How do you use implicit differentiation to find #(dy)/(dx)# given #(x-1)y^2=x+1#?

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Answer 1 · 2017-02-18T08:33:22

#dy/dx= -1/((x-1)^2*y)#

#(x-1)y^2=x+1#

#y^2 =(x+1)/(x-1)#

#2ydy/dx= ((x-1)*1-(x+1)*1)/(x-1)^2#

#= (-2)/(x-1)^2#

#:. dy/dx=(-2)/((x-1)^2*2y)#

#=-1/((x-1)^2*y)#