Calculus Examples

$(1 + \ln (x) + \frac{y}{x}) d x - (1 - \ln (x)) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 1 + \ln (x) + \frac{y}{x}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [1 + \ln (x) + \frac{y}{x}]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [1] + \frac{d}{d y} [\ln (x)] + \frac{d}{d y} [\frac{y}{x}]$

Step 1.2.2

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

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

Step 1.2.3

Since $\ln (x)$ is constant with respect to $y$ , the derivative of $\ln (x)$ with respect to $y$ is $0$ .

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

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 $\frac{1}{x} \frac{d}{d y} [y]$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.4

Combine terms.

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Step 1.4.1

Add $0$ and $0$ .

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

Step 1.4.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (1 - \ln (x))]$

Step 2.2

Differentiate.

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Step 2.2.1

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

$\frac{\partial N}{\partial x} = - \frac{d}{d x} [1 - \ln (x)]$

Step 2.2.2

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

$\frac{\partial N}{\partial x} = - (\frac{d}{d x} [1] + \frac{d}{d x} [- \ln (x)])$

Step 2.2.3

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

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

Step 2.2.4

Add $0$ and $\frac{d}{d x} [- \ln (x)]$ .

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

Step 2.2.5

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

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

Step 2.2.6

Multiply.

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Step 2.2.6.1

Multiply $- 1$ by $- 1$ .

$\frac{\partial N}{\partial x} = 1 \frac{d}{d x} [\ln (x)]$

Step 2.2.6.2

Multiply $\frac{d}{d x} [\ln (x)]$ by $1$ .

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

Step 2.3

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

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

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 $N (x, y)$ .

$f (x, y) = \int - (1 - \ln (x)) d y$

Step 5

Integrate $N (x, y) = - (1 - \ln (x))$ to find $f (x, y)$ .

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Step 5.1

Apply the constant rule.

$f (x, y) = - (1 - \ln (x)) y + C$

Step 5.2

Rewrite $- (1 - \ln (x)) y + C$ as $(- 1 + \ln (x)) y + C$ .

$f (x, y) = (- 1 + \ln (x)) y + C$

Step 6

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

$f (x, y) = (- 1 + \ln (x)) y + g (x)$

Step 7

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

$\frac{\partial f}{\partial x} = 1 + \ln (x) + \frac{y}{x}$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d x} [(- 1 + \ln (x)) y]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

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

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

Step 8.3.6

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

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

Step 8.4

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

$\frac{y}{x} + \frac{d g}{d x} = 1 + \ln (x) + \frac{y}{x}$

Step 8.5

Reorder terms.

$\frac{d g}{d x} + \frac{y}{x} = 1 + \ln (x) + \frac{y}{x}$

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 $\ln (x)$ from both sides of the equation.

$\frac{d g}{d x} + \frac{y}{x} - \ln (x) = 1 + \frac{y}{x}$

Step 9.1.1.2

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

$\frac{d g}{d x} + \frac{y}{x} - \ln (x) - \frac{y}{x} = 1$

Step 9.1.1.3

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

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Step 9.1.1.3.1

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

$\frac{d g}{d x} + 0 - \ln (x) = 1$

Step 9.1.1.3.2

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

$\frac{d g}{d x} - \ln (x) = 1$

Step 9.1.2

Add $\ln (x)$ to both sides of the equation.

$\frac{d g}{d x} = 1 + \ln (x)$

Step 10

Find the antiderivative of $1 + \ln (x)$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 1 + \ln (x)$ .

$\int \frac{d g}{d x} d x = \int 1 + \ln (x) d x$

Step 10.2

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

$g (x) = \int 1 + \ln (x) d x$

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int d x + \int \ln (x) d x$

Step 10.4

Apply the constant rule.

$g (x) = x + C + \int \ln (x) d x$

Step 10.5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = 1$ .

$g (x) = x + C + \ln (x) x - \int x \frac{1}{x} d x$

Step 10.6

Simplify.

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Step 10.6.1

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

$g (x) = x + C + \ln (x) x - \int \frac{x}{x} d x$

Step 10.6.2

Cancel the common factor of $x$ .

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Step 10.6.2.1

Cancel the common factor.

$g (x) = x + C + \ln (x) x - \int \frac{x}{x} d x$

Step 10.6.2.2

Rewrite the expression.

$g (x) = x + C + \ln (x) x - \int d x$

Step 10.7

Apply the constant rule.

$g (x) = x + C + \ln (x) x - (x + C)$

Step 10.8

Simplify.

$g (x) = x + \ln (x) x - x + C$

Step 10.9

Simplify.

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Step 10.9.1

Subtract $x$ from $x$ .

$g (x) = \ln (x) x + 0 + C$

Step 10.9.2

Add $\ln (x) x$ and $0$ .

$g (x) = \ln (x) x + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = (- 1 + \ln (x)) y + g (x)$ .

$f (x, y) = (- 1 + \ln (x)) y + \ln (x) x + C$

Step 12

Simplify $f (x, y) = (- 1 + \ln (x)) y + \ln (x) x + C$ .

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Step 12.1

Simplify each term.

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Step 12.1.1

Apply the distributive property.

$f (x, y) = - 1 y + \ln (x) y + \ln (x) x + C$

Step 12.1.2

Rewrite $- 1 y$ as $- y$ .

$f (x, y) = - y + \ln (x) y + \ln (x) x + C$

Step 12.2

Reorder factors in $f (x, y) = - y + \ln (x) y + \ln (x) x + C$ .

$f (x, y) = - y + y \ln (x) + x \ln (x) + C$