Calculus Examples

$(x^{3} + \frac{y}{x}) d x + (y^{2} + \ln (x)) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [\frac{y}{x}]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 + \frac{1}{x} \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{1}{x} \cdot 1$

Step 1.3.3

Multiply $\frac{1}{x}$ by $1$ .

$\frac{\partial M}{\partial y} = 0 + \frac{1}{x}$

Step 1.4

Add $0$ and $\frac{1}{x}$ .

$\frac{\partial M}{\partial y} = \frac{1}{x}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [\ln (x)]$

Step 2.3

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

$\frac{\partial N}{\partial x} = 0 + \frac{1}{x}$

Step 2.4

Add $0$ and $\frac{1}{x}$ .

$\frac{\partial N}{\partial x} = \frac{1}{x}$

Step 3

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

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Step 3.1

Substitute $\frac{1}{x}$ for $\frac{\partial M}{\partial y}$ and $\frac{1}{x}$ for $\frac{\partial N}{\partial x}$ .

$\frac{1}{x} = \frac{1}{x}$

Step 3.2

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

$\frac{1}{x} = \frac{1}{x}$ is an identity.

Step 4

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

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

Step 5

Integrate $N (x, y) = y^{2} + \ln (x)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [\frac{1}{3} y^{3} + \ln (x) y + g (x)] = x^{3} + \frac{y}{x}$

Step 8.2

Differentiate.

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Step 8.2.1

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

$\frac{d}{d x} [\frac{1}{3} y^{3}] + \frac{d}{d x} [\ln (x) y] + \frac{d}{d x} [g (x)] = x^{3} + \frac{y}{x}$

Step 8.2.2

Since $\frac{1}{3} y^{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3} y^{3}$ with respect to $x$ is $0$ .

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

Step 8.3

Evaluate $\frac{d}{d x} [\ln (x) y]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Combine $y$ and $\frac{1}{x}$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Add $0$ and $\frac{y}{x}$ .

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

Step 8.5.2

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Move all terms containing variables to the left side of the equation.

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Step 9.1.1.1

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

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

Step 9.1.1.2

Subtract $\frac{y}{x}$ from both sides of the equation.

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

Step 9.1.1.3

Combine the opposite terms in $\frac{d g}{d x} + \frac{y}{x} - x^{3} - \frac{y}{x}$ .

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Step 9.1.1.3.1

Subtract $\frac{y}{x}$ from $\frac{y}{x}$ .

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

Step 9.1.1.3.2

Add $\frac{d g}{d x} - x^{3}$ and $0$ .

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

Step 9.1.2

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

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

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

Step 11

Substitute for $g (x)$ in $f (x, y) = \frac{1}{3} y^{3} + \ln (x) y + g (x)$ .

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

Step 12

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

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Step 12.1

Simplify each term.

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Step 12.1.1

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

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

Step 12.1.2

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

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

Step 12.2

Reorder factors in $f (x, y) = \frac{y^{3}}{3} + \ln (x) y + \frac{x^{4}}{4} + C$ .

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