Find dx/dy 4x^2+2y^2=12

$4 x^{2} + 2 y^{2} = 12$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (4 x^{2} + 2 y^{2}) = \frac{d}{d y} (12)$

Step 2

Differentiate the left side of the equation.

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

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

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

Since $4$ is constant with respect to $y$ , the derivative of $4 x^{2}$ with respect to $y$ is $4 \frac{d}{d y} [x^{2}]$ .

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

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 $x$ .

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

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

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

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

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

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

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

Multiply $2$ by $4$ .

$8 x x' + \frac{d}{d y} [2 y^{2}]$

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

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

$8 x x' + 2 \frac{d}{d y} [y^{2}]$

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

$8 x x' + 2 (2 y)$

Multiply $2$ by $2$ .

$8 x x' + 4 y$

Step 3

Since $12$ is constant with respect to $y$ , the derivative of $12$ with respect to $y$ is $0$ .

$0$

Step 4

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

$8 x x' + 4 y = 0$

Step 5

Solve for $x'$ .

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

$8 x x' = - 4 y$

Divide each term in $8 x x' = - 4 y$ by $8 x$ and simplify.

Divide each term in $8 x x' = - 4 y$ by $8 x$ .

$\frac{8 x x'}{8 x} = \frac{- 4 y}{8 x}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x x'}{8 x} = \frac{- 4 y}{8 x}$

Rewrite the expression.

$\frac{x x'}{x} = \frac{- 4 y}{8 x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x x'}{x} = \frac{- 4 y}{8 x}$

Divide $x'$ by $1$ .

$x' = \frac{- 4 y}{8 x}$

Simplify the right side.

Cancel the common factor of $- 4$ and $8$ .

Factor $4$ out of $- 4 y$ .

$x' = \frac{4 (- y)}{8 x}$

Cancel the common factors.

Factor $4$ out of $8 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

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

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

Find dx/dy 4x^2+2y^2=12

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