What is the implicit derivative of #xy+y/sqrt(xy-x) =2 #?

1 Answer
Jul 11, 2018

#y'=(-y+2x*sqrt(xy-x)+y^2-2xy^2*sqrt(xy-x))/(-2x-2x^2sqrt(xy-x)+xy+2x^2ysqrt(xy-x))#

Explanation:

We have

#y+xy'+(y'sqrt(xy-x)-y*1/2*(xy-x)^(-1/2)(y+xy'-1))/(xy-x)=0#
multiplying by #sqrt(xy+x)*(xy-x)# we get

#(y+xy')(xy-x)sqrt(xy-x)+y'(xy-x)-y/2(y+xy'-1)=0#
multiplying by #2#

#2(y+xy')(xy-x)sqrt(xy-x)+2y'(xy-x)-y(y+xy'-1)=0#
#y'(-2x-2x^2sqrt(xy-x)+xy+2x^2ysqrt(xy-x))=-y+2xsqrt(xy-x)+y^2-2xy^2sqrt(xy-x)#

so we get

#y'=(-y+2xsqrt(xy-x)+y^2-2xy^2sqrt(xy-x))/(-2x-2x^2sqrt(xy-x)+xy+2x^2ysqrt(xy-x))#