How do you find #dy/dx# by implicit differentiation of #sqrt(xy)=x-2y#?

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Answer 1 · 2016-12-10T00:56:32

#dy/dx = 1/4#

This is a homogeneous equation and no need for implicit differentiation.

Putting #y = lambda x# into #sqrt(xy)=x-2y# we have

#xsqrt(lambda)=x-2lambda x# or

#(sqrt(lambda)-1+2lambda)x=0# but #x ne 0# so

#sqrt(lambda)-1+2(sqrt(lambda))^2 = 0#

so

#sqrt(lambda) = (-1pmsqrt(1+8))/4=(-1pm3)/4=1/2# so we have

#y/x=lambda = 1/4# and finally

#dy/dx = 1/4#