Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

Multiply $- 1$ by $- 1$ .

$\frac{\partial M}{\partial y} = 0 + \cos (y) + 1 \sin (y)$

Step 1.4.4

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

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

Step 1.5

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

$\frac{\partial N}{\partial x} = (\sin (y) + \cos (y)) \cdot 1$

Step 2.4

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

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

Step 3

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

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Step 3.1

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

$\cos (y) + \sin (y) = \sin (y) + \cos (y)$

Step 3.2

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

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

Step 4

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

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

Step 5

Integrate $N (x, y) = x (\sin (y) + \cos (y))$ to find $f (x, y)$ .

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Step 5.1

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

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

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Multiply $- \cos (y) + \sin (y)$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Reorder terms.

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

Step 9

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

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Step 9.1

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 9.1.1

Add $\cos (y)$ to both sides of the equation.

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

Step 9.1.2

Subtract $\sin (y)$ from both sides of the equation.

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

Step 9.1.3

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

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Subtract $\sin (y)$ from $\sin (y)$ .

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

Step 9.1.3.4

Add $x$ and $0$ .

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

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

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

Rewrite using the commutative property of multiplication.

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

Step 12.3

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

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