[quote="rainy"](3) 定義a^x 和logx (以a為底)互為反函數
可以得到
(lnx)'=1/x
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請稍加說明
上面Errfree大大所寫的就是證明過程,我使用了他的結論
#ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第一題沒得解#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為這只是e的定義#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#好像你要定義π#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#它不就是圓周與直徑之比嗎?#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這只是定義!#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而為什麼e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#的積分還是e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#則可以解,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#設我們不知道哪一個funtion的積分還是它自己,設它是y#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#∫ydx=y#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#兩邊取導數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#y=dy/dx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#移項得#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#dy/y=dx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#兩邊積分,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#In y=x#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#移項得#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#y=e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#的積分等於它自己#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#chongxe 寫到:為什麼lim(1+1/x)^x , x->infinite = e#ed_op#BR#ed_cl##ed_op#BR#ed_cl#為什麼 e^x的微分還是e^x