Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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 $g (y) = x y$ .

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $x$ by $1$ .

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

Step 1.3.7

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Reorder terms.

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

Step 1.4.3

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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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 $g (x) = e^{x y}$ .

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

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 $g (x) = x y$ .

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Step 2.3.2.1

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

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

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 $a$ = $e$ .

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

$\frac{\partial N}{\partial x} = 0 + x (e^{x y} (y \cdot 1)) + e^{x y} \cdot 1$

Step 2.3.6

Multiply $y$ by $1$ .

$\frac{\partial N}{\partial x} = 0 + x (e^{x y} y) + e^{x y} \cdot 1$

Step 2.3.7

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

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

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

Reorder terms.

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

Step 2.4.3

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

Since $y$ is constant with respect to $x$ , move $y$ out of the integral.

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

Step 5.4

Let $u = x y$ . Then $d u = y d x$ , so $\frac{1}{y} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.4.1

Let $u = x y$ . Find $\frac{d u}{d x}$ .

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Step 5.4.1.1

Differentiate $x y$ .

$\frac{d}{d x} [x y]$

Step 5.4.1.2

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

$y \frac{d}{d x} [x]$

Step 5.4.1.3

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

$y \cdot 1$

Step 5.4.1.4

Multiply $y$ by $1$ .

$y$

Step 5.4.2

Rewrite the problem using $u$ and $d u$ .

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

Step 5.5

Combine $e^{u}$ and $\frac{1}{y}$ .

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

Step 5.6

Since $\frac{1}{y}$ is constant with respect to $u$ , move $\frac{1}{y}$ out of the integral.

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

Step 5.7

Simplify.

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Step 5.7.1

Combine $\frac{1}{y}$ and $y$ .

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

Step 5.7.2

Cancel the common factor of $y$ .

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Step 5.7.2.1

Cancel the common factor.

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

Step 5.7.2.2

Rewrite the expression.

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

Step 5.7.3

Multiply $\int e^{u} d u$ by $1$ .

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

Step 5.8

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

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

Step 5.9

Simplify.

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

Step 5.10

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

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

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

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

Step 8.2.2

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

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

Step 8.3

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

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Step 8.3.1

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 $g (y) = x y$ .

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Step 8.3.1.1

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

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

Step 8.3.1.2

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

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

Step 8.3.1.3

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

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

Step 8.3.2

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

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

Step 8.3.3

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

$0 + e^{x y} (x \cdot 1) + \frac{d}{d y} [g (y)] = 2 y + x e^{x y}$

Step 8.3.4

Multiply $x$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

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

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

Step 8.5.2

Reorder terms.

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

Step 8.5.3

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

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

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

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

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

Step 9.1.2

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

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Step 9.1.2.1

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

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

Step 9.1.2.2

Add $2 y$ and $0$ .

$\frac{d g}{d y} = 2 y$

Step 10

Find the antiderivative of $2 y$ to find $g (y)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d y} = 2 y$ .

$\int \frac{d g}{d y} d y = \int 2 y d y$

Step 10.2

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

$g (y) = \int 2 y d y$

Step 10.3

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$g (y) = 2 \int y d y$

Step 10.4

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

$g (y) = 2 (\frac{1}{2} y^{2} + C)$

Step 10.5

Simplify the answer.

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Step 10.5.1

Rewrite $2 (\frac{1}{2} y^{2} + C)$ as $2 (\frac{1}{2}) y^{2} + C$ .

$g (y) = 2 (\frac{1}{2}) y^{2} + C$

Step 10.5.2

Simplify.

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Step 10.5.2.1

Combine $2$ and $\frac{1}{2}$ .

$g (y) = \frac{2}{2} y^{2} + C$

Step 10.5.2.2

Cancel the common factor of $2$ .

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Step 10.5.2.2.1

Cancel the common factor.

$g (y) = \frac{2}{2} y^{2} + C$

Step 10.5.2.2.2

Rewrite the expression.

$g (y) = 1 y^{2} + C$

Step 10.5.2.3

Multiply $y^{2}$ by $1$ .

$g (y) = y^{2} + C$

Step 11

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

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