设y=ln(x+根号下(x^2+a^2)),求dy.
人气:143 ℃ 时间:2019-09-06 05:10:59
解答
∵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′=1+x/√(x^2+a^2)=[x+√(x^2+a^2)]/√(x^2+a^2),
∴y′=1/√(x^2+a^2),
∴dy=[1/√(x^2+a^2)]dx.
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