Question

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

Answer 1

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`