Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

Differentiate.

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Step 1.3.1

By the Sum Rule, the derivative of $2 + y \frac{1}{x^{2}}$ with respect to $y$ is $\frac{d}{d y} [2] + \frac{d}{d y} [y \frac{1}{x^{2}}]$ .

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

Step 1.3.2

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [y \frac{1}{x^{2}}]$

Step 1.4

Evaluate $\frac{d}{d y} [y \frac{1}{x^{2}}]$ .

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [\frac{y}{x^{2}}]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{1}{x^{2}} \frac{d}{d y} [y]$

Step 1.4.3

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

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

Step 1.4.4

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

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

Step 1.5

Add $0$ and $\frac{1}{x^{2}}$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - \frac{d}{d x} [x^{- 1}]$

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $- 1$ .

$\frac{\partial N}{\partial x} = 0 + 1 x^{- 2}$

Step 2.3.4

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

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

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

Integrate $N (x, y) = y - x^{- 1}$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

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

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

By the Sum Rule, the derivative of $\frac{1}{2} y^{2} - x^{- 1} y + g (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2} y^{2}] + \frac{d}{d x} [- x^{- 1} y] + \frac{d}{d x} [g (x)]$ .

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

Step 8.2.2

Since $\frac{1}{2} y^{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} y^{2}$ with respect to $x$ is $0$ .

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Multiply $- 1$ by $- 1$ .

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

Step 8.3.4

Multiply $y$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

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

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

Step 8.5.2

Combine terms.

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Step 8.5.2.1

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

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

Step 8.5.2.2

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

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

Step 8.5.3

Reorder terms.

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

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

Simplify each term.

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Step 9.1.1.1

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

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

Step 9.1.1.2

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

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

Step 9.1.2

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 9.1.2.1

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

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

Step 9.1.2.2

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

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Step 9.1.2.2.1

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

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

Step 9.1.2.2.2

Add $2$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

Apply the constant rule.

$g (x) = 2 x + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = \frac{1}{2} y^{2} - x^{- 1} y + g (x)$ .

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

Step 12

Simplify each term.

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Step 12.1

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

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

Step 12.2

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

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

Step 12.3

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

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