How do you differentiate #xe^y=y-1#?

2 Answers
Oct 9, 2016

#(dy)/(dx)=e^y/(1-xe^y)#

Explanation:

Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#. However, some functions of #y# are written implicitly as functions of #x#, like the one in question above.

So what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.

Using this, differential of #xe^y=y-1# can be derived as follows:

#x xxd/(dy)*e^yxx(dy)/(dx)+1xxe^y=d/(dy)yxx(dy)/(dx)#

or #xe^y(dy)/(dx)+e^y=(dy)/(dx)#

or #(dy)/(dx)(1-xe^y)=e^y#

or #(dy)/(dx)=e^y/(1-xe^y)#

Oct 9, 2016

#e^y/(2-y)#.

Explanation:

#x=e^(-y)(y-1) is from the given solvable-for-x implicit form.

In such problems, use

#x'= (dx)/(dy)=-e^(-y)(y-1)+e^(-y)=(2-y)e^(-y)= 1/(y')#. So,

#(dy)/(dx)=y'=e^y/(2-y)#..