Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = - y$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $- y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d u} [e^{u}] \frac{d}{d y} [- y]$

Step 1.2.2

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

$\frac{\partial M}{\partial y} = e^{u} \frac{d}{d y} [- y]$

Step 1.2.3

Replace all occurrences of $u$ with $- y$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = e^{- y} (- \frac{d}{d y} [y])$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = e^{- y} (- 1 \cdot 1)$

Step 1.3.3

Simplify the expression.

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Step 1.3.3.1

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = e^{- y} \cdot - 1$

Step 1.3.3.2

Move $- 1$ to the left of $e^{- y}$ .

$\frac{\partial M}{\partial y} = - 1 \cdot e^{- y}$

Step 1.3.3.3

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

$\frac{\partial M}{\partial y} = - e^{- y}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - e^{- y} \frac{d}{d x} [x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 - e^{- y} \cdot 1$

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

Subtract $e^{- y}$ from $0$ .

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

Step 3

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

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Step 3.1

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

$- e^{- y} = - e^{- y}$

Step 3.2

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

$- e^{- y} = - e^{- y}$ is an identity.

Step 4

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

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

Step 5

Integrate $M (x, y) = e^{- y}$ to find $f (x, y)$ .

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Step 5.1

Apply the constant rule.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = - y$ .

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Step 8.3.2.1

To apply the Chain Rule, set $u$ as $- y$ .

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

Step 8.3.2.2

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

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

Step 8.3.2.3

Replace all occurrences of $u$ with $- y$ .

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

Step 8.3.3

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

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

Step 8.3.4

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

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

Step 8.3.5

Multiply $- 1$ by $1$ .

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

Step 8.3.6

Move $- 1$ to the left of $e^{- y}$ .

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

Step 8.3.7

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

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

Step 8.5.2

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

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

Step 9

Solve for $\frac{d g}{d y}$ .

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Step 9.1

Move all terms not containing $\frac{d g}{d y}$ to the right side of the equation.

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Step 9.1.1

Add $x e^{- y}$ to both sides of the equation.

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

Step 9.1.2

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

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Step 9.1.2.1

Add $- x e^{- y}$ and $x e^{- y}$ .

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

Step 9.1.2.2

Add $1$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

$g (y) = \int d y$

Step 10.3

Apply the constant rule.

$g (y) = y + C$

Step 11

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

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

Step 12

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

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