How do you find the derivative of y= arctan[(x^2-1)^(1/2)] + "arccsc"(x)?

1 Answer
Sasha P.
Sep 15, 2015

0

Explanation:

y'=1/(((x^2-1)^(1/2))^2+1)1/2(x^2-1)^(-1/2)2x+(arcsin(1/x))'
=1/(x^2-1+1)1/2(2x)/sqrt(x^2-1)+1/sqrt(1-(1/x)^2)(-1/x^2)=
=x/(x^2sqrt(x^2-1))-1/(x^2sqrt((x^2-1)/x^2))=
=1/(xsqrt(x^2-1))-1/(xsqrt(x^2-1))=0

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