What is the implicit derivative of #-2=xy^2-3xy#?

1 Answer
Nov 30, 2015

I found: #(dy)/(dx)=(y(3-y))/(x(2y-3))#

Explanation:

The idea is that you need to derive #y# as well as it represents a function of #x#; so, for example to derive #y^2# you get: #2y(dy)/(dx)#!
Where the #(dy)/(dx)# bit takes into accont the #x# dependence of #y#:

in your case you'll get:
#0=1*y^2+2xy(dy)/(dx)-3y-3x*1(dy)/(dx)#
where I used the Product Rule;
collect: #(dy)/(dx)#:
#(dy)/(dx)[2xy-3x]=3y-y^2#
#(dy)/(dx)=(y(3-y))/(x(2y-3))#