Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $2$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} + \frac{d}{d y} [- 4 y] + \frac{d}{d y} [4 x] + \frac{d}{d y} [- 3]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 \frac{d}{d y} [y] + \frac{d}{d y} [4 x] + \frac{d}{d y} [- 3]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 \cdot 1 + \frac{d}{d y} [4 x] + \frac{d}{d y} [- 3]$

Step 1.4.3

Multiply $- 4$ by $1$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 + \frac{d}{d y} [4 x] + \frac{d}{d y} [- 3]$

Step 1.5

Differentiate using the Constant Rule.

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Step 1.5.1

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

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 + 0 + \frac{d}{d y} [- 3]$

Step 1.5.2

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 + 0 + 0$

Step 1.6

Simplify.

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Step 1.6.1

Combine terms.

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Step 1.6.1.1

Add $6 x y^{2} - 4$ and $0$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4 + 0$

Step 1.6.1.2

Add $6 x y^{2} - 4$ and $0$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} - 4$

Step 1.6.2

Reorder terms.

$\frac{\partial M}{\partial y} = 6 y^{2} x - 4$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x + \frac{d}{d x} [- 4 x]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 6 y^{2} x - 4 \frac{d}{d x} [x]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 6 y^{2} x - 4 \cdot 1$

Step 2.4.3

Multiply $- 4$ by $1$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x - 4$

Step 3

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

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Step 3.1

Substitute $6 y^{2} x - 4$ for $\frac{\partial M}{\partial y}$ and $6 y^{2} x - 4$ for $\frac{\partial N}{\partial x}$ .

$6 y^{2} x - 4 = 6 y^{2} x - 4$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$6 y^{2} x - 4 = 6 y^{2} x - 4$ is an identity.

Step 4

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

$f (x, y) = \int 3 x^{2} y^{2} - 4 x d y$

Step 5

Integrate $N (x, y) = 3 x^{2} y^{2} - 4 x$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 3 x^{2} y^{2} d y + \int - 4 x d y$

Step 5.2

Since $3 x^{2}$ is constant with respect to $y$ , move $3 x^{2}$ out of the integral.

$f (x, y) = 3 x^{2} \int y^{2} d y + \int - 4 x d y$

Step 5.3

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

$f (x, y) = 3 x^{2} (\frac{1}{3} y^{3} + C) + \int - 4 x d y$

Step 5.4

Apply the constant rule.

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

Step 5.5

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

$f (x, y) = 3 x^{2} (\frac{y^{3}}{3} + C) - 4 x y + C$

Step 5.6

Simplify.

$f (x, y) = x^{2} y^{3} - 4 x y + C$

Step 6

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

$f (x, y) = x^{2} y^{3} - 4 x y + g (x)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $- 4$ by $1$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

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

Step 9.1.2

Add $4 y$ to both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Add $- 4 y + 4 x - 3$ and $0$ .

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

Step 9.1.3.4

Add $- 4 y$ and $4 y$ .

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

Step 9.1.3.5

Add $4 x - 3$ and $0$ .

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

Step 10

Find the antiderivative of $4 x - 3$ to find $g (x)$ .

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int 4 x d x + \int - 3 d x$

Step 10.4

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

$g (x) = 4 \int x d x + \int - 3 d x$

Step 10.5

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

$g (x) = 4 (\frac{1}{2} x^{2} + C) + \int - 3 d x$

Step 10.6

Apply the constant rule.

$g (x) = 4 (\frac{1}{2} x^{2} + C) - 3 x + C$

Step 10.7

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

$g (x) = 4 (\frac{x^{2}}{2} + C) - 3 x + C$

Step 10.8

Simplify.

$g (x) = 2 x^{2} - 3 x + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = x^{2} y^{3} - 4 x y + g (x)$ .

$f (x, y) = x^{2} y^{3} - 4 x y + 2 x^{2} - 3 x + C$