Calculus Examples

$(2 y - 2 x y^{3} + 4 x + 6) d x + (2 x - 3 x^{2} y^{2} - 1) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 y - 2 x y^{3} + 4 x + 6$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

By the Sum Rule, the derivative of $2 y - 2 x y^{3} + 4 x + 6$ with respect to $y$ is $\frac{d}{d y} [2 y] + \frac{d}{d y} [- 2 x y^{3}] + \frac{d}{d y} [4 x] + \frac{d}{d y} [6]$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $1$ .

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

Multiply $3$ by $- 2$ .

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

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 $0$ .

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

Step 1.5.2

Since $6$ is constant with respect to $y$ , the derivative of $6$ with respect to $y$ is $0$ .

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

Step 1.6

Simplify.

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Step 1.6.1

Combine terms.

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Step 1.6.1.1

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

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

Step 1.6.1.2

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

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

Step 1.6.2

Reorder terms.

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $1$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $2$ by $- 3$ .

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

Step 2.5

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

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

Step 2.6

Simplify.

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Step 2.6.1

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

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

Step 2.6.2

Reorder terms.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

Integrate $N (x, y) = 2 x - 3 x^{2} y^{2} - 1$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 2 y - 2 x y^{3} + 4 x + 6$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [2 x y - x^{2} y^{3} - y + g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.2

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

$\frac{d}{d x} [2 x y] + \frac{d}{d x} [- x^{2} y^{3}] + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.3

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

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Step 8.3.1

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

$2 y \frac{d}{d x} [x] + \frac{d}{d x} [- x^{2} y^{3}] + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.3.2

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

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

Step 8.3.3

Multiply $2$ by $1$ .

$2 y + \frac{d}{d x} [- x^{2} y^{3}] + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.4

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

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Step 8.4.1

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

$2 y - y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.4.2

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

$2 y - y^{3} (2 x) + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.4.3

Multiply $2$ by $- 1$ .

$2 y - 2 y^{3} x + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.5

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

$2 y - 2 y^{3} x + 0 + \frac{d}{d x} [g (x)] = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.6

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

$2 y - 2 y^{3} x + 0 + \frac{d g}{d x} = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.7

Simplify.

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Step 8.7.1

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

$2 y - 2 y^{3} x + \frac{d g}{d x} = 2 y - 2 x y^{3} + 4 x + 6$

Step 8.7.2

Reorder terms.

$\frac{d g}{d x} - 2 y^{3} x + 2 y = 2 y - 2 x y^{3} + 4 x + 6$

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

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

$\frac{d g}{d x} + 2 y = 2 y - 2 x y^{3} + 4 x + 6 + 2 y^{3} x$

Step 9.1.2

Subtract $2 y$ from both sides of the equation.

$\frac{d g}{d x} = 2 y - 2 x y^{3} + 4 x + 6 + 2 y^{3} x - 2 y$

Step 9.1.3

Combine the opposite terms in $2 y - 2 x y^{3} + 4 x + 6 + 2 y^{3} x - 2 y$ .

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Step 9.1.3.1

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

$\frac{d g}{d x} = 2 y - 2 y^{3} x + 4 x + 6 + 2 y^{3} x - 2 y$

Step 9.1.3.2

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

$\frac{d g}{d x} = 2 y + 4 x + 6 + 0 - 2 y$

Step 9.1.3.3

Add $2 y + 4 x + 6$ and $0$ .

$\frac{d g}{d x} = 2 y + 4 x + 6 - 2 y$

Step 9.1.3.4

Subtract $2 y$ from $2 y$ .

$\frac{d g}{d x} = 4 x + 6 + 0$

Step 9.1.3.5

Add $4 x + 6$ and $0$ .

$\frac{d g}{d x} = 4 x + 6$

Step 10

Find the antiderivative of $4 x + 6$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 4 x + 6$ .

$\int \frac{d g}{d x} d x = \int 4 x + 6 d x$

Step 10.2

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

$g (x) = \int 4 x + 6 d x$

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int 4 x d x + \int 6 d x$

Step 10.4

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

$g (x) = 4 \int x d x + \int 6 d x$

Step 10.5

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

$g (x) = 4 (\frac{1}{2} x^{2} + C) + \int 6 d x$

Step 10.6

Apply the constant rule.

$g (x) = 4 (\frac{1}{2} x^{2} + C) + 6 x + C$

Step 10.7

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

$g (x) = 4 (\frac{x^{2}}{2} + C) + 6 x + C$

Step 10.8

Simplify.

$g (x) = 2 x^{2} + 6 x + C$

Step 11

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

$f (x, y) = 2 x y - x^{2} y^{3} - y + 2 x^{2} + 6 x + C$