Calculus Examples

$(\sin (y) - y \sin (x)) d x + (\cos (x) + x \cos (y) - y) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\sin (y) - y \sin (x)]$

Step 1.2

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

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

Step 1.3

The derivative of $\sin (y)$ with respect to $y$ is $\cos (y)$ .

$\frac{\partial M}{\partial y} = \cos (y) + \frac{d}{d y} [- y \sin (x)]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = \cos (y) - \sin (x) \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = \cos (y) - \sin (x) \cdot 1$

Step 1.4.3

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = \cos (y) - \sin (x)$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\frac{\partial N}{\partial x} = - \sin (x) + \frac{d}{d x} [x \cos (y)] + \frac{d}{d x} [- y]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = - \sin (x) + \cos (y) \frac{d}{d x} [x] + \frac{d}{d x} [- y]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = - \sin (x) + \cos (y) \cdot 1 + \frac{d}{d x} [- y]$

Step 2.4.3

Multiply $\cos (y)$ by $1$ .

$\frac{\partial N}{\partial x} = - \sin (x) + \cos (y) + \frac{d}{d x} [- y]$

Step 2.5

Differentiate using the Constant Rule.

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Step 2.5.1

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

$\frac{\partial N}{\partial x} = - \sin (x) + \cos (y) + 0$

Step 2.5.2

Add $- \sin (x) + \cos (y)$ and $0$ .

$\frac{\partial N}{\partial x} = - \sin (x) + \cos (y)$

Step 3

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

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Step 3.1

Substitute $\cos (y) - \sin (x)$ for $\frac{\partial M}{\partial y}$ and $- \sin (x) + \cos (y)$ for $\frac{\partial N}{\partial x}$ .

$\cos (y) - \sin (x) = - \sin (x) + \cos (y)$

Step 3.2

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

$\cos (y) - \sin (x) = - \sin (x) + \cos (y)$ is an identity.

Step 4

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

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

Step 5

Integrate $M (x, y) = \sin (y) - y \sin (x)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int \sin (y) d x + \int - y \sin (x) d x$

Step 5.2

Apply the constant rule.

$f (x, y) = \sin (y) x + C + \int - y \sin (x) d x$

Step 5.3

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

$f (x, y) = \sin (y) x + C - y \int \sin (x) d x$

Step 5.4

The integral of $\sin (x)$ with respect to $x$ is $- \cos (x)$ .

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

Step 5.5

Simplify.

$f (x, y) = \sin (y) x + y \cos (x) + C$

Step 6

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

$f (x, y) = \sin (y) x + y \cos (x) + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = \cos (x) + x \cos (y) - y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

The derivative of $\sin (y)$ with respect to $y$ is $\cos (y)$ .

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

Step 8.4

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

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Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

Multiply $\cos (x)$ by $1$ .

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

Step 8.5

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

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

Step 8.6

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Subtract $x \cos (y)$ from both sides of the equation.

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

Step 9.1.2

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

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

Step 9.1.3

Combine the opposite terms in $\cos (x) + x \cos (y) - y - x \cos (y) - \cos (x)$ .

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Step 9.1.3.1

Subtract $x \cos (y)$ from $x \cos (y)$ .

$\frac{d g}{d y} = \cos (x) - y + 0 - \cos (x)$

Step 9.1.3.2

Add $\cos (x) - y$ and $0$ .

$\frac{d g}{d y} = \cos (x) - y - \cos (x)$

Step 9.1.3.3

Subtract $\cos (x)$ from $\cos (x)$ .

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

Step 9.1.3.4

Add $- y$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

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

Step 10.4

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

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

Step 10.5

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

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

Step 11

Substitute for $g (y)$ in $f (x, y) = \sin (y) x + y \cos (x) + g (y)$ .

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

Step 12

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

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Step 12.1

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

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

Step 12.2

Reorder factors in $f (x, y) = \sin (y) x + y \cos (x) - \frac{y^{2}}{2} + C$ .

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