Question

If \( \sqrt{3}(x^2 + y^2) = 4xy \), then find \( \frac{dy}{dx} \) at \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \).

Hint:

You can often plug in the coordinates right after differentiation instead of rearranging to save time.
Ensure you apply the chain rule correctly to every term containing \( y \).

Answer 1

Concept:
• Implicit Differentiation: Differentiate every term with respect to \( x \), treating \( y \) as a function of \( x \) (using the chain rule).
• Product Rule: \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \).

Step 1: Differentiate the equation with respect to \( x \)
Given: \( \sqrt{3}x^2 + \sqrt{3}y^2 = 4xy \) Differentiating both sides: \[ \sqrt{3}(2x) + \sqrt{3}\left( 2y \frac{dy}{dx} \right) = 4 \left[ x \frac{dy}{dx} + y(1) \right] \] \[ 2\sqrt{3}x + 2\sqrt{3}y \frac{dy}{dx} = 4x \frac{dy}{dx} + 4y \]

Step 2: Rearrange to solve for \( \frac{dy}{dx} \)
Gather all \( \frac{dy}{dx} \) terms on one side: \[ 2\sqrt{3}y \frac{dy}{dx} - 4x \frac{dy}{dx} = 4y - 2\sqrt{3}x \] Factor out the derivative: \[ \frac{dy}{dx} (2\sqrt{3}y - 4x) = 4y - 2\sqrt{3}x \] \[ \frac{dy}{dx} = \frac{4y - 2\sqrt{3}x}{2\sqrt{3}y - 4x} \]

Step 3: Substitute the coordinates \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \)
\[ \frac{dy}{dx} = \frac{4(\frac{\sqrt{3}}{2}) - 2\sqrt{3}(\frac{1}{2})}{2\sqrt{3}(\frac{\sqrt{3}}{2}) - 4(\frac{1}{2})} \] Simplify numerator: \( 2\sqrt{3} - \sqrt{3} = \sqrt{3} \) Simplify denominator: \( 3 - 2 = 1 \) \[ \frac{dy}{dx} = \frac{\sqrt{3}}{1} = \sqrt{3} \]