Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

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

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

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

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

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Step 2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = | x |$ .

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Step 2.3.1.1

To apply the Chain Rule, set $u$ as $| x |$ .

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

Step 2.3.1.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{1}{u} \frac{d}{d x} [| x |]$

Step 2.3.1.3

Replace all occurrences of $u$ with $| x |$ .

$\frac{\partial N}{\partial x} = 0 + \frac{1}{| x |} \frac{d}{d x} [| x |]$

Step 2.3.2

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

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

Step 2.3.3

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

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

Step 2.3.4

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\partial N}{\partial x} = 0 + \frac{x}{| x \cdot x |}$

Step 2.3.5

Raise $x$ to the power of $1$ .

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

Step 2.3.6

Raise $x$ to the power of $1$ .

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

Step 2.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3.8

Add $1$ and $1$ .

$\frac{\partial N}{\partial x} = 0 + \frac{x}{| x^{2} |}$

Step 2.4

Simplify.

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Step 2.4.1

Add $0$ and $\frac{x}{| x^{2} |}$ .

$\frac{\partial N}{\partial x} = \frac{x}{| x^{2} |}$

Step 2.4.2

Remove non-negative terms from the absolute value.

$\frac{\partial N}{\partial x} = \frac{x}{x^{2}}$

Step 2.4.3

Cancel the common factor of $x$ and $x^{2}$ .

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Step 2.4.3.1

Raise $x$ to the power of $1$ .

$\frac{\partial N}{\partial x} = \frac{x^{1}}{x^{2}}$

Step 2.4.3.2

Factor $x$ out of $x^{1}$ .

$\frac{\partial N}{\partial x} = \frac{x \cdot 1}{x^{2}}$

Step 2.4.3.3

Cancel the common factors.

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Step 2.4.3.3.1

Factor $x$ out of $x^{2}$ .

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

Step 2.4.3.3.2

Cancel the common factor.

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

Step 2.4.3.3.3

Rewrite the expression.

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

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 $M (x, y)$ .

$f (x, y) = \int \frac{y}{x} d x$

Step 5

Integrate $M (x, y) = \frac{y}{x}$ to find $f (x, y)$ .

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Step 5.1

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

$f (x, y) = y \int \frac{1}{x} d x$

Step 5.2

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

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

Step 5.3

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial y} = y^{2} + \ln (| x |)$

Step 8

Differentiate $f$ with respect to $y$ .

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Rewrite.

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

Step 9.1.2

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

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Step 9.1.2.1

Differentiate $M$ with respect to $y$ .

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

Step 9.1.2.2

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

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

Step 9.1.2.3

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

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

Step 9.1.2.4

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

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

Step 9.1.3

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

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Step 9.1.3.1

Differentiate $N$ with respect to $x$ .

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

Step 9.1.3.2

Differentiate.

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Step 9.1.3.2.1

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 9.1.3.2.2

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 9.1.3.3

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

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Step 9.1.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = | x |$ .

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Step 9.1.3.3.1.1

To apply the Chain Rule, set $u$ as $| x |$ .

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

Step 9.1.3.3.1.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{1}{u} \frac{d}{d x} [| x |]$

Step 9.1.3.3.1.3

Replace all occurrences of $u$ with $| x |$ .

$\frac{\partial N}{\partial x} = 0 + \frac{1}{| x |} \frac{d}{d x} [| x |]$

Step 9.1.3.3.2

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

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

Step 9.1.3.3.3

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

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

Step 9.1.3.3.4

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\partial N}{\partial x} = 0 + \frac{x}{| x \cdot x |}$

Step 9.1.3.3.5

Raise $x$ to the power of $1$ .

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

Step 9.1.3.3.6

Raise $x$ to the power of $1$ .

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

Step 9.1.3.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 9.1.3.3.8

Add $1$ and $1$ .

$\frac{\partial N}{\partial x} = 0 + \frac{x}{| x^{2} |}$

Step 9.1.3.4

Simplify.

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Step 9.1.3.4.1

Add $0$ and $\frac{x}{| x^{2} |}$ .

$\frac{\partial N}{\partial x} = \frac{x}{| x^{2} |}$

Step 9.1.3.4.2

Remove non-negative terms from the absolute value.

$\frac{\partial N}{\partial x} = \frac{x}{x^{2}}$

Step 9.1.3.4.3

Cancel the common factor of $x$ and $x^{2}$ .

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Step 9.1.3.4.3.1

Raise $x$ to the power of $1$ .

$\frac{\partial N}{\partial x} = \frac{x^{1}}{x^{2}}$

Step 9.1.3.4.3.2

Factor $x$ out of $x^{1}$ .

$\frac{\partial N}{\partial x} = \frac{x \cdot 1}{x^{2}}$

Step 9.1.3.4.3.3

Cancel the common factors.

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Step 9.1.3.4.3.3.1

Factor $x$ out of $x^{2}$ .

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

Step 9.1.3.4.3.3.2

Cancel the common factor.

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

Step 9.1.3.4.3.3.3

Rewrite the expression.

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

Step 9.1.4

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

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Step 9.1.4.1

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 9.1.4.2

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 9.1.5

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

$f (x, y) = \int \frac{y}{x} d x$

Step 9.1.6

Integrate $M (x, y) = \frac{y}{x}$ to find $f (x, y)$ .

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Step 9.1.6.1

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

$f (x, y) = y \int \frac{1}{x} d x$

Step 9.1.6.2

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

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

Step 9.1.6.3

Simplify.

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

Step 9.1.7

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

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

Step 9.1.8

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

$\frac{\partial f}{\partial y} = y^{2} + \ln (| x |)$

Step 9.1.9

Simplify the left side.

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Step 9.1.9.1

Simplify $\frac{d}{d y} \cdot (y \ln (| x |) + g (y))$ .

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Step 9.1.9.1.1

Cancel the common factor of $d$ .

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Step 9.1.9.1.1.1

Cancel the common factor.

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

Step 9.1.9.1.1.2

Rewrite the expression.

$\frac{1}{y} \cdot (y \ln (| x |) + g (y)) = y^{2} + \ln (| x |)$

Step 9.1.9.1.2

Multiply $\frac{1}{y}$ by $y \ln (| x |) + g y$ .

$\frac{y \ln (| x |) + g y}{y} = y^{2} + \ln (| x |)$

Step 9.1.9.1.3

Factor $y$ out of $y \ln (| x |) + g y$ .

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Step 9.1.9.1.3.1

Factor $y$ out of $y \ln (| x |)$ .

$\frac{y \ln (| x |) + g y}{y} = y^{2} + \ln (| x |)$

Step 9.1.9.1.3.2

Factor $y$ out of $g y$ .

$\frac{y \ln (| x |) + y g}{y} = y^{2} + \ln (| x |)$

Step 9.1.9.1.3.3

Factor $y$ out of $y \ln (| x |) + y g$ .

$\frac{y (\ln (| x |) + g)}{y} = y^{2} + \ln (| x |)$

Step 9.1.9.1.4

Cancel the common factor of $y$ .

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Step 9.1.9.1.4.1

Cancel the common factor.

$\frac{y (\ln (| x |) + g)}{y} = y^{2} + \ln (| x |)$

Step 9.1.9.1.4.2

Divide $\ln (| x |) + g$ by $1$ .

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

Step 9.1.10

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| x |) - \ln (| x |) = - g + y^{2}$

Step 9.1.11

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| x |}{| x |}) = - g + y^{2}$

Step 9.1.12

Cancel the common factor of $| x |$ .

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Step 9.1.12.1

Cancel the common factor.

$\ln (\frac{| x |}{| x |}) = - g + y^{2}$

Step 9.1.12.2

Rewrite the expression.

$\ln (1) = - g + y^{2}$

Step 9.1.13

The natural logarithm of $1$ is $0$ .

$0 = - g + y^{2}$

Step 10

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

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Step 10.1

Integrate both sides of $0 = - g + y^{2}$ .

$\int 0 d y = \int - g + y^{2} d y$

Step 10.2

Evaluate $\int 0 d y$ .

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

Step 10.3

Split the single integral into multiple integrals.

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

Step 10.4

Apply the constant rule.

$g (y) = - g y + C + \int y^{2} d y$

Step 10.5

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

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

Step 10.6

Simplify.

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

Step 11

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

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

Step 12

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

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