hhww81
幼苗
共回答了9个问题采纳率:100% 举报
(a^x)'
=lim(h->0) [a^(x+h) - a^x]/h
=lim(h->0) a^x*[a^h - 1]/h
① ∵ (h->0) [a^h - 1] h*lna
=lim(h->0) a^x*[h*lna]/h
= a^x*lna
② ∵ 令:t=[a^h - 1] 则:h->0 时,t->0,h=ln(1+t)] *(1/lna )
=lim(t->0) a^x* t/[ln(1+t)] *1/lna ]
=lim(t->0) a^x*lna* t/ln(1+t)
= a^x*lna
(loga x)' (x>0)
=[(1/lna)*lnx]'
=(1/lna)*[lnx]'
=(1/lna)* lim(h->0) [ln(x+h) - lnx]/h
=(1/lna)* lim(h->0) [ln(1+h/x)]/h
∵ (h->0时,h/x ->0 ,ln(1+h/x) h/x
=(1/lna)* lim(h->0) [h/x]/h
= 1/lna*(1/x)
= 1/xlna
1年前
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