lim(h->0)[e^(x+h)-e^x]/h
=lim(h->0)e^x[e^(h)-1]/h
=lim(h->0)e^x*h/h
=e^x
如果是a^x
a^x=e^xlna,同理可证;
lim(h->0)[log(a,x+h)-log(a,x)]/h
=lim(h->0)[log(a,1+h/x)]/h
=lim(h->0)[log(a,(1+h/x)^(1/h))]
=[log(a,e^(1/x))]
=1/x*log(a,e)
=1/(xlna)
任何一本高数书上都有啊!首先按导数的定义求对数函数的的导数,再根据反函数的性质求指数函数的导数就行啦。希望能帮到你