Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 1 + e^{x} y + x e^{x} y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $e^{x}$ by $1$ .

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

Multiply $x$ by $1$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Add $0$ and $e^{x}$ .

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

Step 1.5.2

Reorder terms.

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

Step 1.5.3

Reorder factors in $e^{x} x + e^{x}$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

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

Step 2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $e^{x}$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

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Step 2.5.1

Add $x e^{x} + e^{x}$ and $0$ .

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

Step 2.5.2

Reorder terms.

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

Step 2.5.3

Reorder factors in $e^{x} x + e^{x}$ .

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

Step 3

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

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Step 3.1

Substitute $x e^{x} + e^{x}$ for $\frac{\partial M}{\partial y}$ and $x e^{x} + e^{x}$ for $\frac{\partial N}{\partial x}$ .

$x e^{x} + e^{x} = x e^{x} + e^{x}$

Step 3.2

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

$x e^{x} + e^{x} = x e^{x} + e^{x}$ is an identity.

Step 4

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

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

Step 5

Integrate $N (x, y) = x e^{x} + 2$ to find $f (x, y)$ .

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Step 5.1

Apply the constant rule.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 1 + e^{x} y + x e^{x} y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

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

Step 8.3.4

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 8.3.5

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

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

Step 8.3.6

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

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

Step 8.3.7

Multiply $e^{x}$ by $1$ .

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

Step 8.3.8

Add $x e^{x} + e^{x}$ and $0$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Apply the distributive property.

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

Step 8.5.2

Reorder terms.

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

Step 8.5.3

Reorder factors in $\frac{d g}{d x} + e^{x} x y + e^{x} y$ .

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

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 $1 + e^{x} y + x e^{x} y$ .

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

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

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

Step 9.1.2.2

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

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

Step 9.1.2.3

Combine the opposite terms in $1 + y e^{x} + x y e^{x} - x y e^{x} - y e^{x}$ .

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Step 9.1.2.3.1

Subtract $x y e^{x}$ from $x y e^{x}$ .

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

Step 9.1.2.3.2

Add $1 + y e^{x}$ and $0$ .

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

Step 9.1.2.3.3

Subtract $y e^{x}$ from $y e^{x}$ .

$\frac{d g}{d x} = 1 + 0$

Step 9.1.2.3.4

Add $1$ and $0$ .

$\frac{d g}{d x} = 1$

Step 10

Find the antiderivative of $1$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 1$ .

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

Step 10.2

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

$g (x) = \int d x$

Step 10.3

Apply the constant rule.

$g (x) = x + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = (x e^{x} + 2) y + g (x)$ .

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

Step 12

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

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Step 12.1

Apply the distributive property.

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

Step 12.2

Reorder factors in $f (x, y) = x e^{x} y + 2 y + x + C$ .

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