How do you differentiate #y=xcsc^-1x#?

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Answer 1 · 2017-12-10T05:52:31

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

Let #csc^(-1)x=t#

then #csct=x# or #sint=1/x# and #t=sin^(-1)(1/x)#

as #d/(dx)sin^(-1)u=1/sqrt(1-u^2)#

#d/(dx)csc^(-1)x=d/(dx)sin^(-1)(1/x)#

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

Hence if #y=xcsc^(-1)x#

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

= #csc^(-1)x+x^2/sqrt(x^2-1)#