What is the derivative of #sqrt( x^2-1)#?

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Answer 1 · 2015-12-27T17:09:18

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

This is equivalent to #(x^2-1)^(1/2)#.

According to the chain rule:

#d/dx(u^(1/2))=1/2u^(-1/2) * (du)/dx#

Thus,

#d/dx((x^2-1)^(1/2))=1/2(x^2-1)^(-1/2)d/dx(x^2-1)#

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

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