Calculus Examples

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

Step 1

Find

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

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

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Step 1.1

Differentiate

$M$ with respect to

$y$ .

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

Step 1.2

Since

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

$y$ , the derivative of

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

$y$ is

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

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

Step 1.3

Differentiate using the Power Rule which states that

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

$n y^{n - 1}$ where

$n = 2$ .

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

Step 1.4

Simplify the expression.

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Step 1.4.1

Move

$2$ to the left of

$e^{2 x}$ .

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

Step 1.4.2

Reorder factors in

$2 e^{2 x} y$ .

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

Step 2

Find

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

$N (x, y) = e^{2 x} y - 2 y$ .

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Step 2.1

Differentiate

$N$ with respect to

$x$ .

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

Step 2.2

By the Sum Rule, the derivative of

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

$x$ is

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

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

Step 2.3

Evaluate

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

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Step 2.3.1

Since

$y$ is constant with respect to

$x$ , the derivative of

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

$x$ is

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

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

Step 2.3.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = e^{x}$ and

$g (x) = 2 x$ .

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Step 2.3.2.1

To apply the Chain Rule, set

$u$ as

$2 x$ .

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

Step 2.3.2.2

Differentiate using the Exponential Rule which states that

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

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

$a$ =

$e$ .

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

Step 2.3.2.3

Replace all occurrences of

$u$ with

$2 x$ .

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

Step 2.3.3

Since

$2$ is constant with respect to

$x$ , the derivative of

$2 x$ with respect to

$x$ is

$2 \frac{d}{d x} [x]$ .

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

Step 2.3.4

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.3.5

Multiply

$2$ by

$1$ .

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

Step 2.3.6

Move

$2$ to the left of

$e^{2 x}$ .

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

Step 2.3.7

Move

$2$ to the left of

$y$ .

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

Step 2.4

Since

$- 2 y$ is constant with respect to

$x$ , the derivative of

$- 2 y$ with respect to

$x$ is

$0$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Add

$2 y e^{2 x}$ and

$0$ .

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

Step 2.5.2

Reorder the factors of

$2 y e^{2 x}$ .

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

Step 2.5.3

Reorder factors in

$2 e^{2 x} y$ .

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

Step 3

Check that

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

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Step 3.1

Substitute

$2 y e^{2 x}$ for

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

$2 y e^{2 x}$ for

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

$2 y e^{2 x} = 2 y e^{2 x}$

Step 3.2

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

$2 y e^{2 x} = 2 y e^{2 x}$ is an identity.

Step 4

Set

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

$N (x, y)$ .

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

Step 5

Integrate

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

$f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Since

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

$y$ , move

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

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

Step 5.3

By the Power Rule, the integral of

$y$ with respect to

$y$ is

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

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

Step 5.4

Since

$- 2$ is constant with respect to

$y$ , move

$- 2$ out of the integral.

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

Step 5.5

By the Power Rule, the integral of

$y$ with respect to

$y$ is

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

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

Step 5.6

Simplify.

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

Step 5.7

Simplify.

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Step 5.7.1

Combine

$\frac{1}{2}$ and

$e^{2 x}$ .

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

Step 5.7.2

Combine

$\frac{e^{2 x}}{2}$ and

$y^{2}$ .

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

Step 5.7.3

Combine

$\frac{1}{2}$ and

$y^{2}$ .

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

Step 5.7.4

Combine

$- 2$ and

$\frac{y^{2}}{2}$ .

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

Step 5.7.5

Cancel the common factor of

$- 2$ and

$2$ .

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Step 5.7.5.1

Factor

$2$ out of

$- 2 y^{2}$ .

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

Step 5.7.5.2

Cancel the common factors.

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Step 5.7.5.2.1

Factor

$2$ out of

$2$ .

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

Step 5.7.5.2.2

Cancel the common factor.

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

Step 5.7.5.2.3

Rewrite the expression.

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

Step 5.7.5.2.4

Divide

$- y^{2}$ by

$1$ .

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

Step 5.8

Reorder terms.

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

Step 6

Since the integral of

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

$C$ with

$g (x)$ .

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

Step 7

Set

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

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

Step 8

Find

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

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Step 8.1

Differentiate

$f$ with respect to

$x$ .

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

Step 8.2

By the Sum Rule, the derivative of

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

$x$ is

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

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

Step 8.3

Evaluate

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

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Step 8.3.1

Combine

$\frac{1}{2}$ and

$e^{2 x}$ .

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

Step 8.3.2

Combine

$\frac{e^{2 x}}{2}$ and

$y^{2}$ .

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

Step 8.3.3

Since

$\frac{y^{2}}{2}$ is constant with respect to

$x$ , the derivative of

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

$x$ is

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

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

Step 8.3.4

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = e^{x}$ and

$g (x) = 2 x$ .

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Step 8.3.4.1

To apply the Chain Rule, set

$u$ as

$2 x$ .

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

Step 8.3.4.2

Differentiate using the Exponential Rule which states that

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

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

$a$ =

$e$ .

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

Step 8.3.4.3

Replace all occurrences of

$u$ with

$2 x$ .

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

Step 8.3.5

Since

$2$ is constant with respect to

$x$ , the derivative of

$2 x$ with respect to

$x$ is

$2 \frac{d}{d x} [x]$ .

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

Step 8.3.6

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 1$ .

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

Step 8.3.7

Multiply

$2$ by

$1$ .

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

Step 8.3.8

Move

$2$ to the left of

$e^{2 x}$ .

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

Step 8.3.9

Combine

$2$ and

$\frac{y^{2}}{2}$ .

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

Step 8.3.10

Combine

$\frac{2 y^{2}}{2}$ and

$e^{2 x}$ .

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

Step 8.3.11

Cancel the common factor of

$2$ .

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Step 8.3.11.1

Cancel the common factor.

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

Step 8.3.11.2

Divide

$y^{2} e^{2 x}$ by

$1$ .

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

Step 8.4

Since

$- y^{2}$ is constant with respect to

$x$ , the derivative of

$- y^{2}$ with respect to

$x$ is

$0$ .

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

Step 8.5

Differentiate using the function rule which states that the derivative of

$g (x)$ is

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

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

Step 8.6

Simplify.

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Step 8.6.1

Add

$y^{2} e^{2 x}$ and

$0$ .

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

Step 8.6.2

Reorder terms.

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

Step 8.6.3

Reorder factors in

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

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

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

Reorder factors in

$e^{2 x} y^{2}$ .

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

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

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

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

Step 9.1.2.2

Subtract

$y^{2} e^{2 x}$ from

$y^{2} e^{2 x}$ .

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

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

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

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

Step 12

Simplify

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

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Step 12.1

Simplify each term.

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Step 12.1.1

Combine

$\frac{1}{2}$ and

$e^{2 x}$ .

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

Step 12.1.2

Combine

$\frac{e^{2 x}}{2}$ and

$y^{2}$ .

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

Step 12.2

Reorder factors in

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

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