Calculus Examples

(sin (y) + x) d x + (x cos (y)) d y = 0

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

Step 1

Find

\frac{\partial M}{\partial y}

$\frac{\partial M}{\partial y}$ where

M (x, y) = sin (y) + x

$M (x, y) = \sin (y) + x$ .

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Step 1.1

Differentiate

M

$M$ with respect to

y

$y$ .

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

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

Step 1.2

By the Sum Rule, the derivative of

sin (y) + x

$\sin (y) + x$ with respect to

y

$y$ is

\frac{d}{d y} [sin (y)] + \frac{d}{d y} [x]

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

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

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

Step 1.3

The derivative of

sin (y)

$\sin (y)$ with respect to

y

$y$ is

cos (y)

$\cos (y)$ .

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

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since

x

$x$ is constant with respect to

y

$y$ , the derivative of

x

$x$ with respect to

y

$y$ is

0

$0$ .

\frac{\partial M}{\partial y} = cos (y) + 0

$\frac{\partial M}{\partial y} = \cos (y) + 0$

Step 1.4.2

Add

cos (y)

$\cos (y)$ and

0

$0$ .

\frac{\partial M}{\partial y} = cos (y)

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

\frac{\partial M}{\partial y} = cos (y)

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

\frac{\partial M}{\partial y} = cos (y)

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

Step 2

Find

\frac{\partial N}{\partial x}

$\frac{\partial N}{\partial x}$ where

N (x, y) = x cos (y)

$N (x, y) = x \cos (y)$ .

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Step 2.1

Differentiate

N

$N$ with respect to

x

$x$ .

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

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

Step 2.2

Since

cos (y)

$\cos (y)$ is constant with respect to

x

$x$ , the derivative of

x cos (y)

$x \cos (y)$ with respect to

x

$x$ is

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

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

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

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

Step 2.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

\frac{\partial N}{\partial x} = cos (y) \cdot 1

$\frac{\partial N}{\partial x} = \cos (y) \cdot 1$

Step 2.4

Multiply

cos (y)

$\cos (y)$ by

1

$1$ .

\frac{\partial N}{\partial x} = cos (y)

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

\frac{\partial N}{\partial x} = cos (y)

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

Step 3

Check that

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

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

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Step 3.1

Substitute

cos (y)

$\cos (y)$ for

\frac{\partial M}{\partial y}

$\frac{\partial M}{\partial y}$ and

cos (y)

$\cos (y)$ for

\frac{\partial N}{\partial x}

$\frac{\partial N}{\partial x}$ .

cos (y) = cos (y)

$\cos (y) = \cos (y)$

Step 3.2

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

cos (y) = cos (y)

$\cos (y) = \cos (y)$ is an identity.

cos (y) = cos (y)

$\cos (y) = \cos (y)$ is an identity.

Step 4

Set

f (x, y)

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

N (x, y)

$N (x, y)$ .

f (x, y) = \int x cos (y) d y

$f (x, y) = \int x \cos (y) d y$

Step 5

Integrate

N (x, y) = x cos (y)

$N (x, y) = x \cos (y)$ to find

f (x, y)

$f (x, y)$ .

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Step 5.1

Since

x

$x$ is constant with respect to

y

$y$ , move

x

$x$ out of the integral.

f (x, y) = x \int cos (y) d y

$f (x, y) = x \int \cos (y) d y$

Step 5.2

The integral of

cos (y)

$\cos (y)$ with respect to

y

$y$ is

sin (y)

$\sin (y)$ .

f (x, y) = x (sin (y) + C)

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

Step 5.3

Simplify.

f (x, y) = x sin (y) + C

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

f (x, y) = x sin (y) + C

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

Step 6

Since the integral of

g (x)

$g (x)$ will contain an integration constant, we can replace

C

$C$ with

g (x)

$g (x)$ .

f (x, y) = x sin (y) + g (x)

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

Step 7

Set

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

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

\frac{\partial f}{\partial x} = sin (y) + x

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

Step 8

Find

\frac{\partial f}{\partial x}

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

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Step 8.1

Differentiate

f

$f$ with respect to

x

$x$ .

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

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

Step 8.2

By the Sum Rule, the derivative of

x sin (y) + g (x)

$x \sin (y) + g (x)$ with respect to

x

$x$ is

\frac{d}{d x} [x sin (y)] + \frac{d}{d x} [g (x)]

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

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

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

Step 8.3

Evaluate

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

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

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Step 8.3.1

Since

sin (y)

$\sin (y)$ is constant with respect to

x

$x$ , the derivative of

x sin (y)

$x \sin (y)$ with respect to

x

$x$ is

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

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

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

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

Step 8.3.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

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

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

Step 8.3.3

Multiply

sin (y)

$\sin (y)$ by

1

$1$ .

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

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

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

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

Step 8.4

Differentiate using the function rule which states that the derivative of

$g (x)$ is

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

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

Step 8.5

Reorder terms.

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

Step 9

Solve for

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

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Step 9.1

Move all terms not containing

$\frac{d g}{d x}$ to the right side of the equation.

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Step 9.1.1

Subtract

$\sin (y)$ from both sides of the equation.

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

Step 9.1.2

Combine the opposite terms in

$\sin (y) + x - \sin (y)$ .

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Step 9.1.2.1

Subtract

$\sin (y)$ from

$\sin (y)$ .

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

Step 9.1.2.2

Add

$x$ and

$0$ .

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

Step 10

Find the antiderivative of

$x$ to find

$g (x)$ .

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Step 10.1

Integrate both sides of

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

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

Step 10.2

Evaluate

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

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

Step 10.3

By the Power Rule, the integral of

$x$ with respect to

$x$ is

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

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

Step 11

Substitute for

$g (x)$ in

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

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

Step 12

Combine

$\frac{1}{2}$ and

$x^{2}$ .

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

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