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

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Answer 1 · 2017-01-04T08:48:05

#(dy)/(dx)=2/(xsqrt(x^4-1))#

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

#=>x^2=secy#

differentiate wrt #x#

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

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

now substitute back using:

#secy=x^2#

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

#tany=sqrt(x^4-1)#

#(dy)/(dx)=(2x)/(x^2sqrt(x^4-1))=2/(xsqrt(x^4-1))#