Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.4

Since $e^{x}$ is constant with respect to $y$ , the derivative of $e^{x}$ with respect to $y$ is $0$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Add $2 x e^{y}$ and $0$ .

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

Step 1.5.2

Reorder the factors of $2 x e^{y}$ .

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

Step 1.5.3

Reorder factors in $2 e^{y} x$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = (x^{2} + 1) e^{y}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Since $e^{y}$ is constant with respect to $x$ , the derivative of $(x^{2} + 1) e^{y}$ with respect to $x$ is $e^{y} \frac{d}{d x} [x^{2} + 1]$ .

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

Step 2.3

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 2.5

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

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

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $2 x$ and $0$ .

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

Step 2.6.2

Move $2$ to the left of $e^{y}$ .

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

Step 2.6.3

Reorder factors in $2 e^{y} x$ .

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

Step 3

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

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Step 3.1

Substitute $2 x e^{y}$ for $\frac{\partial M}{\partial y}$ and $2 x e^{y}$ for $\frac{\partial N}{\partial x}$ .

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

Step 3.2

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

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

Step 4

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

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

Step 5

Integrate $N (x, y) = (x^{2} + 1) e^{y}$ to find $f (x, y)$ .

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Step 5.1

Since $x^{2} + 1$ is constant with respect to $y$ , move $x^{2} + 1$ out of the integral.

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

Step 5.2

The integral of $e^{y}$ with respect to $y$ is $e^{y}$ .

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

Step 5.3

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

By the Sum Rule, the derivative of $(x^{2} + 1) e^{y} + g (x)$ with respect to $x$ is $\frac{d}{d x} [(x^{2} + 1) e^{y}] + \frac{d}{d x} [g (x)]$ .

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

Step 8.3

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

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Step 8.3.1

Since $e^{y}$ is constant with respect to $x$ , the derivative of $(x^{2} + 1) e^{y}$ with respect to $x$ is $e^{y} \frac{d}{d x} [x^{2} + 1]$ .

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

Step 8.3.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 8.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 8.3.4

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

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

Step 8.3.5

Add $2 x$ and $0$ .

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

Step 8.3.6

Move $2$ to the left of $e^{y}$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

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

Step 8.5.2

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

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

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

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

Step 9.1.2

Combine the opposite terms in $2 x e^{y} + e^{x} - 2 x e^{y}$ .

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Step 9.1.2.1

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

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

Step 9.1.2.2

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

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

Step 10

Find the antiderivative of $e^{x}$ to find $g (x)$ .

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Step 10.1

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

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

Step 10.2

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

$g (x) = \int e^{x} d x$

Step 10.3

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

$g (x) = e^{x} + C$

Step 11

Substitute for $g (x)$ in $f (x, y) = (x^{2} + 1) e^{y} + g (x)$ .

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

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

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