Calculus Examples

$(2 x y^{3} + y \cos (x)) d x + (3 x^{2} y^{2} + \sin (x)) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y^{3} + y \cos (x)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y^{3} + y \cos (x)]$

Step 1.2

By the Sum Rule, the derivative of $2 x y^{3} + y \cos (x)$ with respect to $y$ is $\frac{d}{d y} [2 x y^{3}] + \frac{d}{d y} [y \cos (x)]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y^{3}] + \frac{d}{d y} [y \cos (x)]$

Step 1.3

Evaluate $\frac{d}{d y} [2 x y^{3}]$ .

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Step 1.3.1

Since $2 x$ is constant with respect to $y$ , the derivative of $2 x y^{3}$ with respect to $y$ is $2 x \frac{d}{d y} [y^{3}]$ .

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y^{3}] + \frac{d}{d y} [y \cos (x)]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

$\frac{\partial M}{\partial y} = 2 x (3 y^{2}) + \frac{d}{d y} [y \cos (x)]$

Step 1.3.3

Multiply $3$ by $2$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} + \frac{d}{d y} [y \cos (x)]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 6 x y^{2} + \cos (x) \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 6 x y^{2} + \cos (x) \cdot 1$

Step 1.4.3

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

$\frac{\partial M}{\partial y} = 6 x y^{2} + \cos (x)$

Step 1.5

Reorder terms.

$\frac{\partial M}{\partial y} = 6 y^{2} x + \cos (x)$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 x^{2} y^{2} + \sin (x)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2} y^{2} + \sin (x)]$

Step 2.2

By the Sum Rule, the derivative of $3 x^{2} y^{2} + \sin (x)$ with respect to $x$ is $\frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [\sin (x)]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [\sin (x)]$

Step 2.3

Evaluate $\frac{d}{d x} [3 x^{2} y^{2}]$ .

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Step 2.3.1

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 x^{2} y^{2}$ with respect to $x$ is $3 y^{2} \frac{d}{d x} [x^{2}]$ .

$\frac{\partial N}{\partial x} = 3 y^{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [\sin (x)]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 3 y^{2} (2 x) + \frac{d}{d x} [\sin (x)]$

Step 2.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x + \frac{d}{d x} [\sin (x)]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 6 y^{2} x + \cos (x)$

Step 3

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

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Step 3.1

Substitute $6 y^{2} x + \cos (x)$ for $\frac{\partial M}{\partial y}$ and $6 y^{2} x + \cos (x)$ for $\frac{\partial N}{\partial x}$ .

$6 y^{2} x + \cos (x) = 6 y^{2} x + \cos (x)$

Step 3.2

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

$6 y^{2} x + \cos (x) = 6 y^{2} x + \cos (x)$ is an identity.

Step 4

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

$f (x, y) = \int 3 x^{2} y^{2} + \sin (x) d y$

Step 5

Integrate $N (x, y) = 3 x^{2} y^{2} + \sin (x)$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int 3 x^{2} y^{2} d y + \int \sin (x) d y$

Step 5.2

Since $3 x^{2}$ is constant with respect to $y$ , move $3 x^{2}$ out of the integral.

$f (x, y) = 3 x^{2} \int y^{2} d y + \int \sin (x) d y$

Step 5.3

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

$f (x, y) = 3 x^{2} (\frac{1}{3} y^{3} + C) + \int \sin (x) d y$

Step 5.4

Apply the constant rule.

$f (x, y) = 3 x^{2} (\frac{1}{3} y^{3} + C) + \sin (x) y + C$

Step 5.5

Combine $\frac{1}{3}$ and $y^{3}$ .

$f (x, y) = 3 x^{2} (\frac{y^{3}}{3} + C) + \sin (x) y + C$

Step 5.6

Simplify.

$f (x, y) = x^{2} y^{3} + \sin (x) y + C$

Step 6

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

$f (x, y) = x^{2} y^{3} + \sin (x) y + g (x)$

Step 7

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

$\frac{\partial f}{\partial x} = 2 x y^{3} + y \cos (x)$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x^{2} y^{3} + \sin (x) y + g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.2

By the Sum Rule, the derivative of $x^{2} y^{3} + \sin (x) y + g (x)$ with respect to $x$ is $\frac{d}{d x} [x^{2} y^{3}] + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [x^{2} y^{3}] + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.3

Evaluate $\frac{d}{d x} [x^{2} y^{3}]$ .

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Step 8.3.1

Since $y^{3}$ is constant with respect to $x$ , the derivative of $x^{2} y^{3}$ with respect to $x$ is $y^{3} \frac{d}{d x} [x^{2}]$ .

$y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.3.2

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

$y^{3} (2 x) + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.3.3

Move $2$ to the left of $y^{3}$ .

$2 y^{3} x + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.4

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

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Step 8.4.1

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

$2 y^{3} x + y \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.4.2

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

$2 y^{3} x + y \cos (x) + \frac{d}{d x} [g (x)] = 2 x y^{3} + y \cos (x)$

Step 8.5

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

$2 y^{3} x + y \cos (x) + \frac{d g}{d x} = 2 x y^{3} + y \cos (x)$

Step 8.6

Reorder terms.

$\frac{d g}{d x} + 2 y^{3} x + y \cos (x) = 2 x y^{3} + y \cos (x)$

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 $2 y^{3} x$ from both sides of the equation.

$\frac{d g}{d x} + y \cos (x) = 2 x y^{3} + y \cos (x) - 2 y^{3} x$

Step 9.1.2

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

$\frac{d g}{d x} = 2 x y^{3} + y \cos (x) - 2 y^{3} x - y \cos (x)$

Step 9.1.3

Combine the opposite terms in $2 x y^{3} + y \cos (x) - 2 y^{3} x - y \cos (x)$ .

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Step 9.1.3.1

Reorder the factors in the terms $2 x y^{3}$ and $- 2 y^{3} x$ .

$\frac{d g}{d x} = 2 y^{3} x + y \cos (x) - 2 y^{3} x - y \cos (x)$

Step 9.1.3.2

Subtract $2 y^{3} x$ from $2 y^{3} x$ .

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

Step 9.1.3.3

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

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

Step 9.1.3.4

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

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

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 $f (x, y) = x^{2} y^{3} + \sin (x) y + g (x)$ .

$f (x, y) = x^{2} y^{3} + \sin (x) y + C$

Step 12

Reorder factors in $f (x, y) = x^{2} y^{3} + \sin (x) y + C$ .

$f (x, y) = x^{2} y^{3} + y \sin (x) + C$