How do you implicitly differentiate #11=-xy^2-(xy)/(2x-y)#?

Question

William Abdalla · Accepted Answer

#dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)# #11=-xy^2-(xy)/(2x-y)# #11(2x-y)=-xy^2(2x-y)-xy# #22x-11y=-2x^2y^2+xy^3-xy# #22-11dy/dx=-4xy^2-4x^2y dy/dx+y^3+3xy^2 dy/dx-y-xdy/dx# #4x^2y dy/dx-3xy^2 dy/dx+xdy/dx-11dy/dx=y^3-4xy^2-y-22# #dy/dx(4x^2y -3xy^2 +x-11)=y^3-4xy^2-y-22# #dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)#

Samantha Adkison · Answer

undefined To implicitly differentiate the given equation, you need to differentiate each term with respect to $ x $ and apply the chain rule wherever necessary.

Given the equation: $ 11 = -xy^2 - \frac{xy}{2x - y} $

Differentiating both sides with respect to $ x $, you get:

$ \frac{d}{dx}[11] = \frac{d}{dx}[-xy^2] - \frac{d}{dx}\left[\frac{xy}{2x - y}\right] $

Simplify each term:

$ 0 = -y^2 - x \frac{d}{dx}[y^2] - \left[\frac{(2x - y)\frac{d}{dx}[xy] - xy \frac{d}{dx}[2x - y]}{(2x - y)^2}\right] $

Now, apply the chain rule and product rule where necessary:

$ 0 = -y^2 - x(2y \frac{dy}{dx}) - \left[\frac{(2x - y)(y + x\frac{dy}{dx}) - xy(2 - \frac{dy}{dx})}{(2x - y)^2}\right] $

Simplify and solve for $ \frac{dy}{dx} $:

$ 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(2xy - y^2 + xy + x^2\frac{dy}{dx}) - (2xy - xy^2)}{(2x - y)^2}\right] $

$ 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(2xy - y^2 + xy + x^2\frac{dy}{dx}) - 2xy + xy^2}{(2x - y)^2}\right] $

$ 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(x^2\frac{dy}{dx} - y^2)}{(2x - y)^2}\right] $

$ 0 = -y^2 - 2xy\frac{dy}{dx} - \frac{x^2\frac{dy}{dx} - y^2}{(2x - y)^2} $

Now, solve for $ \frac{dy}{dx} $.