# Find \frac{dy}{dx} by implicit differentiation y^2 + 3x^2 = xy.

## Question:

Find {eq}\frac{dy}{dx} {/eq} by implicit differentiation {eq}y^2 + 3x^2 = xy. {/eq}

Taking the derivative of both sides yields $$2y\dfrac{dy}{dx}+6x=x\dfrac{dy}{dx}+y,$$ by the product rule. Then we can isolate {eq}\dfrac{dy}{dx} {/eq} to get \begin{align} 2y\dfrac{dy}{dx}+6x=x\dfrac{dy}{dx}+y&\Longrightarrow 2y\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-6x \\ &\Longrightarrow \dfrac{dy}{dx}(2y-x)=y-6x \\ &\Longrightarrow \dfrac{dy}{dx}=\dfrac{y-6x}{2y-x}. \end{align}