How do you implicitly differentiate # sqrt(xy) = x-y/(x-1)#?

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Answer 1 · 2018-07-27T19:46:56

#(dy)/(dx)=(2sqrt(xy)(x-1)^2+2ysqrt(xy)-y(x-1)^2)/((x^2-x+2sqrt(xy))(x-1))#

Here ,

#sqrt(xy)=x-y/(x-1)#

Diff. each term w.r.t. #x#, (using product,quotient and chain rules)

#1/(2sqrt(xy))*d/(dx)(xy)=1-((x-1)y_1-y(1))/(x-1)^2#

#=>1/(2sqrt(xy))[xy_1+y*1]=1-y_1/(x-1)+y/(x-1)^2#

#=>(xy_1)/(2sqrt(xy))+y/(2sqrt(xy))=1-y_1/(x-1)+y/(x-1)^2#

#=>(xy_1)/(2sqrt(xy))+y_1/(x-1)=1+y/(x-1)^2-y/(2sqrt(xy))#

#y_1{x/(2sqrt(xy))+1/(x-1)}=(2sqrt(xy)(x-1)^2+2ysqrt(xy)-y(x- 1)^2)/(2sqrt(xy)(x-1)^2)#

#y_1{(x^2-x+2sqrt(xy))/(2sqrt(xy)(x-1))}#=#(2sqrt(xy)(x- 1)^2+2ysqrt(xy)-y(x-1)^2)/(2sqrt(xy)(x-1)^2)#

#y_1(x^2-x+2sqrt(xy))=(2sqrt(xy)(x-1)^2+2ysqrt(xy)-y(x-1)^2)/((x-1))#

#:.y_1=(2sqrt(xy)(x-1)^2+2ysqrt(xy)-y(x-1)^2)/((x^2-x+2sqrt(xy))(x-1))#

Where, #y_1=(dy)/(dx)#