Calculus Examples

$x^{2} y + y^{2} x = - 2$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

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

Step 2.2.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x^{2} y' + y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2} x]$

Step 2.2.3

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

$x^{2} y' + y (2 x) + \frac{d}{d x} [y^{2} x]$

Step 2.2.4

Move $2$ to the left of $y$ .

$x^{2} y' + 2 y x + \frac{d}{d x} [y^{2} x]$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$x^{2} y' + 2 y x + y^{2} \frac{d}{d x} [x] + x \frac{d}{d x} [y^{2}]$

Step 2.3.2

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

$x^{2} y' + 2 y x + y^{2} \cdot 1 + x \frac{d}{d x} [y^{2}]$

Step 2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

Tap for more steps...

Step 2.3.3.1

To apply the Chain Rule, set $u$ as $y$ .

$x^{2} y' + 2 y x + y^{2} \cdot 1 + x (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

Step 2.3.3.2

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

$x^{2} y' + 2 y x + y^{2} \cdot 1 + x (2 u \frac{d}{d x} [y])$

Step 2.3.3.3

Replace all occurrences of $u$ with $y$ .

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

Step 2.3.4

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x^{2} y' + 2 y x + y^{2} \cdot 1 + x (2 y y')$

Step 2.3.5

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

$x^{2} y' + 2 y x + y^{2} + x (2 y y')$

Step 2.3.6

Move $2$ to the left of $x$ .

$x^{2} y' + 2 y x + y^{2} + 2 x y y'$

Step 2.4

Reorder terms.

$y^{2} + x^{2} y' + 2 x y + 2 x y y'$

Step 3

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

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$y^{2} + x^{2} y' + 2 x y + 2 x y y' = 0$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Move all terms not containing $y'$ to the right side of the equation.

Tap for more steps...

Step 5.1.1

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

$x^{2} y' + 2 x y + 2 x y y' = - y^{2}$

Step 5.1.2

Subtract $2 x y$ from both sides of the equation.

$x^{2} y' + 2 x y y' = - y^{2} - 2 x y$

Step 5.2

Factor $x y'$ out of $x^{2} y' + 2 x y y'$ .

Tap for more steps...

Step 5.2.1

Factor $x y'$ out of $x^{2} y'$ .

$x y' x + 2 x y y' = - y^{2} - 2 x y$

Step 5.2.2

Factor $x y'$ out of $2 x y y'$ .

$x y' x + x y' (2 y) = - y^{2} - 2 x y$

Step 5.2.3

Factor $x y'$ out of $x y' x + x y' (2 y)$ .

$x y' (x + 2 y) = - y^{2} - 2 x y$

Step 5.3

Divide each term in $x y' (x + 2 y) = - y^{2} - 2 x y$ by $x (x + 2 y)$ and simplify.

Tap for more steps...

Step 5.3.1

Divide each term in $x y' (x + 2 y) = - y^{2} - 2 x y$ by $x (x + 2 y)$ .

$\frac{x y' (x + 2 y)}{x (x + 2 y)} = \frac{- y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.3.2.1.1

Cancel the common factor.

$\frac{x y' (x + 2 y)}{x (x + 2 y)} = \frac{- y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y' (x + 2 y)}{x + 2 y} = \frac{- y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.2.2

Cancel the common factor of $x + 2 y$ .

Tap for more steps...

Step 5.3.2.2.1

Cancel the common factor.

$\frac{y' (x + 2 y)}{x + 2 y} = \frac{- y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.1

Simplify each term.

Tap for more steps...

Step 5.3.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.3.1.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.3.3.1.2.1

Cancel the common factor.

$y' = - \frac{y^{2}}{x (x + 2 y)} + \frac{- 2 x y}{x (x + 2 y)}$

Step 5.3.3.1.2.2

Rewrite the expression.

$y' = - \frac{y^{2}}{x (x + 2 y)} + \frac{- 2 y}{x + 2 y}$

Step 5.3.3.1.3

Move the negative in front of the fraction.

$y' = - \frac{y^{2}}{x (x + 2 y)} - \frac{2 y}{x + 2 y}$

Step 5.3.3.2

To write $- \frac{2 y}{x + 2 y}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$y' = - \frac{y^{2}}{x (x + 2 y)} - \frac{2 y}{x + 2 y} \cdot \frac{x}{x}$

Step 5.3.3.3

Write each expression with a common denominator of $x (x + 2 y)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 5.3.3.3.1

Multiply $\frac{2 y}{x + 2 y}$ by $\frac{x}{x}$ .

$y' = - \frac{y^{2}}{x (x + 2 y)} - \frac{2 y x}{(x + 2 y) x}$

Step 5.3.3.3.2

Reorder the factors of $(x + 2 y) x$ .

$y' = - \frac{y^{2}}{x (x + 2 y)} - \frac{2 y x}{x (x + 2 y)}$

Step 5.3.3.4

Combine the numerators over the common denominator.

$y' = \frac{- y^{2} - 2 y x}{x (x + 2 y)}$

Step 5.3.3.5

Factor $y$ out of $- y^{2} - 2 y x$ .

Tap for more steps...

Step 5.3.3.5.1

Factor $y$ out of $- y^{2}$ .

$y' = \frac{y (- y) - 2 y x}{x (x + 2 y)}$

Step 5.3.3.5.2

Factor $y$ out of $- 2 y x$ .

$y' = \frac{y (- y) + y (- 2 x)}{x (x + 2 y)}$

Step 5.3.3.5.3

Factor $y$ out of $y (- y) + y (- 2 x)$ .

$y' = \frac{y (- y - 2 x)}{x (x + 2 y)}$

Step 5.3.3.6

Factor $- 1$ out of $- y$ .

$y' = \frac{y (- (y) - 2 x)}{x (x + 2 y)}$

Step 5.3.3.7

Factor $- 1$ out of $- 2 x$ .

$y' = \frac{y (- (y) - (2 x))}{x (x + 2 y)}$

Step 5.3.3.8

Factor $- 1$ out of $- (y) - (2 x)$ .

$y' = \frac{y (- (y + 2 x))}{x (x + 2 y)}$

Step 5.3.3.9

Simplify the expression.

Tap for more steps...

Step 5.3.3.9.1

Rewrite $- (y + 2 x)$ as $- 1 (y + 2 x)$ .

$y' = \frac{y (- 1 (y + 2 x))}{x (x + 2 y)}$

Step 5.3.3.9.2

Move the negative in front of the fraction.

$y' = - \frac{y (y + 2 x)}{x (x + 2 y)}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

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