Questions: Use implicit differentiation to find dy/dx. 2xy - y^2 = 1

Transcript text: Use implicit differentiation to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. \[ 2 x y-y^{2}=1 \]

Solution

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiation, follow these steps:

Differentiate both sides of the equation $2xy - y^2 = 1$ with respect to $x$.
Apply the product rule to the term $2xy$.
Collect all terms involving $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation.
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Step 1: Differentiate Both Sides with Respect to $x$

Given the equation: \[ 2xy - y^2 = 1 \]

Differentiate both sides with respect to $x$: \[ \frac{d}{dx}(2xy - y^2) = \frac{d}{dx}(1) \]

Step 2: Apply the Product Rule and Chain Rule

Using the product rule on $2xy$ and the chain rule on $y^2$: \[ 2 \left( y + x \frac{dy}{dx} \right) - 2y \frac{dy}{dx} = 0 \]

Step 3: Collect Terms Involving $\frac{dy}{dx}$

Rearrange the equation to collect terms involving $\frac{dy}{dx}$: \[ 2y + 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0 \]

Step 4: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: \[ 2y + (2x - 2y) \frac{dy}{dx} = 0 \]

Isolate $\frac{dy}{dx}$: \[ (2x - 2y) \frac{dy}{dx} = -2y \]

\[ \frac{dy}{dx} = \frac{-2y}{2x - 2y} \]

Simplify the expression: \[ \frac{dy}{dx} = \frac{y}{y - x} \]