Calculus Examples

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

Step 1

Find

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

$M (x, y) = y \cos (x) + 2 x e^{y}$ .

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Step 1.1

Differentiate

$M$ with respect to

$y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \cos (x) + 2 x e^{y}]$

Step 1.2

By the Sum Rule, the derivative of

$y \cos (x) + 2 x e^{y}$ with respect to

$y$ is

$\frac{d}{d y} [y \cos (x)] + \frac{d}{d y} [2 x e^{y}]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \cos (x)] + \frac{d}{d y} [2 x e^{y}]$

Step 1.3

Evaluate

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

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Step 1.3.1

Since

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

$y$ , the derivative of

$y \cos (x)$ with respect to

$y$ is

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

$\frac{\partial M}{\partial y} = \cos (x) \frac{d}{d y} [y] + \frac{d}{d y} [2 x e^{y}]$

Step 1.3.2

Differentiate using the Power Rule which states that

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

$n y^{n - 1}$ where

$n = 1$ .

$\frac{\partial M}{\partial y} = \cos (x) \cdot 1 + \frac{d}{d y} [2 x e^{y}]$

Step 1.3.3

Multiply

$\cos (x)$ by

$1$ .

$\frac{\partial M}{\partial y} = \cos (x) + \frac{d}{d y} [2 x e^{y}]$

Step 1.4

Evaluate

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

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Step 1.4.1

Since

$2 x$ is constant with respect to

$y$ , the derivative of

$2 x e^{y}$ with respect to

$y$ is

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

$\frac{\partial M}{\partial y} = \cos (x) + 2 x \frac{d}{d y} [e^{y}]$

Step 1.4.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d y} [a^{y}]$ is

$a^{y} \ln (a)$ where

$a$ =

$e$ .

$\frac{\partial M}{\partial y} = \cos (x) + 2 x e^{y}$

Step 1.5

Simplify.

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Step 1.5.1

Reorder terms.

$\frac{\partial M}{\partial y} = 2 e^{y} x + \cos (x)$

Step 1.5.2

Reorder factors in

$2 e^{y} x + \cos (x)$ .

$\frac{\partial M}{\partial y} = 2 x e^{y} + \cos (x)$

Step 2

Find

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

$N (x, y) = \sin (x) + x^{2} e^{y} - 1$ .

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Step 2.1

Differentiate

$N$ with respect to

$x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [\sin (x) + x^{2} e^{y} - 1]$

Step 2.2

By the Sum Rule, the derivative of

$\sin (x) + x^{2} e^{y} - 1$ with respect to

$x$ is

$\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [x^{2} e^{y}] + \frac{d}{d x} [- 1]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [x^{2} e^{y}] + \frac{d}{d x} [- 1]$

Step 2.3

The derivative of

$\sin (x)$ with respect to

$x$ is

$\cos (x)$ .

$\frac{\partial N}{\partial x} = \cos (x) + \frac{d}{d x} [x^{2} e^{y}] + \frac{d}{d x} [- 1]$

Step 2.4

Evaluate

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

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Step 2.4.1

Since

$e^{y}$ is constant with respect to

$x$ , the derivative of

$x^{2} e^{y}$ with respect to

$x$ is

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

$\frac{\partial N}{\partial x} = \cos (x) + e^{y} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$

Step 2.4.2

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 2$ .

$\frac{\partial N}{\partial x} = \cos (x) + e^{y} (2 x) + \frac{d}{d x} [- 1]$

Step 2.4.3

Move

$2$ to the left of

$e^{y}$ .

$\frac{\partial N}{\partial x} = \cos (x) + 2 e^{y} x + \frac{d}{d x} [- 1]$

Step 2.5

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- 1$ with respect to

$x$ is

$0$ .

$\frac{\partial N}{\partial x} = \cos (x) + 2 e^{y} x + 0$

Step 2.6

Simplify.

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Step 2.6.1

Add

$\cos (x) + 2 e^{y} x$ and

$0$ .

$\frac{\partial N}{\partial x} = \cos (x) + 2 e^{y} x$

Step 2.6.2

Reorder terms.

$\frac{\partial N}{\partial x} = 2 e^{y} x + \cos (x)$

Step 2.6.3

Reorder factors in

$2 e^{y} x + \cos (x)$ .

$\frac{\partial N}{\partial x} = 2 x e^{y} + \cos (x)$

Step 3

Check that

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

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Step 3.1

Substitute

$2 x e^{y} + \cos (x)$ for

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

$2 x e^{y} + \cos (x)$ for

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

$2 x e^{y} + \cos (x) = 2 x e^{y} + \cos (x)$

Step 3.2

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

$2 x e^{y} + \cos (x) = 2 x e^{y} + \cos (x)$ is an identity.

Step 4

Set

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

$N (x, y)$ .

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

Step 5

Integrate

$N (x, y) = \sin (x) + x^{2} e^{y} - 1$ to find

$f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

Since

$x^{2}$ is constant with respect to

$y$ , move

$x^{2}$ out of the integral.

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

Step 5.4

The integral of

$e^{y}$ with respect to

$y$ is

$e^{y}$ .

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

Step 5.5

Apply the constant rule.

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

Step 5.6

Simplify.

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

Step 6

Since the integral of

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

$C$ with

$g (x)$ .

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

Step 7

Set

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

$\frac{\partial f}{\partial x} = y \cos (x) + 2 x e^{y}$

Step 8

Find

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

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Step 8.1

Differentiate

$f$ with respect to

$x$ .

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

Step 8.2

By the Sum Rule, the derivative of

$\sin (x) y + x^{2} e^{y} - y + g (x)$ with respect to

$x$ is

$\frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [x^{2} e^{y}] + \frac{d}{d x} [- y] + \frac{d}{d x} [g (x)]$ .

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

Step 8.3

Evaluate

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

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Step 8.3.1

Since

$y$ is constant with respect to

$x$ , the derivative of

$\sin (x) y$ with respect to

$x$ is

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

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

Step 8.3.2

The derivative of

$\sin (x)$ with respect to

$x$ is

$\cos (x)$ .

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

Step 8.4

Evaluate

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

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Step 8.4.1

Since

$e^{y}$ is constant with respect to

$x$ , the derivative of

$x^{2} e^{y}$ with respect to

$x$ is

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

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

Step 8.4.2

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 2$ .

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

Step 8.4.3

Move

$2$ to the left of

$e^{y}$ .

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

Step 8.5

Since

$- y$ is constant with respect to

$x$ , the derivative of

$- y$ with respect to

$x$ is

$0$ .

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

Step 8.6

Differentiate using the function rule which states that the derivative of

$g (x)$ is

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

$y \cos (x) + 2 e^{y} x + 0 + \frac{d g}{d x} = y \cos (x) + 2 x e^{y}$

Step 8.7

Simplify.

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Step 8.7.1

Add

$y \cos (x) + 2 e^{y} x$ and

$0$ .

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

Step 8.7.2

Reorder terms.

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

Step 8.7.3

Reorder factors in

$\frac{d g}{d x} + y \cos (x) + 2 e^{y} x$ .

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

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

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

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

Step 9.1.2

Subtract

$2 x e^{y}$ from both sides of the equation.

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

Step 9.1.3

Combine the opposite terms in

$y \cos (x) + 2 x e^{y} - y \cos (x) - 2 x e^{y}$ .

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Step 9.1.3.1

Subtract

$y \cos (x)$ from

$y \cos (x)$ .

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

Step 9.1.3.2

Add

$2 x e^{y}$ and

$0$ .

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

Step 9.1.3.3

Subtract

$2 x e^{y}$ from

$2 x e^{y}$ .

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

Step 10

Find the antiderivative of

$0$ to find

$g (x)$ .

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Step 10.1

Integrate both sides of

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

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

Step 10.2

Evaluate

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

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

Step 10.3

The integral of

$0$ with respect to

$x$ is

$0$ .

$g (x) = 0 + C$

Step 10.4

Add

$0$ and

$C$ .

$g (x) = C$

Step 11

Substitute for

$g (x)$ in

$f (x, y) = \sin (x) y + x^{2} e^{y} - y + g (x)$ .

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

Step 12

Reorder factors in

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

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