Calculus Examples

$(y - \frac{1}{y}) d x + (x + \frac{x}{y^{2}}) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [- \frac{1}{y}]$ .

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Step 1.3.1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = - 1$ and $g (y) = \frac{1}{y}$ .

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

Step 1.3.2

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

$\frac{\partial M}{\partial y} = 1 - \frac{d}{d y} [y^{- 1}] + \frac{1}{y} \frac{d}{d y} [- 1]$

Step 1.3.3

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

$\frac{\partial M}{\partial y} = 1 - - y^{- 2} + \frac{1}{y} \frac{d}{d y} [- 1]$

Step 1.3.4

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

$\frac{\partial M}{\partial y} = 1 - - y^{- 2} + \frac{1}{y} \cdot 0$

Step 1.3.5

Multiply $- 1$ by $- 1$ .

$\frac{\partial M}{\partial y} = 1 + 1 y^{- 2} + \frac{1}{y} \cdot 0$

Step 1.3.6

Multiply $y^{- 2}$ by $1$ .

$\frac{\partial M}{\partial y} = 1 + y^{- 2} + \frac{1}{y} \cdot 0$

Step 1.3.7

Multiply $\frac{1}{y}$ by $0$ .

$\frac{\partial M}{\partial y} = 1 + y^{- 2} + 0$

Step 1.3.8

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

$\frac{\partial M}{\partial y} = 1 + y^{- 2}$

Step 1.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\partial M}{\partial y} = 1 + \frac{1}{y^{2}}$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = \frac{1}{y^{2}} + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 1 + \frac{d}{d x} [\frac{x}{y^{2}}]$

Step 2.3

Evaluate $\frac{d}{d x} [\frac{x}{y^{2}}]$ .

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Step 2.3.1

Since $\frac{1}{y^{2}}$ is constant with respect to $x$ , the derivative of $\frac{x}{y^{2}}$ with respect to $x$ is $\frac{1}{y^{2}} \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = 1 + \frac{1}{y^{2}} \frac{d}{d x} [x]$

Step 2.3.2

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

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

Step 2.3.3

Multiply $\frac{1}{y^{2}}$ by $1$ .

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

Step 2.4

Reorder terms.

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

Step 3

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

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Step 3.1

Substitute $\frac{1}{y^{2}} + 1$ for $\frac{\partial M}{\partial y}$ and $\frac{1}{y^{2}} + 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{1}{y^{2}} + 1 = \frac{1}{y^{2}} + 1$

Step 3.2

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

$\frac{1}{y^{2}} + 1 = \frac{1}{y^{2}} + 1$ is an identity.

Step 4

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

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

Step 5

Integrate $M (x, y) = y - \frac{1}{y}$ to find $f (x, y)$ .

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Step 5.1

Apply the constant rule.

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

Step 6

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

$f (x, y) = (y - \frac{1}{y}) x + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = - 1$ and $g (y) = \frac{1}{y}$ .

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

Step 8.3.5

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

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

Step 8.3.6

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

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

Step 8.3.7

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

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

Step 8.3.8

Multiply $- 1$ by $- 1$ .

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

Step 8.3.9

Multiply $y^{- 2}$ by $1$ .

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

Step 8.3.10

Multiply $\frac{1}{y}$ by $0$ .

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

Step 8.3.11

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

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

Step 8.4

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

$x (1 + y^{- 2}) + \frac{d g}{d y} = x + \frac{x}{y^{2}}$

Step 8.5

Simplify.

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Step 8.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 8.5.2

Apply the distributive property.

$x \cdot 1 + x \frac{1}{y^{2}} + \frac{d g}{d y} = x + \frac{x}{y^{2}}$

Step 8.5.3

Combine terms.

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Step 8.5.3.1

Multiply $x$ by $1$ .

$x + x \frac{1}{y^{2}} + \frac{d g}{d y} = x + \frac{x}{y^{2}}$

Step 8.5.3.2

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

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

Step 8.5.4

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Subtract $x$ from both sides of the equation.

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

Step 9.1.2

Subtract $\frac{x}{y^{2}}$ from both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

Subtract $x$ from $x$ .

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

Step 9.1.3.2

Add $\frac{d g}{d y} + \frac{x}{y^{2}}$ and $0$ .

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

Step 9.1.3.3

Subtract $\frac{x}{y^{2}}$ from $\frac{x}{y^{2}}$ .

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

Step 9.1.3.4

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

$\frac{d g}{d y} = 0$

Step 10

Find the antiderivative of $0$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 10.2

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

$g (y) = \int 0 d y$

Step 10.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 10.4

Add $0$ and $C$ .

$g (y) = C$

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

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

$f (x, y) = y x - \frac{x}{y} + C$