Calculus Examples

$(1 + \sin (y)) d x + (x \cos (y) - 2 y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 1 + \sin (y)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

The derivative of $\sin (y)$ with respect to $y$ is $\cos (y)$ .

$\frac{\partial M}{\partial y} = 0 + \cos (y)$

Step 1.4

Add $0$ and $\cos (y)$ .

$\frac{\partial M}{\partial y} = \cos (y)$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$\frac{\partial N}{\partial x} = \cos (y) \cdot 1 + \frac{d}{d x} [- 2 y]$

Step 2.3.3

Multiply $\cos (y)$ by $1$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = \cos (y) + 0$

Step 2.4.2

Add $\cos (y)$ and $0$ .

$\frac{\partial N}{\partial x} = \cos (y)$

Step 3

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

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Step 3.1

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

$\cos (y) = \cos (y)$

Step 3.2

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

$\cos (y) = \cos (y)$ is an identity.

Step 4

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

$f (x, y) = \int 1 + \sin (y) d x$

Step 5

Integrate $M (x, y) = 1 + \sin (y)$ to find $f (x, y)$ .

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Step 5.1

Apply the constant rule.

$f (x, y) = (1 + \sin (y)) x + C$

Step 6

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

$f (x, y) = (1 + \sin (y)) x + g (y)$

Step 7

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

$\frac{\partial f}{\partial y} = x \cos (y) - 2 y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d y} [(1 + \sin (y)) x]$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

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

$x (0 + \frac{d}{d y} [\sin (y)]) + \frac{d}{d y} [g (y)] = x \cos (y) - 2 y$

Step 8.3.4

The derivative of $\sin (y)$ with respect to $y$ is $\cos (y)$ .

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

Step 8.3.5

Add $0$ and $\cos (y)$ .

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

Step 8.4

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

$x \cos (y) + \frac{d g}{d y} = x \cos (y) - 2 y$

Step 8.5

Reorder terms.

$\frac{d g}{d y} + x \cos (y) = x \cos (y) - 2 y$

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 $x \cos (y)$ from both sides of the equation.

$\frac{d g}{d y} = x \cos (y) - 2 y - x \cos (y)$

Step 9.1.2

Combine the opposite terms in $x \cos (y) - 2 y - x \cos (y)$ .

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Step 9.1.2.1

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

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

Step 9.1.2.2

Add $- 2 y$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

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

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

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 $- 2$ .

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

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 $2$ .

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

Step 10.5.2.2.2.2

Cancel the common factor.

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

Step 10.5.2.2.2.3

Rewrite the expression.

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

Step 10.5.2.2.2.4

Divide $- 1$ by $1$ .

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

Step 11

Substitute for $g (y)$ in $f (x, y) = (1 + \sin (y)) x + g (y)$ .

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

Step 12

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

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Step 12.1

Simplify each term.

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Step 12.1.1

Apply the distributive property.

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

Step 12.1.2

Multiply $x$ by $1$ .

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

Step 12.2

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

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