How do you implicitly differentiate #7=1-e^y/(xy)#?

1 Answer
Apr 18, 2016

I found: #(dy)/(dx)=y/(x(y-1))#

Explanation:

We need to remember that #y# will be itself a function of #x# so we need to derive it accordingly.
For example, if you have #y^2# the derivative will be:
#2y*(dy)/(dx)#
to consider its dependence from #x#.
In our case we have:
#0=0-(xye^y*(dy)/(dx)-e^y(y+x(dy)/(dx)))/(xy)^2#
rearranging:
#(dy)/(dx)[xye^y-xe^y]=ye^y#
#(dy)/(dx)=(ye^y)/(xe^y(y-1))=y/(x(y-1))#