Find dx/dy x^2+4xy+y^2=13

$x^{2} + 4 x y + y^{2} = 13$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x^{2} + 4 x y + y^{2}) = \frac{d}{d y} (13)$

Step 2

Differentiate the left side of the equation.

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

$\frac{d}{d y} [x^{2}] + \frac{d}{d y} [4 x y] + \frac{d}{d y} [y^{2}]$

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

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

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

$\frac{d}{d u} [u^{2}] \frac{d}{d y} [x] + \frac{d}{d y} [4 x y] + \frac{d}{d y} [y^{2}]$

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

$2 u \frac{d}{d y} [x] + \frac{d}{d y} [4 x y] + \frac{d}{d y} [y^{2}]$

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

$2 x \frac{d}{d y} [x] + \frac{d}{d y} [4 x y] + \frac{d}{d y} [y^{2}]$

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

$2 x x' + \frac{d}{d y} [4 x y] + \frac{d}{d y} [y^{2}]$

Evaluate $\frac{d}{d y} [4 x y]$ .

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

$2 x x' + 4 \frac{d}{d y} [x y] + \frac{d}{d y} [y^{2}]$

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 $f (y) = x$ and $g (y) = y$ .

$2 x x' + 4 (x \frac{d}{d y} [y] + y \frac{d}{d y} [x]) + \frac{d}{d y} [y^{2}]$

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

$2 x x' + 4 (x \cdot 1 + y \frac{d}{d y} [x]) + \frac{d}{d y} [y^{2}]$

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

$2 x x' + 4 (x \cdot 1 + y x') + \frac{d}{d y} [y^{2}]$

Multiply $x$ by $1$ .

$2 x x' + 4 (x + y x') + \frac{d}{d y} [y^{2}]$

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

$2 x x' + 4 (x + y x') + 2 y$

Simplify.

Apply the distributive property.

$2 x x' + 4 x + 4 (y x') + 2 y$

Remove unnecessary parentheses.

$2 x x' + 4 x + 4 y x' + 2 y$

Reorder terms.

$2 x x' + 4 y x' + 4 x + 2 y$

Step 3

Since $13$ is constant with respect to $y$ , the derivative of $13$ with respect to $y$ is $0$ .

$0$

Step 4

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

$2 x x' + 4 y x' + 4 x + 2 y = 0$

Step 5

Solve for $x'$ .

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

Subtract $4 x$ from both sides of the equation.

$2 x x' + 4 y x' + 2 y = - 4 x$

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

$2 x x' + 4 y x' = - 4 x - 2 y$

Factor $2 x'$ out of $2 x x' + 4 y x'$ .

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

$2 x' x + 4 y x' = - 4 x - 2 y$

Factor $2 x'$ out of $4 y x'$ .

$2 x' x + 2 x' (2 y) = - 4 x - 2 y$

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

$2 x' (x + 2 y) = - 4 x - 2 y$

Divide each term in $2 x' (x + 2 y) = - 4 x - 2 y$ by $2 (x + 2 y)$ and simplify.

Divide each term in $2 x' (x + 2 y) = - 4 x - 2 y$ by $2 (x + 2 y)$ .

$\frac{2 x' (x + 2 y)}{2 (x + 2 y)} = \frac{- 4 x}{2 (x + 2 y)} + \frac{- 2 y}{2 (x + 2 y)}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x' (x + 2 y)}{2 (x + 2 y)} = \frac{- 4 x}{2 (x + 2 y)} + \frac{- 2 y}{2 (x + 2 y)}$

Rewrite the expression.

$\frac{x' (x + 2 y)}{x + 2 y} = \frac{- 4 x}{2 (x + 2 y)} + \frac{- 2 y}{2 (x + 2 y)}$

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

Cancel the common factor.

$\frac{x' (x + 2 y)}{x + 2 y} = \frac{- 4 x}{2 (x + 2 y)} + \frac{- 2 y}{2 (x + 2 y)}$

Divide $x'$ by $1$ .

$x' = \frac{- 4 x}{2 (x + 2 y)} + \frac{- 2 y}{2 (x + 2 y)}$

Simplify the right side.

Simplify each term.

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

Factor $2$ out of $- 4 x$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Simplify terms.

Combine the numerators over the common denominator.

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

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

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

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

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

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

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

Simplify the expression.

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

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

Move the negative in front of the fraction.

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

Step 6

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

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

Find dx/dy x^2+4xy+y^2=13

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