Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $\cos (x)$ is constant with respect to $y$ , the derivative of $\cos (x)$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [\ln (y)]$

Step 1.3

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

$\frac{\partial M}{\partial y} = 0 + \frac{1}{y}$

Step 1.4

Add $0$ and $\frac{1}{y}$ .

$\frac{\partial M}{\partial y} = \frac{1}{y}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $\frac{1}{y}$ by $1$ .

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

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 $0$ .

$\frac{\partial N}{\partial x} = \frac{1}{y} + 0$

Step 2.4.2

Add $\frac{1}{y}$ and $0$ .

$\frac{\partial N}{\partial x} = \frac{1}{y}$

Step 3

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

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Step 3.1

Substitute $\frac{1}{y}$ for $\frac{\partial M}{\partial y}$ and $\frac{1}{y}$ for $\frac{\partial N}{\partial x}$ .

$\frac{1}{y} = \frac{1}{y}$

Step 3.2

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

$\frac{1}{y} = \frac{1}{y}$ is an identity.

Step 4

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

$f (x, y) = \int \cos (x) + \ln (y) d x$

Step 5

Integrate $M (x, y) = \cos (x) + \ln (y)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int \cos (x) d x + \int \ln (y) d x$

Step 5.2

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

$f (x, y) = \sin (x) + C + \int \ln (y) d x$

Step 5.3

Apply the constant rule.

$f (x, y) = \sin (x) + C + \ln (y) x + C$

Step 5.4

Simplify.

$f (x, y) = \sin (x) + \ln (y) x + C$

Step 6

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

$f (x, y) = \sin (x) + \ln (y) x + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

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

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

Step 8.2.2

Since $\sin (x)$ is constant with respect to $y$ , the derivative of $\sin (x)$ with respect to $y$ is $0$ .

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Add $0$ and $\frac{x}{y}$ .

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

Step 8.5.2

Reorder terms.

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

Step 9

Solve for $\frac{d g}{d y}$ .

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Step 9.1

Solve for $\frac{d g}{d y}$ .

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Step 9.1.1

Move all terms containing variables to the left side of the equation.

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Step 9.1.1.1

Subtract $\frac{x}{y}$ from both sides of the equation.

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

Step 9.1.1.2

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

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

Step 9.1.1.3

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

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Step 9.1.1.3.1

Subtract $\frac{x}{y}$ from $\frac{x}{y}$ .

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

Step 9.1.1.3.2

Add $\frac{d g}{d y}$ and $0$ .

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

Step 9.1.2

Add $e^{y}$ to both sides of the equation.

$\frac{d g}{d y} = e^{y}$

Step 10

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

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Step 10.1

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

$\int \frac{d g}{d y} d y = \int e^{y} d y$

Step 10.2

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

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

Step 10.3

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

$g (y) = e^{y} + C$

Step 11

Substitute for $g (y)$ in $f (x, y) = \sin (x) + \ln (y) x + g (y)$ .

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

Step 12

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

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