Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

By the Sum Rule, the derivative of $2 x e^{3 y} + e^{x}$ with respect to $y$ is $\frac{d}{d y} [2 x e^{3 y}] + \frac{d}{d y} [e^{x}]$ .

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

Step 1.3

Evaluate $\frac{d}{d y} [2 x e^{3 y}]$ .

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Step 1.3.1

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

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

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 $g (y) = 3 y$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $3 y$ .

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

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 $a$ = $e$ .

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

Step 1.3.2.3

Replace all occurrences of $u$ with $3 y$ .

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

Multiply $3$ by $1$ .

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

Step 1.3.6

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

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

Step 1.3.7

Multiply $3$ by $2$ .

$\frac{\partial M}{\partial y} = 6 x e^{3 y} + \frac{d}{d y} [e^{x}]$

Step 1.4

Since $e^{x}$ is constant with respect to $y$ , the derivative of $e^{x}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 6 x e^{3 y} + 0$

Step 1.5

Simplify.

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Step 1.5.1

Add $6 x e^{3 y}$ and $0$ .

$\frac{\partial M}{\partial y} = 6 x e^{3 y}$

Step 1.5.2

Reorder the factors of $6 x e^{3 y}$ .

$\frac{\partial M}{\partial y} = 6 e^{3 y} x$

Step 1.5.3

Reorder factors in $6 e^{3 y} x$ .

$\frac{\partial M}{\partial y} = 6 x e^{3 y}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

By the Sum Rule, the derivative of $3 x^{2} e^{3 y} - y^{2}$ with respect to $x$ is $\frac{d}{d x} [3 x^{2} e^{3 y}] + \frac{d}{d x} [- y^{2}]$ .

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

Step 2.3

Evaluate $\frac{d}{d x} [3 x^{2} e^{3 y}]$ .

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 6 e^{3 y} x + \frac{d}{d x} [- y^{2}]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 6 e^{3 y} x + 0$

Step 2.5

Simplify.

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Step 2.5.1

Add $6 e^{3 y} x$ and $0$ .

$\frac{\partial N}{\partial x} = 6 e^{3 y} x$

Step 2.5.2

Reorder factors in $6 e^{3 y} x$ .

$\frac{\partial N}{\partial x} = 6 x e^{3 y}$

Step 3

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

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Step 3.1

Substitute $6 x e^{3 y}$ for $\frac{\partial M}{\partial y}$ and $6 x e^{3 y}$ for $\frac{\partial N}{\partial x}$ .

$6 x e^{3 y} = 6 x e^{3 y}$

Step 3.2

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

$6 x e^{3 y} = 6 x e^{3 y}$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Since $2 e^{3 y}$ is constant with respect to $x$ , move $2 e^{3 y}$ out of the integral.

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

Step 5.3

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

$f (x, y) = 2 e^{3 y} (\frac{1}{2} x^{2} + C) + \int e^{x} d x$

Step 5.4

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

$f (x, y) = 2 e^{3 y} (\frac{1}{2} x^{2} + C) + e^{x} + C$

Step 5.5

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

$f (x, y) = 2 e^{3 y} (\frac{x^{2}}{2} + C) + e^{x} + C$

Step 5.6

Simplify.

$f (x, y) = e^{3 y} x^{2} + e^{x} + C$

Step 6

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

$f (x, y) = e^{3 y} x^{2} + e^{x} + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

By the Sum Rule, the derivative of $e^{3 y} x^{2} + e^{x} + g (y)$ with respect to $y$ is $\frac{d}{d y} [e^{3 y} x^{2}] + \frac{d}{d y} [e^{x}] + \frac{d}{d y} [g (y)]$ .

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

Step 8.3

Evaluate $\frac{d}{d y} [e^{3 y} x^{2}]$ .

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Step 8.3.1

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

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

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 $g (y) = 3 y$ .

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Step 8.3.2.1

To apply the Chain Rule, set $u$ as $3 y$ .

$x^{2} (\frac{d}{d u} [e^{u}] \frac{d}{d y} [3 y]) + \frac{d}{d y} [e^{x}] + \frac{d}{d y} [g (y)] = 3 x^{2} e^{3 y} - y^{2}$

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 $a$ = $e$ .

$x^{2} (e^{u} \frac{d}{d y} [3 y]) + \frac{d}{d y} [e^{x}] + \frac{d}{d y} [g (y)] = 3 x^{2} e^{3 y} - y^{2}$

Step 8.3.2.3

Replace all occurrences of $u$ with $3 y$ .

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

Multiply $3$ by $1$ .

$x^{2} (e^{3 y} \cdot 3) + \frac{d}{d y} [e^{x}] + \frac{d}{d y} [g (y)] = 3 x^{2} e^{3 y} - y^{2}$

Step 8.3.6

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

$x^{2} (3 \cdot e^{3 y}) + \frac{d}{d y} [e^{x}] + \frac{d}{d y} [g (y)] = 3 x^{2} e^{3 y} - y^{2}$

Step 8.3.7

Move $3$ to the left of $x^{2}$ .

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

Step 8.4

Since $e^{x}$ is constant with respect to $y$ , the derivative of $e^{x}$ with respect to $y$ is $0$ .

$3 x^{2} e^{3 y} + 0 + \frac{d}{d y} [g (y)] = 3 x^{2} e^{3 y} - y^{2}$

Step 8.5

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

$3 x^{2} e^{3 y} + 0 + \frac{d g}{d y} = 3 x^{2} e^{3 y} - y^{2}$

Step 8.6

Simplify.

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Step 8.6.1

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

$3 x^{2} e^{3 y} + \frac{d g}{d y} = 3 x^{2} e^{3 y} - y^{2}$

Step 8.6.2

Reorder terms.

$\frac{d g}{d y} + 3 e^{3 y} x^{2} = 3 x^{2} e^{3 y} - y^{2}$

Step 8.6.3

Reorder factors in $\frac{d g}{d y} + 3 e^{3 y} x^{2}$ .

$\frac{d g}{d y} + 3 x^{2} e^{3 y} = 3 x^{2} e^{3 y} - y^{2}$

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 $3 x^{2} e^{3 y}$ from both sides of the equation.

$\frac{d g}{d y} = 3 x^{2} e^{3 y} - y^{2} - 3 x^{2} e^{3 y}$

Step 9.1.2

Combine the opposite terms in $3 x^{2} e^{3 y} - y^{2} - 3 x^{2} e^{3 y}$ .

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Step 9.1.2.1

Subtract $3 x^{2} e^{3 y}$ from $3 x^{2} e^{3 y}$ .

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

Step 9.1.2.2

Add $- y^{2}$ and $0$ .

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

Step 10

Find the antiderivative of $- y^{2}$ to find $g (y)$ .

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Step 10.1

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

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

Step 10.2

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

$g (y) = \int - y^{2} d y$

Step 10.3

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$g (y) = - \int y^{2} d y$

Step 10.4

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

$g (y) = - (\frac{1}{3} y^{3} + C)$

Step 10.5

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

$g (y) = - \frac{1}{3} y^{3} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = e^{3 y} x^{2} + e^{x} + g (y)$ .

$f (x, y) = e^{3 y} x^{2} + e^{x} - \frac{1}{3} y^{3} + C$

Step 12

Simplify $f (x, y) = e^{3 y} x^{2} + e^{x} - \frac{1}{3} y^{3} + C$ .

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Step 12.1

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

$f (x, y) = e^{3 y} x^{2} + e^{x} - \frac{y^{3}}{3} + C$

Step 12.2

Reorder factors in $f (x, y) = e^{3 y} x^{2} + e^{x} - \frac{y^{3}}{3} + C$ .

$f (x, y) = x^{2} e^{3 y} + e^{x} - \frac{y^{3}}{3} + C$