Question

find dy/dx of 2x^2-xy+y^2+3y-4=0​

Answer 1

Step-by-step explanation:

Given function is 2x² - xy + y² + 3y - 4 = 0

Differentiating both sides w.r.t x, we get,

$4x - (y + x\frac{dy}{dx} ) + 2y \frac{dy}{dx} + 3 \frac{dy}{dx} = 0$

$= > 4x - y - x\frac{dy}{dx} + 2y \frac{dy}{dx} + 3 \frac{dy}{dx} = 0$

$= > 4x - y + \frac{dy}{dx} (2y + 3 - x) = 0$

$= > \frac{dy}{dx} = \frac{y - 4x}{2y + 3 - x}$