Calculus Examples

$(4 x^{3} y - 15 x^{2} - y) d x + (x^{4} + 3 y^{2} - x) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 4 x^{3} y - 15 x^{2} - y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [4 x^{3} y - 15 x^{2} - y]$

Step 1.2

By the Sum Rule, the derivative of $4 x^{3} y - 15 x^{2} - y$ with respect to $y$ is $\frac{d}{d y} [4 x^{3} y] + \frac{d}{d y} [- 15 x^{2}] + \frac{d}{d y} [- y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [4 x^{3} y] + \frac{d}{d y} [- 15 x^{2}] + \frac{d}{d y} [- y]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 4 x^{3} \frac{d}{d y} [y] + \frac{d}{d y} [- 15 x^{2}] + \frac{d}{d y} [- y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 4 x^{3} \cdot 1 + \frac{d}{d y} [- 15 x^{2}] + \frac{d}{d y} [- y]$

Step 1.3.3

Multiply $4$ by $1$ .

$\frac{\partial M}{\partial y} = 4 x^{3} + \frac{d}{d y} [- 15 x^{2}] + \frac{d}{d y} [- y]$

Step 1.4

Since $- 15 x^{2}$ is constant with respect to $y$ , the derivative of $- 15 x^{2}$ with respect to $y$ is $0$ .

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

Step 1.5

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

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Step 1.5.1

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

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

Step 1.5.2

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

$\frac{\partial M}{\partial y} = 4 x^{3} + 0 - 1 \cdot 1$

Step 1.5.3

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = 4 x^{3} + 0 - 1$

Step 1.6

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

$\frac{\partial M}{\partial y} = 4 x^{3} - 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $0$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = 4 x^{3} + 0 - 1$

Step 2.4

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

$\frac{\partial N}{\partial x} = 4 x^{3} - 1$

Step 3

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

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Step 3.1

Substitute $4 x^{3} - 1$ for $\frac{\partial M}{\partial y}$ and $4 x^{3} - 1$ for $\frac{\partial N}{\partial x}$ .

$4 x^{3} - 1 = 4 x^{3} - 1$

Step 3.2

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

$4 x^{3} - 1 = 4 x^{3} - 1$ is an identity.

Step 4

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

$f (x, y) = \int 4 x^{3} y - 15 x^{2} - y d x$

Step 5

Integrate $M (x, y) = 4 x^{3} y - 15 x^{2} - y$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 4 x^{3} y d x + \int - 15 x^{2} d x + \int - y d x$

Step 5.2

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

$f (x, y) = 4 y \int x^{3} d x + \int - 15 x^{2} d x + \int - y d x$

Step 5.3

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

$f (x, y) = 4 y (\frac{1}{4} x^{4} + C) + \int - 15 x^{2} d x + \int - y d x$

Step 5.4

Since $- 15$ is constant with respect to $x$ , move $- 15$ out of the integral.

$f (x, y) = 4 y (\frac{1}{4} x^{4} + C) - 15 \int x^{2} d x + \int - y d x$

Step 5.5

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

$f (x, y) = 4 y (\frac{1}{4} x^{4} + C) - 15 (\frac{1}{3} x^{3} + C) + \int - y d x$

Step 5.6

Apply the constant rule.

$f (x, y) = 4 y (\frac{1}{4} x^{4} + C) - 15 (\frac{1}{3} x^{3} + C) - y x + C$

Step 5.7

Simplify.

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Step 5.7.1

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

$f (x, y) = 4 y (\frac{x^{4}}{4} + C) - 15 (\frac{1}{3} x^{3} + C) - y x + C$

Step 5.7.2

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

$f (x, y) = 4 y (\frac{x^{4}}{4} + C) - 15 (\frac{x^{3}}{3} + C) - y x + C$

Step 5.8

Simplify.

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

Step 6

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

$f (x, y) = y x^{4} - 5 x^{3} - y x + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

By the Sum Rule, the derivative of $y x^{4} - 5 x^{3} - y x + g (y)$ with respect to $y$ is $\frac{d}{d y} [y x^{4}] + \frac{d}{d y} [- 5 x^{3}] + \frac{d}{d y} [- y x] + \frac{d}{d y} [g (y)]$ .

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

$x^{4} \cdot 1 + \frac{d}{d y} [- 5 x^{3}] + \frac{d}{d y} [- y x] + \frac{d}{d y} [g (y)] = x^{4} + 3 y^{2} - x$

Step 8.3.3

Multiply $x^{4}$ by $1$ .

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

Step 8.4

Since $- 5 x^{3}$ is constant with respect to $y$ , the derivative of $- 5 x^{3}$ with respect to $y$ is $0$ .

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

Step 8.5

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

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Step 8.5.1

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

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

Step 8.5.2

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

$x^{4} + 0 - x \cdot 1 + \frac{d}{d y} [g (y)] = x^{4} + 3 y^{2} - x$

Step 8.5.3

Multiply $- 1$ by $1$ .

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

Step 8.6

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

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

Step 8.7

Simplify.

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Step 8.7.1

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

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

Step 8.7.2

Reorder terms.

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

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

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

Step 9.1.2

Add $x$ to both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Add $- x$ and $x$ .

$\frac{d g}{d y} = 3 y^{2} + 0$

Step 9.1.3.4

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

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

Step 10

Find the antiderivative of $3 y^{2}$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = 3 y^{2}$ .

$\int \frac{d g}{d y} d y = \int 3 y^{2} d y$

Step 10.2

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

$g (y) = \int 3 y^{2} d y$

Step 10.3

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

$g (y) = 3 \int y^{2} d y$

Step 10.4

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

$g (y) = 3 (\frac{1}{3} y^{3} + C)$

Step 10.5

Simplify the answer.

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Step 10.5.1

Rewrite $3 (\frac{1}{3} y^{3} + C)$ as $3 (\frac{1}{3}) y^{3} + C$ .

$g (y) = 3 (\frac{1}{3}) y^{3} + C$

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

$g (y) = \frac{3}{3} y^{3} + C$

Step 10.5.2.2

Cancel the common factor of $3$ .

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Step 10.5.2.2.1

Cancel the common factor.

$g (y) = \frac{3}{3} y^{3} + C$

Step 10.5.2.2.2

Rewrite the expression.

$g (y) = 1 y^{3} + C$

Step 10.5.2.3

Multiply $y^{3}$ by $1$ .

$g (y) = y^{3} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = y x^{4} - 5 x^{3} - y x + g (y)$ .

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