Calculus Examples

$(9 x^{2} + y - 1) d x - (4 y - x) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 9 x^{2} + y - 1$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [9 x^{2} + y - 1]$

Step 1.2

By the Sum Rule, the derivative of $9 x^{2} + y - 1$ with respect to $y$ is $\frac{d}{d y} [9 x^{2}] + \frac{d}{d y} [y] + \frac{d}{d y} [- 1]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [9 x^{2}] + \frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 1.3

Since $9 x^{2}$ is constant with respect to $y$ , the derivative of $9 x^{2}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 1.4

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

$\frac{\partial M}{\partial y} = 0 + 1 + \frac{d}{d y} [- 1]$

Step 1.5

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

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

Step 1.6

Combine terms.

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Step 1.6.1

Add $0$ and $1$ .

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

Step 1.6.2

Add $1$ and $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

$\frac{\partial N}{\partial x} = - (0 + \frac{d}{d x} [- x])$

Step 2.5

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 2.6

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

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

Step 2.7

Multiply.

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Step 2.7.1

Multiply $- 1$ by $- 1$ .

$\frac{\partial N}{\partial x} = 1 \frac{d}{d x} [x]$

Step 2.7.2

Multiply $\frac{d}{d x} [x]$ by $1$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x]$

Step 2.8

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

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

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 $N (x, y)$ .

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

Step 5

Integrate $N (x, y) = - (4 y - x)$ 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.

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

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = 9 x^{2} + y - 1$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [- (2 y^{2} - x y) + g (x)] = 9 x^{2} + y - 1$

Step 8.2

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

$\frac{d}{d x} [- (2 y^{2} - x y)] + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3

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

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Step 8.3.1

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

$- \frac{d}{d x} [2 y^{2} - x y] + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.2

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

$- (\frac{d}{d x} [2 y^{2}] + \frac{d}{d x} [- x y]) + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.3

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

$- (0 + \frac{d}{d x} [- x y]) + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.4

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

$- (0 - y \frac{d}{d x} [x]) + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.5

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

$- (0 - y \cdot 1) + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.6

Multiply $- 1$ by $1$ .

$- (0 - y) + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.7

Subtract $y$ from $0$ .

$- - y + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.8

Multiply $- 1$ by $- 1$ .

$1 y + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.3.9

Multiply $y$ by $1$ .

$y + \frac{d}{d x} [g (x)] = 9 x^{2} + y - 1$

Step 8.4

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

$y + \frac{d g}{d x} = 9 x^{2} + y - 1$

Step 8.5

Reorder terms.

$\frac{d g}{d x} + y = 9 x^{2} + y - 1$

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 $y$ from both sides of the equation.

$\frac{d g}{d x} = 9 x^{2} + y - 1 - y$

Step 9.1.2

Combine the opposite terms in $9 x^{2} + y - 1 - y$ .

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Step 9.1.2.1

Subtract $y$ from $y$ .

$\frac{d g}{d x} = 9 x^{2} + 0 - 1$

Step 9.1.2.2

Add $9 x^{2}$ and $0$ .

$\frac{d g}{d x} = 9 x^{2} - 1$

Step 10

Find the antiderivative of $9 x^{2} - 1$ to find $g (x)$ .

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = 9 x^{2} - 1$ .

$\int \frac{d g}{d x} d x = \int 9 x^{2} - 1 d x$

Step 10.2

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

$g (x) = \int 9 x^{2} - 1 d x$

Step 10.3

Split the single integral into multiple integrals.

$g (x) = \int 9 x^{2} d x + \int - 1 d x$

Step 10.4

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$g (x) = 9 \int x^{2} d x + \int - 1 d x$

Step 10.5

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

$g (x) = 9 (\frac{1}{3} x^{3} + C) + \int - 1 d x$

Step 10.6

Apply the constant rule.

$g (x) = 9 (\frac{1}{3} x^{3} + C) - x + C$

Step 10.7

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

$g (x) = 9 (\frac{x^{3}}{3} + C) - x + C$

Step 10.8

Simplify.

$g (x) = 3 x^{3} - x + C$

Step 11

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

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

Step 12

Simplify each term.

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Step 12.1

Apply the distributive property.

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

Step 12.2

Multiply $2$ by $- 1$ .

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

Step 12.3

Multiply $- (- x y)$ .

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Step 12.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.3.2

Multiply $x$ by $1$ .

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