Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

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

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

Step 1.5.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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 $g (x) = 2 x$ .

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Differentiate.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Simplify the expression.

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Step 2.4.3.1

Multiply $2$ by $1$ .

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

Step 2.4.3.2

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

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

Step 2.5

Simplify.

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the answer.

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Step 5.3.1

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

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

Step 5.3.2

Simplify.

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.3

Reorder terms.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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 $g (x) = 2 x$ .

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Step 8.3.4.1

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

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

Step 8.3.4.2

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

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

Step 8.3.4.3

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

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

Step 8.3.5

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

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

Step 8.3.6

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

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

Step 8.3.7

Multiply $2$ by $1$ .

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

Step 8.3.8

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

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

Step 8.3.9

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

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

Step 8.3.10

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

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

Step 8.3.11

Cancel the common factor of $2$ .

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Step 8.3.11.1

Cancel the common factor.

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

Step 8.3.11.2

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

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

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

Step 8.5.2

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

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

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 $e^{2 x} y^{2} - 2 x$ .

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

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

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

Step 9.1.2.2

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

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Step 9.1.2.2.1

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

$\frac{d g}{d x} = - 2 x + 0$

Step 9.1.2.2.2

Add $- 2 x$ and $0$ .

$\frac{d g}{d x} = - 2 x$

Step 10

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

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Step 10.1

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

$\int \frac{d g}{d x} d x = \int - 2 x d x$

Step 10.2

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

$g (x) = \int - 2 x d x$

Step 10.3

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

$g (x) = - 2 \int x d x$

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

Step 10.5.2.2

Cancel the common factor of $- 2$ and $2$ .

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Step 10.5.2.2.1

Factor $2$ out of $- 2$ .

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

Step 10.5.2.2.2

Cancel the common factors.

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Step 10.5.2.2.2.1

Factor $2$ out of $2$ .

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

Step 10.5.2.2.2.2

Cancel the common factor.

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

Step 10.5.2.2.2.3

Rewrite the expression.

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

Step 10.5.2.2.2.4

Divide $- 1$ by $1$ .

$g (x) = - x^{2} + C$

Step 11

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

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

Step 12

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

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Step 12.1

Simplify each term.

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Step 12.1.1

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

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

Step 12.1.2

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

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

Step 12.2

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

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