x→0时,lim{lncos2x / lncos3x} =lim{ (lncos2x )' / (lncos3x)' }

coolggwei2022-10-04 11:39:541条回答

x→0时,lim{lncos2x / lncos3x} =lim{ (lncos2x )' / (lncos3x)' } =lim{ 2tan2x / 3tan3x}我想问下
(lncos2x )'=2tan2x的具体步骤

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香香的心路历程共回答了18个问题 | 采纳率88.9%: (lncos2x )'=1/cos2x *(-sin2x)*2=-2tan2x
(lncos3x)'=1/cos3x *(-sin3x)*3=-3tan3x; 1年前