Calculus Examples

$y (2 x y^{2} - 3) d x + (3 x^{2} y^{2} - 3 x + 4 y) d y = 0$

Step 1

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

Tap for more steps...

Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = 2 x y^{2} - 3$ .

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

Step 1.3

Differentiate.

Tap for more steps...

Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = y (4 x y + \frac{d}{d y} [- 3]) + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.3.5

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = y (4 x y + 0) + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.3.6

Add $4 x y$ and $0$ .

$\frac{\partial M}{\partial y} = y (4 x y) + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.4

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = 4 x (y^{1} y) + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.5

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = 4 x (y^{1} y^{1}) + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\partial M}{\partial y} = 4 x y^{1 + 1} + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.7

Add $1$ and $1$ .

$\frac{\partial M}{\partial y} = 4 x y^{2} + (2 x y^{2} - 3) \frac{d}{d y} [y]$

Step 1.8

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

$\frac{\partial M}{\partial y} = 4 x y^{2} + (2 x y^{2} - 3) \cdot 1$

Step 1.9

Simplify by adding terms.

Tap for more steps...

Step 1.9.1

Multiply $2 x y^{2} - 3$ by $1$ .

$\frac{\partial M}{\partial y} = 4 x y^{2} + 2 x y^{2} - 3$

Step 1.9.2

Add $4 x y^{2}$ and $2 x y^{2}$ .

$\frac{\partial M}{\partial y} = 6 x y^{2} - 3$

Step 1.9.3

Reorder terms.

$\frac{\partial M}{\partial y} = 6 y^{2} x - 3$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 x^{2} y^{2} - 3 x + 4 y$ .

Tap for more steps...

Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2} y^{2} - 3 x + 4 y]$

Step 2.2

By the Sum Rule, the derivative of $3 x^{2} y^{2} - 3 x + 4 y$ with respect to $x$ is $\frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4 y]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4 y]$

Step 2.3

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

Tap for more steps...

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

$\frac{\partial N}{\partial x} = 3 y^{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4 y]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 3 y^{2} (2 x) + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4 y]$

Step 2.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4 y]$

Step 2.4

Evaluate $\frac{d}{d x} [- 3 x]$ .

Tap for more steps...

Step 2.4.1

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

$\frac{\partial N}{\partial x} = 6 y^{2} x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [4 y]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 6 y^{2} x - 3 \cdot 1 + \frac{d}{d x} [4 y]$

Step 2.4.3

Multiply $- 3$ by $1$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x - 3 + \frac{d}{d x} [4 y]$

Step 2.5

Differentiate using the Constant Rule.

Tap for more steps...

Step 2.5.1

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

$\frac{\partial N}{\partial x} = 6 y^{2} x - 3 + 0$

Step 2.5.2

Add $6 y^{2} x - 3$ and $0$ .

$\frac{\partial N}{\partial x} = 6 y^{2} x - 3$

Step 3

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

Tap for more steps...

Step 3.1

Substitute $6 y^{2} x - 3$ for $\frac{\partial M}{\partial y}$ and $6 y^{2} x - 3$ for $\frac{\partial N}{\partial x}$ .

$6 y^{2} x - 3 = 6 y^{2} x - 3$

Step 3.2

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

$6 y^{2} x - 3 = 6 y^{2} x - 3$ is an identity.

Step 4

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

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

Step 5

Integrate $M (x, y) = y (2 x y^{2} - 3)$ to find $f (x, y)$ .

Tap for more steps...

Step 5.1

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

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

Step 5.2

Split the single integral into multiple integrals.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial y} = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8

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

Tap for more steps...

Step 8.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [y (y^{2} x^{2} - 3 x) + g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.2

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

$\frac{d}{d y} [y (y^{2} x^{2} - 3 x)] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3

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

Tap for more steps...

Step 8.3.1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = y^{2} x^{2} - 3 x$ .

$y \frac{d}{d y} [y^{2} x^{2} - 3 x] + (y^{2} x^{2} - 3 x) \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.2

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

$y (\frac{d}{d y} [y^{2} x^{2}] + \frac{d}{d y} [- 3 x]) + (y^{2} x^{2} - 3 x) \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.3

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

$y (x^{2} \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 3 x]) + (y^{2} x^{2} - 3 x) \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.4

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

$y (x^{2} (2 y) + \frac{d}{d y} [- 3 x]) + (y^{2} x^{2} - 3 x) \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.5

Since $- 3 x$ is constant with respect to $y$ , the derivative of $- 3 x$ with respect to $y$ is $0$ .

$y (x^{2} (2 y) + 0) + (y^{2} x^{2} - 3 x) \frac{d}{d y} [y] + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.6

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

$y (x^{2} (2 y) + 0) + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.7

Move $2$ to the left of $x^{2}$ .

$y (2 \cdot x^{2} y + 0) + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.8

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

$y (2 x^{2} y) + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.9

Raise $y$ to the power of $1$ .

$2 x^{2} (y^{1} y) + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.10

Raise $y$ to the power of $1$ .

$2 x^{2} (y^{1} y^{1}) + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 x^{2} y^{1 + 1} + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.12

Add $1$ and $1$ .

$2 x^{2} y^{2} + (y^{2} x^{2} - 3 x) \cdot 1 + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.13

Multiply $y^{2} x^{2} - 3 x$ by $1$ .

$2 x^{2} y^{2} + y^{2} x^{2} - 3 x + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.14

Add $2 x^{2} y^{2}$ and $y^{2} x^{2}$ .

Tap for more steps...

Step 8.3.14.1

Reorder $y^{2}$ and $x^{2}$ .

$2 x^{2} y^{2} + x^{2} y^{2} - 3 x + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.3.14.2

Add $2 x^{2} y^{2}$ and $x^{2} y^{2}$ .

$3 x^{2} y^{2} - 3 x + \frac{d}{d y} [g (y)] = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.4

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

$3 x^{2} y^{2} - 3 x + \frac{d g}{d y} = 3 x^{2} y^{2} - 3 x + 4 y$

Step 8.5

Reorder terms.

$\frac{d g}{d y} + 3 x^{2} y^{2} - 3 x = 3 x^{2} y^{2} - 3 x + 4 y$

Step 9

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

Tap for more steps...

Step 9.1

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

Tap for more steps...

Step 9.1.1

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

$\frac{d g}{d y} - 3 x = 3 x^{2} y^{2} - 3 x + 4 y - 3 x^{2} y^{2}$

Step 9.1.2

Add $3 x$ to both sides of the equation.

$\frac{d g}{d y} = 3 x^{2} y^{2} - 3 x + 4 y - 3 x^{2} y^{2} + 3 x$

Step 9.1.3

Combine the opposite terms in $3 x^{2} y^{2} - 3 x + 4 y - 3 x^{2} y^{2} + 3 x$ .

Tap for more steps...

Step 9.1.3.1

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

$\frac{d g}{d y} = - 3 x + 4 y + 0 + 3 x$

Step 9.1.3.2

Add $- 3 x + 4 y$ and $0$ .

$\frac{d g}{d y} = - 3 x + 4 y + 3 x$

Step 9.1.3.3

Add $- 3 x$ and $3 x$ .

$\frac{d g}{d y} = 4 y + 0$

Step 9.1.3.4

Add $4 y$ and $0$ .

$\frac{d g}{d y} = 4 y$

Step 10

Find the antiderivative of $4 y$ to find $g (y)$ .

Tap for more steps...

Step 10.1

Integrate both sides of $\frac{d g}{d y} = 4 y$ .

$\int \frac{d g}{d y} d y = \int 4 y d y$

Step 10.2

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

$g (y) = \int 4 y d y$

Step 10.3

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

$g (y) = 4 \int y d y$

Step 10.4

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

$g (y) = 4 (\frac{1}{2} y^{2} + C)$

Step 10.5

Simplify the answer.

Tap for more steps...

Step 10.5.1

Rewrite $4 (\frac{1}{2} y^{2} + C)$ as $4 (\frac{1}{2}) y^{2} + C$ .

$g (y) = 4 (\frac{1}{2}) y^{2} + C$

Step 10.5.2

Simplify.

Tap for more steps...

Step 10.5.2.1

Combine $4$ and $\frac{1}{2}$ .

$g (y) = \frac{4}{2} y^{2} + C$

Step 10.5.2.2

Cancel the common factor of $4$ and $2$ .

Tap for more steps...

Step 10.5.2.2.1

Factor $2$ out of $4$ .

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

Step 10.5.2.2.2

Cancel the common factors.

Tap for more steps...

Step 10.5.2.2.2.1

Factor $2$ out of $2$ .

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

Step 10.5.2.2.2.2

Cancel the common factor.

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

Step 10.5.2.2.2.3

Rewrite the expression.

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

Step 10.5.2.2.2.4

Divide $2$ by $1$ .

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

Step 11

Substitute for $g (y)$ in $f (x, y) = y (y^{2} x^{2} - 3 x) + g (y)$ .

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

Step 12

Simplify each term.

Tap for more steps...

Step 12.1

Apply the distributive property.

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

Step 12.2

Multiply $y$ by $y^{2}$ by adding the exponents.

Tap for more steps...

Step 12.2.1

Move $y^{2}$ .

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

Step 12.2.2

Multiply $y^{2}$ by $y$ .

Tap for more steps...

Step 12.2.2.1

Raise $y$ to the power of $1$ .

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

Step 12.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 12.2.3

Add $2$ and $1$ .

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

Step 12.3

Rewrite using the commutative property of multiplication.

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