply the distributive property.

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (x^{4} (2 x) + x^{4} (2 y) + x^{4} \cdot - 1) d y = 0$

Step 6.9

Simplify.

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Step 6.9.1

Rewrite using the commutative property of multiplication.

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{4} x + x^{4} (2 y) + x^{4} \cdot - 1) d y = 0$

Step 6.9.2

Rewrite using the commutative property of multiplication.

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{4} x + 2 x^{4} y + x^{4} \cdot - 1) d y = 0$

Step 6.9.3

Move $- 1$ to the left of $x^{4}$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{4} x + 2 x^{4} y - 1 \cdot x^{4}) d y = 0$

Step 6.10

Simplify each term.

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Step 6.10.1

Multiply $x^{4}$ by $x$ by adding the exponents.

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Step 6.10.1.1

Move $x$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 (x \cdot x^{4}) + 2 x^{4} y - 1 \cdot x^{4}) d y = 0$

Step 6.10.1.2

Multiply $x$ by $x^{4}$ .

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Step 6.10.1.2.1

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

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 (x^{1} x^{4}) + 2 x^{4} y - 1 \cdot x^{4}) d y = 0$

Step 6.10.1.2.2

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

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{1 + 4} + 2 x^{4} y - 1 \cdot x^{4}) d y = 0$

Step 6.10.1.3

Add $1$ and $4$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{5} + 2 x^{4} y - 1 \cdot x^{4}) d y = 0$

Step 6.10.2

Rewrite $- 1 x^{4}$ as $- x^{4}$ .

$(4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}) d x + (2 x^{5} + 2 x^{4} y - x^{4}) d y = 0$

Step 7

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

$f (x, y) = \int 2 x^{5} + 2 x^{4} y - x^{4} d y$

Step 8

Integrate $N (x, y) = 2 x^{5} + 2 x^{4} y - x^{4}$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

$f (x, y) = \int 2 x^{5} d y + \int 2 x^{4} y d y + \int - x^{4} d y$

Step 8.2

Apply the constant rule.

$f (x, y) = 2 x^{5} y + C + \int 2 x^{4} y d y + \int - x^{4} d y$

Step 8.3

Since $2 x^{4}$ is constant with respect to $y$ , move $2 x^{4}$ out of the integral.

$f (x, y) = 2 x^{5} y + C + 2 x^{4} \int y d y + \int - x^{4} d y$

Step 8.4

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

$f (x, y) = 2 x^{5} y + C + 2 x^{4} (\frac{1}{2} y^{2} + C) + \int - x^{4} d y$

Step 8.5

Apply the constant rule.

$f (x, y) = 2 x^{5} y + C + 2 x^{4} (\frac{1}{2} y^{2} + C) - x^{4} y + C$

Step 8.6

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

$f (x, y) = 2 x^{5} y + C + 2 x^{4} (\frac{y^{2}}{2} + C) - x^{4} y + C$

Step 8.7

Simplify.

$f (x, y) = 2 x^{5} y + x^{4} y^{2} - x^{4} y + C$

Step 9

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

$f (x, y) = 2 x^{5} y + x^{4} y^{2} - x^{4} y + g (x)$

Step 10

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

$\frac{\partial f}{\partial x} = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [2 x^{5} y + x^{4} y^{2} - x^{4} y + g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.2

By the Sum Rule, the derivative of $2 x^{5} y + x^{4} y^{2} - x^{4} y + g (x)$ with respect to $x$ is $\frac{d}{d x} [2 x^{5} y] + \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [2 x^{5} y] + \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.3

Evaluate $\frac{d}{d x} [2 x^{5} y]$ .

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Step 11.3.1

Since $2 y$ is constant with respect to $x$ , the derivative of $2 x^{5} y$ with respect to $x$ is $2 y \frac{d}{d x} [x^{5}]$ .

$2 y \frac{d}{d x} [x^{5}] + \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$2 y (5 x^{4}) + \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.3.3

Multiply $5$ by $2$ .

$10 y x^{4} + \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.4

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

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Step 11.4.1

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

$10 y x^{4} + y^{2} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.4.2

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

$10 y x^{4} + y^{2} (4 x^{3}) + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.4.3

Move $4$ to the left of $y^{2}$ .

$10 y x^{4} + 4 y^{2} x^{3} + \frac{d}{d x} [- x^{4} y] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.5

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

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Step 11.5.1

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

$10 y x^{4} + 4 y^{2} x^{3} - y \frac{d}{d x} [x^{4}] + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.5.2

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

$10 y x^{4} + 4 y^{2} x^{3} - y (4 x^{3}) + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.5.3

Multiply $4$ by $- 1$ .

$10 y x^{4} + 4 y^{2} x^{3} - 4 y x^{3} + \frac{d}{d x} [g (x)] = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.6

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

$10 y x^{4} + 4 y^{2} x^{3} - 4 y x^{3} + \frac{d g}{d x} = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 11.7

Reorder terms.

$\frac{d g}{d x} + 10 x^{4} y + 4 x^{3} y^{2} - 4 x^{3} y = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3}$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $10 x^{4} y$ from both sides of the equation.

$\frac{d g}{d x} + 4 x^{3} y^{2} - 4 x^{3} y = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3} - 10 x^{4} y$

Step 12.1.2

Subtract $4 x^{3} y^{2}$ from both sides of the equation.

$\frac{d g}{d x} - 4 x^{3} y = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3} - 10 x^{4} y - 4 x^{3} y^{2}$

Step 12.1.3

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

$\frac{d g}{d x} = 4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3} - 10 x^{4} y - 4 x^{3} y^{2} + 4 x^{3} y$

Step 12.1.4

Combine the opposite terms in $4 y^{2} x^{3} + 10 x^{4} y - 4 y x^{3} + 8 x^{3} - 10 x^{4} y - 4 x^{3} y^{2} + 4 x^{3} y$ .

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Step 12.1.4.1

Subtract $10 x^{4} y$ from $10 x^{4} y$ .

$\frac{d g}{d x} = 4 y^{2} x^{3} - 4 y x^{3} + 8 x^{3} + 0 - 4 x^{3} y^{2} + 4 x^{3} y$

Step 12.1.4.2

Add $4 y^{2} x^{3} - 4 y x^{3} + 8 x^{3}$ and $0$ .

$\frac{d g}{d x} = 4 y^{2} x^{3} - 4 y x^{3} + 8 x^{3} - 4 x^{3} y^{2} + 4 x^{3} y$

Step 12.1.4.3

Reorder the factors in the terms $4 y^{2} x^{3}$ and $- 4 x^{3} y^{2}$ .

$\frac{d g}{d x} = 4 x^{3} y^{2} - 4 y x^{3} + 8 x^{3} - 4 x^{3} y^{2} + 4 x^{3} y$

Step 12.1.4.4

Subtract $4 x^{3} y^{2}$ from $4 x^{3} y^{2}$ .

$\frac{d g}{d x} = - 4 y x^{3} + 8 x^{3} + 0 + 4 x^{3} y$

Step 12.1.4.5

Add $- 4 y x^{3} + 8 x^{3}$ and $0$ .

$\frac{d g}{d x} = - 4 y x^{3} + 8 x^{3} + 4 x^{3} y$

Step 12.1.4.6

Reorder the factors in the terms $- 4 y x^{3}$ and $4 x^{3} y$ .

$\frac{d g}{d x} = - 4 x^{3} y + 8 x^{3} + 4 x^{3} y$

Step 12.1.4.7

Add $- 4 x^{3} y$ and $4 x^{3} y$ .

$\frac{d g}{d x} = 8 x^{3} + 0$

Step 12.1.4.8

Add $8 x^{3}$ and $0$ .

$\frac{d g}{d x} = 8 x^{3}$

Step 13

Find the antiderivative of $8 x^{3}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = 8 x^{3}$ .

$\int \frac{d g}{d x} d x = \int 8 x^{3} d x$

Step 13.2

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

$g (x) = \int 8 x^{3} d x$

Step 13.3

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$g (x) = 8 \int x^{3} d x$

Step 13.4

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$g (x) = 8 (\frac{1}{4} x^{4} + C)$

Step 13.5

Simplify the answer.

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Step 13.5.1

Rewrite $8 (\frac{1}{4} x^{4} + C)$ as $8 (\frac{1}{4}) x^{4} + C$ .

$g (x) = 8 (\frac{1}{4}) x^{4} + C$

Step 13.5.2

Simplify.

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Step 13.5.2.1

Combine $8$ and $\frac{1}{4}$ .

$g (x) = \frac{8}{4} x^{4} + C$

Step 13.5.2.2

Cancel the common factor of $8$ and $4$ .

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Step 13.5.2.2.1

Factor $4$ out of $8$ .

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

Step 13.5.2.2.2

Cancel the common factors.

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Step 13.5.2.2.2.1

Factor $4$ out of $4$ .

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

Step 13.5.2.2.2.2

Cancel the common factor.

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

Step 13.5.2.2.2.3

Rewrite the expression.

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

Step 13.5.2.2.2.4

Divide $2$ by $1$ .

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

Step 14

Substitute for $g (x)$ in $f (x, y) = 2 x^{5} y + x^{4} y^{2} - x^{4} y + g (x)$ .

$f (x, y) = 2 x^{5} y + x^{4} y^{2} - x^{4} y + 2 x^{4} + C$