Calculus Examples

$3 x^{2} e^{y} d x + (x^{3} e^{y} - 1) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 3 x^{2} e^{y}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x^{2} e^{y}]$

Step 1.2

Since $3 x^{2}$ is constant with respect to $y$ , the derivative of $3 x^{2} e^{y}$ with respect to $y$ is $3 x^{2} \frac{d}{d y} [e^{y}]$ .

$\frac{\partial M}{\partial y} = 3 x^{2} \frac{d}{d y} [e^{y}]$

Step 1.3

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

$\frac{\partial M}{\partial y} = 3 x^{2} e^{y}$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x^{3} e^{y} - 1$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{3} e^{y} - 1]$

Step 2.2

By the Sum Rule, the derivative of $x^{3} e^{y} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{3} e^{y}] + \frac{d}{d x} [- 1]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{3} e^{y}] + \frac{d}{d x} [- 1]$

Step 2.3

Evaluate $\frac{d}{d x} [x^{3} e^{y}]$ .

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Step 2.3.1

Since $e^{y}$ is constant with respect to $x$ , the derivative of $x^{3} e^{y}$ with respect to $x$ is $e^{y} \frac{d}{d x} [x^{3}]$ .

$\frac{\partial N}{\partial x} = e^{y} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 1]$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$\frac{\partial N}{\partial x} = e^{y} (3 x^{2}) + \frac{d}{d x} [- 1]$

Step 2.3.3

Move $3$ to the left of $e^{y}$ .

$\frac{\partial N}{\partial x} = 3 e^{y} x^{2} + \frac{d}{d x} [- 1]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 3 e^{y} x^{2} + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $3 e^{y} x^{2}$ and $0$ .

$\frac{\partial N}{\partial x} = 3 e^{y} x^{2}$

Step 2.5.2

Reorder factors in $3 e^{y} x^{2}$ .

$\frac{\partial N}{\partial x} = 3 x^{2} e^{y}$

Step 3

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

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Step 3.1

Substitute $3 x^{2} e^{y}$ for $\frac{\partial M}{\partial y}$ and $3 x^{2} e^{y}$ for $\frac{\partial N}{\partial x}$ .

$3 x^{2} e^{y} = 3 x^{2} e^{y}$

Step 3.2

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

$3 x^{2} e^{y} = 3 x^{2} e^{y}$ is an identity.

Step 4

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

$f (x, y) = \int 3 x^{2} e^{y} d x$

Step 5

Integrate $M (x, y) = 3 x^{2} e^{y}$ to find $f (x, y)$ .

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Step 5.1

Since $3 e^{y}$ is constant with respect to $x$ , move $3 e^{y}$ out of the integral.

$f (x, y) = 3 e^{y} \int x^{2} d x$

Step 5.2

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$f (x, y) = 3 e^{y} (\frac{1}{3} x^{3} + C)$

Step 5.3

Simplify the answer.

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Step 5.3.1

Rewrite $3 e^{y} (\frac{1}{3} x^{3} + C)$ as $3 e^{y} \frac{1}{3} x^{3} + C$ .

$f (x, y) = 3 e^{y} \frac{1}{3} x^{3} + C$

Step 5.3.2

Simplify.

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Step 5.3.2.1

Combine $\frac{1}{3}$ and $3$ .

$f (x, y) = \frac{3}{3} e^{y} x^{3} + C$

Step 5.3.2.2

Combine $\frac{3}{3}$ and $e^{y}$ .

$f (x, y) = \frac{3 e^{y}}{3} x^{3} + C$

Step 5.3.2.3

Cancel the common factor of $3$ .

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Step 5.3.2.3.1

Cancel the common factor.

$f (x, y) = \frac{3 e^{y}}{3} x^{3} + C$

Step 5.3.2.3.2

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

$f (x, y) = e^{y} x^{3} + C$

Step 6

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

$f (x, y) = e^{y} x^{3} + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = x^{3} e^{y} - 1$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [e^{y} x^{3} + g (y)] = x^{3} e^{y} - 1$

Step 8.2

By the Sum Rule, the derivative of $e^{y} x^{3} + g (y)$ with respect to $y$ is $\frac{d}{d y} [e^{y} x^{3}] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [e^{y} x^{3}] + \frac{d}{d y} [g (y)] = x^{3} e^{y} - 1$

Step 8.3

Evaluate $\frac{d}{d y} [e^{y} x^{3}]$ .

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Step 8.3.1

Since $x^{3}$ is constant with respect to $y$ , the derivative of $e^{y} x^{3}$ with respect to $y$ is $x^{3} \frac{d}{d y} [e^{y}]$ .

$x^{3} \frac{d}{d y} [e^{y}] + \frac{d}{d y} [g (y)] = x^{3} e^{y} - 1$

Step 8.3.2

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

$x^{3} e^{y} + \frac{d}{d y} [g (y)] = x^{3} e^{y} - 1$

Step 8.4

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

$x^{3} e^{y} + \frac{d g}{d y} = x^{3} e^{y} - 1$

Step 8.5

Simplify.

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Step 8.5.1

Reorder terms.

$\frac{d g}{d y} + e^{y} x^{3} = x^{3} e^{y} - 1$

Step 8.5.2

Reorder factors in $\frac{d g}{d y} + e^{y} x^{3}$ .

$\frac{d g}{d y} + x^{3} e^{y} = x^{3} e^{y} - 1$

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 $x^{3} e^{y}$ from both sides of the equation.

$\frac{d g}{d y} = x^{3} e^{y} - 1 - x^{3} e^{y}$

Step 9.1.2

Combine the opposite terms in $x^{3} e^{y} - 1 - x^{3} e^{y}$ .

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Step 9.1.2.1

Subtract $x^{3} e^{y}$ from $x^{3} e^{y}$ .

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

Step 9.1.2.2

Subtract $1$ from $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

$g (y) = \int - 1 d y$

Step 10.3

Apply the constant rule.

$g (y) = - y + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = e^{y} x^{3} + g (y)$ .

$f (x, y) = e^{y} x^{3} - y + C$

Step 12

Reorder factors in $f (x, y) = e^{y} x^{3} - y + C$ .

$f (x, y) = x^{3} e^{y} - y + C$