How do you differentiate #5xy + y^3 = 2x + 3y#?

1 Answer
Aug 4, 2015

I found: #(dy)/(dx)=(2-5y)/(5x+3y^2-3)#

Explanation:

You can use implicit differentiation rememberig that #y# represents a function of #x# and needs to be differentiated accordingly;
for example: if you have #y^2# you differentiate it to get:
#2y*(dy)/(dx)# where you use #(dy)/(dx)# to take into account the dependence with #x#.
In your case you get:
#5y+5x(dy)/(dx)+3y^2(dy)/(dx)=2+3(dy)/(dx)#
collect #(dy)/(dx)#:
#(dy)/(dx)(5x+3y^2-3)=2-5y#
and:
#(dy)/(dx)=(2-5y)/(5x+3y^2-3)#