How do you differentiate #y=sec^-1(x^2-x)#?

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Answer 1 · 2018-01-06T09:46:21

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

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

#=>x^2-x=secy#

differentiate #wrt" "x#

#2x-1=secytany(dy)/(dx)#

#(dy)/(dx)=(2x-1)/(secytany)--(1)#

now

#1+tan^2x=sec^2x=>tany=sqrt((sec^2y-1)#

so substituting back into #(1)#

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

which can be simplified further as required