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 $y=xcsc^(-1)x$

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

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

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