Calculus Examples

$(y + 2 x y^{3}) d x + (1 + 3 x^{2} y^{2} + x) d y = 0$

Step 1

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

Tap for more steps...

Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

Tap for more steps...

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $2$ .

$\frac{\partial M}{\partial y} = 1 + 6 x y^{2}$

Step 1.4

Reorder terms.

$\frac{\partial M}{\partial y} = 6 y^{2} x + 1$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 1 + 3 x^{2} y^{2} + x$ .

Tap for more steps...

Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

Tap for more steps...

Step 2.2.1

By the Sum Rule, the derivative of $1 + 3 x^{2} y^{2} + x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [x]$ .

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [3 x^{2} y^{2}] + \frac{d}{d x} [x]$

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

$\frac{\partial N}{\partial x} = 0 + 3 y^{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]$

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $3$ .

$\frac{\partial N}{\partial x} = 0 + 6 y^{2} x + \frac{d}{d x} [x]$

Step 2.4

Differentiate using the Power Rule.

Tap for more steps...

Step 2.4.1

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

$\frac{\partial N}{\partial x} = 0 + 6 y^{2} x + 1$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 6 y^{2} x + 1$

Step 3

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

Tap for more steps...

Step 3.1

Substitute $6 y^{2} x + 1$ for $\frac{\partial M}{\partial y}$ and $6 y^{2} x + 1$ for $\frac{\partial N}{\partial x}$ .

$6 y^{2} x + 1 = 6 y^{2} x + 1$

Step 3.2

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

$6 y^{2} x + 1 = 6 y^{2} x + 1$ is an identity.

Step 4

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

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

Step 5

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

Tap for more steps...

Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

Apply the constant rule.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 5.6

Simplify.

Tap for more steps...

Step 5.6.1

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

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

Step 5.6.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.6.2.1

Cancel the common factor.

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

Step 5.6.2.2

Rewrite the expression.

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

Step 5.6.3

Multiply $y^{3}$ by $1$ .

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial y} = 1 + 3 x^{2} y^{2} + x$

Step 8

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

Tap for more steps...

Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

Tap for more steps...

Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Multiply $x$ by $1$ .

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

Step 8.4

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

Tap for more steps...

Step 8.4.1

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

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

Step 8.4.2

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

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

Step 8.4.3

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

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

Step 8.5

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

$x + 3 x^{2} y^{2} + \frac{d g}{d y} = 1 + 3 x^{2} y^{2} + x$

Step 8.6

Reorder terms.

$\frac{d g}{d y} + 3 x^{2} y^{2} + x = 1 + 3 x^{2} y^{2} + x$

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.

$\frac{d g}{d y} + x = 1 + 3 x^{2} y^{2} + x - 3 x^{2} y^{2}$

Step 9.1.2

Subtract $x$ from both sides of the equation.

$\frac{d g}{d y} = 1 + 3 x^{2} y^{2} + x - 3 x^{2} y^{2} - x$

Step 9.1.3

Combine the opposite terms in $1 + 3 x^{2} y^{2} + x - 3 x^{2} y^{2} - x$ .

Tap for more steps...

Step 9.1.3.1

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

$\frac{d g}{d y} = 1 + x + 0 - x$

Step 9.1.3.2

Add $1 + x$ and $0$ .

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

Step 9.1.3.3

Subtract $x$ from $x$ .

$\frac{d g}{d y} = 1 + 0$

Step 9.1.3.4

Add $1$ and $0$ .

$\frac{d g}{d y} = 1$

Step 10

Find the antiderivative of $1$ to find $g (y)$ .

Tap for more steps...

Step 10.1

Integrate both sides of $\frac{d g}{d y} = 1$ .

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

Step 10.2

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

$g (y) = \int d y$

Step 10.3

Apply the constant rule.

$g (y) = y + C$

Step 11

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

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