导数公式表
请帮我证明(x^a)'=ax^(a-1),a为常数且a≠0
(sinx)'=cosx
人气:293 ℃ 时间:2020-05-12 05:42:55
解答
令:T=x^a
则 lnT=ln(x^a)
lnT=alnx
两边同时求导
1/T*T'=a*1/x
T'=a*1/x*T
T'=ax^(a-1)
得证
(sinx)'=(sin(x+o)-sinx)/o
=(sinxcoso+cosxsino-sinx)/o
然后用极限o趋近0时,coso=1;sino/o=1
得上式=cosx
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