Calculus Examples

$(2 x + \frac{y}{x}) d x + (4 y + \ln (x)) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $2 x$ is constant with respect to $y$ , the derivative of $2 x$ 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 $N (x, y) = 4 y + \ln (x)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Since $4 y$ is constant with respect to $x$ , the derivative of $4 y$ 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 $N (x, y)$ .

$f (x, y) = \int 4 y + \ln (x) d y$

Step 5

Integrate $N (x, y) = 4 y + \ln (x)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 4 y d y + \int \ln (x) d y$

Step 5.2

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

$f (x, y) = 4 \int y d y + \int \ln (x) d y$

Step 5.3

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

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

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

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

Step 8.2.2

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

$0 + \frac{d}{d x} [\ln (x) y] + \frac{d}{d x} [g (x)] = 2 x + \frac{y}{x}$

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

$0 + y \frac{d}{d x} [\ln (x)] + \frac{d}{d x} [g (x)] = 2 x + \frac{y}{x}$

Step 8.3.2

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

$0 + y \frac{1}{x} + \frac{d}{d x} [g (x)] = 2 x + \frac{y}{x}$

Step 8.3.3

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

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

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

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

Step 8.5.2

Reorder terms.

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

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 $2 x$ from both sides of the equation.

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

Step 9.1.1.2

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

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

Step 9.1.1.3

Combine the opposite terms in $\frac{d g}{d x} + \frac{y}{x} - 2 x - \frac{y}{x}$ .

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Step 9.1.1.3.1

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

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

Step 9.1.1.3.2

Add $\frac{d g}{d x} - 2 x$ and $0$ .

$\frac{d g}{d x} - 2 x = 0$

Step 9.1.2

Add $2 x$ to both sides of the equation.

$\frac{d g}{d x} = 2 x$

Step 10

Find the antiderivative of $2 x$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 2 x$ .

$\int \frac{d g}{d x} d x = \int 2 x d x$

Step 10.2

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

$g (x) = \int 2 x d x$

Step 10.3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$g (x) = 2 \int x d x$

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

Rewrite $2 (\frac{1}{2} x^{2} + C)$ as $2 (\frac{1}{2}) x^{2} + C$ .

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

Step 10.5.2.2

Cancel the common factor of $2$ .

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Step 10.5.2.2.1

Cancel the common factor.

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

Step 10.5.2.2.2

Rewrite the expression.

$g (x) = 1 x^{2} + C$

Step 10.5.2.3

Multiply $x^{2}$ by $1$ .

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

Step 11

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

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

Step 12

Reorder factors in $f (x, y) = 2 y^{2} + \ln (x) y + x^{2} + C$ .

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