Calculus Examples

$(2 x y + \sin (x)) d x + (x^{2} + \cos (y)) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $1$ .

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 2 x + 0$

Step 1.4.2

Add $2 x$ and $0$ .

$\frac{\partial M}{\partial y} = 2 x$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

$\frac{\partial N}{\partial x} = 2 x + 0$

Step 2.5

Add $2 x$ and $0$ .

$\frac{\partial N}{\partial x} = 2 x$

Step 3

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

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Step 3.1

Substitute $2 x$ for $\frac{\partial M}{\partial y}$ and $2 x$ for $\frac{\partial N}{\partial x}$ .

$2 x = 2 x$

Step 3.2

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

$2 x = 2 x$ is an identity.

Step 4

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

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

Step 5

Integrate $N (x, y) = x^{2} + \cos (y)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

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

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

Step 5.4

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 2 x y + \sin (x)$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Move $2$ to the left of $y$ .

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

Simplify.

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Step 8.6.1

Add $2 y x$ and $0$ .

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

Step 8.6.2

Reorder terms.

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

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

Subtract $2 x y$ from both sides of the equation.

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

Step 9.1.2

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

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Step 9.1.2.1

Subtract $2 x y$ from $2 x y$ .

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

Step 9.1.2.2

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

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

Step 10

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

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Step 10.1

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

$\int \frac{d g}{d x} d x = \int \sin (x) d x$

Step 10.2

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

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

Step 10.3

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

$g (x) = - \cos (x) + C$

Step 11

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

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