Question

设y=ln(x+根号下(x^2+a^2)),求dy.

Answer 1

∵y＝ln［x＋√（x^2＋a^2）］,∴e^y＝x＋√（x^2＋a^2）,∴（e^y－x）^2＝x^2＋a^2,∴2（e^y－x）（e^y－x）′＝2x,∴［x＋√（x^2＋a^2）－x］［（e^y）y′－1］＝x,∴（e^y）y′－1＝x/√（x^2＋a^2）,∴（e^y）y...