Calculus Examples

$(2 x - y) d x - (x + 4 y) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x - y]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x] + \frac{d}{d y} [- y]$

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 - \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 - 1 \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

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

Step 1.4

Subtract $1$ from $0$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - (x + 4 y)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x + 4 y)]$

Step 2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x + 4 y)$ with respect to $x$ is $- \frac{d}{d x} [x + 4 y]$ .

$\frac{\partial N}{\partial x} = - \frac{d}{d x} [x + 4 y]$

Step 2.3

By the Sum Rule, the derivative of $x + 4 y$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4 y]$ .

$\frac{\partial N}{\partial x} = - (\frac{d}{d x} [x] + \frac{d}{d x} [4 y])$

Step 2.4

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

$\frac{\partial N}{\partial x} = - (1 + \frac{d}{d x} [4 y])$

Step 2.5

Since $4 y$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = - (1 + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = - 1 \cdot 1$

Step 2.6.2

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

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

$- 1 = - 1$

Step 3.2

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

$- 1 = - 1$ is an identity.

Step 4

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

$f (x, y) = \int - (x + 4 y) d y$

Step 5

Integrate $N (x, y) = - (x + 4 y)$ to find $f (x, y)$ .

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Step 5.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$f (x, y) = - \int x + 4 y d y$

Step 5.2

Split the single integral into multiple integrals.

$f (x, y) = - (\int x d y + \int 4 y d y)$

Step 5.3

Apply the constant rule.

$f (x, y) = - (x y + C + \int 4 y d y)$

Step 5.4

Since $4$ is constant with respect to $y$ , move $4$ out of the integral.

$f (x, y) = - (x y + C + 4 \int y d y)$

Step 5.5

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

$f (x, y) = - (x y + C + 4 (\frac{1}{2} y^{2} + C))$

Step 5.6

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

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

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

Step 8.3

Evaluate $\frac{d}{d x} [- (x y + 2 y^{2})]$ .

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Step 8.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x y + 2 y^{2})$ with respect to $x$ is $- \frac{d}{d x} [x y + 2 y^{2}]$ .

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 8.3.4

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

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

Step 8.3.5

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

$- (y \cdot 1 + 0) + \frac{d}{d x} [g (x)] = 2 x - y$

Step 8.3.6

Multiply $y$ by $1$ .

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

Step 8.3.7

Add $y$ and $0$ .

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

Step 8.4

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

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

Step 8.5

Reorder terms.

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

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 $y$ to both sides of the equation.

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

Step 9.1.2

Combine the opposite terms in $2 x - y + y$ .

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Step 9.1.2.1

Add $- y$ and $y$ .

$\frac{d g}{d x} = 2 x + 0$

Step 9.1.2.2

Add $2 x$ and $0$ .

$\frac{d g}{d x} = 2 x$

Step 10

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

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Step 10.1

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

$\int \frac{d g}{d x} d x = \int 2 x d x$

Step 10.2

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

$g (x) = \int 2 x d x$

Step 10.3

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

$g (x) = 2 \int x d x$

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

Step 10.5.2.2

Cancel the common factor of $2$ .

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Step 10.5.2.2.1

Cancel the common factor.

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

Step 10.5.2.2.2

Rewrite the expression.

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

Step 10.5.2.3

Multiply $x^{2}$ by $1$ .

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

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

Multiply $2$ by $- 1$ .

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