Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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

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

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 $g (y) = 2 x y$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $2 x y$ .

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

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 $a$ = $e$ .

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

Step 1.3.2.3

Replace all occurrences of $u$ with $2 x y$ .

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

$\frac{\partial M}{\partial y} = 0 + y (e^{2 x y} (2 x \cdot 1)) + e^{2 x y} \cdot 1$

Step 1.3.6

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 + y (e^{2 x y} (2 x)) + e^{2 x y} \cdot 1$

Step 1.3.7

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Reorder terms.

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

Step 1.4.3

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Multiply $x$ by $1$ .

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

Step 2.3

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

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

Step 2.4

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 $g (x) = 2 x y$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $2 x y$ .

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

Step 2.4.2

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

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

Step 2.4.3

Replace all occurrences of $u$ with $2 x y$ .

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

Step 2.5

Differentiate.

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $2$ by $1$ .

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

Step 2.5.4

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

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

Step 2.5.5

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

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

Step 2.6

Simplify.

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Step 2.6.1

Reorder terms.

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

Step 2.6.2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

Multiply $x$ by $1$ .

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

Step 5.2

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

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

Step 5.3

Let $u = 2 x y$ . Then $d u = 2 x d y$ , so $\frac{1}{2 x} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.3.1

Let $u = 2 x y$ . Find $\frac{d u}{d y}$ .

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Step 5.3.1.1

Differentiate $2 x y$ .

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

Step 5.3.1.2

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

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

Step 5.3.1.3

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

$2 x \cdot 1$

Step 5.3.1.4

Multiply $2$ by $1$ .

$2 x$

Step 5.3.2

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

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

Step 5.4

Combine $e^{u}$ and $\frac{1}{2 x}$ .

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

Step 5.5

Since $\frac{1}{2 x}$ is constant with respect to $u$ , move $\frac{1}{2 x}$ out of the integral.

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

Step 5.6

Simplify.

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Step 5.6.1

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

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

Step 5.6.2

Cancel the common factor of $x$ .

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Step 5.6.2.1

Cancel the common factor.

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

Step 5.6.2.2

Rewrite the expression.

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

Step 5.7

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

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

Step 5.8

Simplify.

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

Step 5.9

Replace all occurrences of $u$ with $2 x y$ .

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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 $g (x) = 2 x y$ .

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Step 8.3.2.1

To apply the Chain Rule, set $u$ as $2 x y$ .

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

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 $a$ = $e$ .

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

Step 8.3.2.3

Replace all occurrences of $u$ with $2 x y$ .

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

Multiply $2$ by $1$ .

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

Step 8.3.6

Combine $e^{2 x y}$ and $\frac{1}{2}$ .

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

Step 8.3.7

Combine $2$ and $\frac{e^{2 x y}}{2}$ .

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

Step 8.3.8

Combine $\frac{2 e^{2 x y}}{2}$ and $y$ .

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

Step 8.3.9

Cancel the common factor of $2$ .

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Step 8.3.9.1

Cancel the common factor.

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

Step 8.3.9.2

Divide $e^{2 x y} y$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

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

Step 8.5.2

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

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

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

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

Step 9.1.2

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

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Step 9.1.2.1

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

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

Step 9.1.2.2

Add $x$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

$g (x) = \int x d x$

Step 10.3

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

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

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Combine $\frac{1}{2}$ and $e^{2 x y}$ .

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

Step 12.2

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

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