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
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