Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [y \ln (x)]$ .

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $\ln (x)$ by $1$ .

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

Step 1.4

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

$\frac{\partial M}{\partial y} = \ln (x) + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

Evaluate $\frac{d}{d x} [x \ln (x)]$ .

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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 $g (x) = \ln (x)$ .

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

Step 2.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Cancel the common factor of $x$ .

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Step 2.3.5.1

Cancel the common factor.

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

Step 2.3.5.2

Rewrite the expression.

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

Step 2.3.6

Multiply $\ln (x)$ by $1$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 1 + \ln (x) + 0$

Step 2.4.2

Add $1 + \ln (x)$ and $0$ .

$\frac{\partial N}{\partial x} = 1 + \ln (x)$

Step 3

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

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Step 3.1

Substitute $\ln (x) + 1$ for $\frac{\partial M}{\partial y}$ and $1 + \ln (x)$ for $\frac{\partial N}{\partial x}$ .

$\ln (x) + 1 = 1 + \ln (x)$

Step 3.2

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

$\ln (x) + 1 = 1 + \ln (x)$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = y \ln (x) + y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d x} [x \ln (x) y]$ .

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Step 8.3.1

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

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

Step 8.3.2

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 $g (x) = \ln (x)$ .

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

Step 8.3.3

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 8.3.4

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

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

Step 8.3.5

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

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

Step 8.3.6

Cancel the common factor of $x$ .

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Step 8.3.6.1

Cancel the common factor.

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

Step 8.3.6.2

Rewrite the expression.

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

Step 8.3.7

Multiply $\ln (x)$ by $1$ .

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

Step 8.4

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

$y (1 + \ln (x)) + 0 + \frac{d}{d x} [g (x)] = y \ln (x) + y$

Step 8.5

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

$y (1 + \ln (x)) + 0 + \frac{d g}{d x} = y \ln (x) + y$

Step 8.6

Simplify.

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Step 8.6.1

Apply the distributive property.

$y \cdot 1 + y \ln (x) + 0 + \frac{d g}{d x} = y \ln (x) + y$

Step 8.6.2

Combine terms.

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Step 8.6.2.1

Multiply $y$ by $1$ .

$y + y \ln (x) + 0 + \frac{d g}{d x} = y \ln (x) + y$

Step 8.6.2.2

Add $y + y \ln (x)$ and $0$ .

$y + y \ln (x) + \frac{d g}{d x} = y \ln (x) + y$

Step 8.6.3

Reorder terms.

$\frac{d g}{d x} + y \ln (x) + y = y \ln (x) + y$

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

Move all the terms containing a logarithm to the left side of the equation.

$y \ln (x) - y \ln (x) = - \frac{d g}{d x} - y + y$

Step 9.1.2

Subtract $y \ln (x)$ from $y \ln (x)$ .

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

Step 9.1.3

Combine the opposite terms in $- \frac{d g}{d x} - y + y$ .

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Step 9.1.3.1

Add $- y$ and $y$ .

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

Step 9.1.3.2

Add $- \frac{d g}{d x}$ and $0$ .

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

Step 9.1.4

Since $\frac{d g}{d x}$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 9.1.5

Divide each term in $- \frac{d g}{d x} = 0$ by $- 1$ and simplify.

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Step 9.1.5.1

Divide each term in $- \frac{d g}{d x} = 0$ by $- 1$ .

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

Step 9.1.5.2

Simplify the left side.

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Step 9.1.5.2.1

Dividing two negative values results in a positive value.

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

Step 9.1.5.2.2

Divide $\frac{d g}{d x}$ by $1$ .

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

Step 9.1.5.3

Simplify the right side.

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Step 9.1.5.3.1

Divide $0$ by $- 1$ .

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

Step 10

Find the antiderivative of $0$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 0$ .

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

Step 10.2

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

$g (x) = \int 0 d x$

Step 10.3

The integral of $0$ with respect to $x$ is $0$ .

$g (x) = 0 + C$

Step 10.4

Add $0$ and $C$ .

$g (x) = C$

Step 11

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

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

Step 12

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

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