Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $3$ .

$\frac{\partial M}{\partial y} = 9 x^{2} y^{2} + \frac{d}{d y} [y^{4}]$

Step 1.4

Differentiate using the Power Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 9 x^{2} y^{2} + 4 y^{3}$

Step 1.4.2

Reorder terms.

$\frac{\partial M}{\partial y} = 4 y^{3} + 9 x^{2} y^{2}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $3$ by $3$ .

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [4 x y^{3}]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + 0 + \frac{d}{d x} [4 x y^{3}]$

Step 2.5

Evaluate $\frac{d}{d x} [4 x y^{3}]$ .

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Step 2.5.1

Since $4 y^{3}$ is constant with respect to $x$ , the derivative of $4 x y^{3}$ with respect to $x$ is $4 y^{3} \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + 0 + 4 y^{3} \frac{d}{d x} [x]$

Step 2.5.2

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

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + 0 + 4 y^{3} \cdot 1$

Step 2.5.3

Multiply $4$ by $1$ .

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + 0 + 4 y^{3}$

Step 2.6

Simplify.

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Step 2.6.1

Add $9 y^{2} x^{2}$ and $0$ .

$\frac{\partial N}{\partial x} = 9 y^{2} x^{2} + 4 y^{3}$

Step 2.6.2

Reorder terms.

$\frac{\partial N}{\partial x} = 4 y^{3} + 9 x^{2} y^{2}$

Step 3

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

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Step 3.1

Substitute $4 y^{3} + 9 x^{2} y^{2}$ for $\frac{\partial M}{\partial y}$ and $4 y^{3} + 9 x^{2} y^{2}$ for $\frac{\partial N}{\partial x}$ .

$4 y^{3} + 9 x^{2} y^{2} = 4 y^{3} + 9 x^{2} y^{2}$

Step 3.2

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

$4 y^{3} + 9 x^{2} y^{2} = 4 y^{3} + 9 x^{2} y^{2}$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Move $3$ to the left of $x^{3}$ .

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Move $4$ to the left of $x$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

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

Step 9.1.2

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

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

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

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

Step 9.1.3.4

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

$\frac{d g}{d y} = y^{4} + 0$

Step 9.1.3.5

Add $y^{4}$ and $0$ .

$\frac{d g}{d y} = y^{4}$

Step 10

Find the antiderivative of $y^{4}$ to find $g (y)$ .

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Step 10.1

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

$\int \frac{d g}{d y} d y = \int y^{4} d y$

Step 10.2

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

$g (y) = \int y^{4} d y$

Step 10.3

By the Power Rule, the integral of $y^{4}$ with respect to $y$ is $\frac{1}{5} y^{5}$ .

$g (y) = \frac{1}{5} y^{5} + C$

Step 11

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

$f (x, y) = y^{3} x^{3} + y^{4} x + \frac{1}{5} y^{5} + C$

Step 12

Combine $\frac{1}{5}$ and $y^{5}$ .

$f (x, y) = y^{3} x^{3} + y^{4} x + \frac{y^{5}}{5} + C$