What is the implicit derivative of #5=y/(x-y)#?

1 Answer
Nov 20, 2015

#(dy)/(dx)=y/x#

Explanation:

Quotient Rule for Derivatives:
#d(u/v)/dx = (v*(du)/(dx)-u*(dv)/(dx))/(v^2)#

With #u = y# and #v=(x-y)#
#color(white)("XXX")(du)/(dx) = (dy)/(dx)#
and
#color(white)("XXX")(dv)/(dx) = (d(x-y))/(dx) = 1-(dy)/(dx)#

Therefore
#color(white)("XXX")(d(y/(x-y)))/(dx) = ((x-y)(dy)/(dx) - y(1-(dy)/(dx)))/((x-y)^2)#

#color(white)("XXXXXXXX")=(x(dy)/(dx)-y)/((x-y)^2)#

#(d(5))/(dx)=0#

So, since #5= y/(x-y)#

#rArrcolor(white)("XXX")(d(5))/(dx) = (d(y/(x-y)))/(dx)#

#rArrcolor(white)("XXX")0 = (x(dy)/(dx)-y)/((x-y)^2)#

#rArrcolor(white)("XXX")x(dy)/(dx)-y=0#

#rArrcolor(white)("XXX")(dy)/(dx) = y/x#