y=ln(x- √(x^2-1))y' =[1/(x- √(x^2-1))]. d/dx (x- √(x^2-1)) =[1/(x- √(x^2-1))]. (1- x/√(x^2-1) ) = -1/√(x^2-1)
y'=1/[x-√(x^2-1)] × [1-x/√(x^2-1)]=1/[x-√(x^2-1)] × [(√(x^2-1)-x)/√(x^2-1)]=-1/√(x^2-1)
