设y=√(x2-1)lnx-arctan√(x2-1),求y(√5)．

设y=√(x²-1)lnx-arctan√(x²-1),求y(√5)．
【正确答案】：y'=2xlnx/2√(x²-1)+√(x²-1)•1/x-1/{1+[√(x²-1)]²}•2x/[2√(x²-1)] =xlnx/√(x²-1)+√(x²-1)/x-1/[x√(x²-1)] y'(√5)=(√5ln√5)/2+2/√5-1/(2√5)=(√5/4)ln5+(3/10)√5