respect to x.
f(x)=e^(-x)
麻煩幫忙解答=)
更新1:
我想問點解由 e^(-x)lim (h->0) [e^(-h)-1]/h 跳到下一步既？ =e^(-x)lim (h->0) [-e^(-h)] 可唔可以解一解？
Math stats問題
2009-01-31 12:23 am
differentiate from first principles, the following functions with
respect to x.
f(x)=e^(-x)
麻煩幫忙解答=)
respect to x.
f(x)=e^(-x)
麻煩幫忙解答=)
回答 (2)
2009-01-31 4:52 am
✔ 最佳答案
myisland8132兄：Maths & Stats冇教L'Hospital Rule。
仲有，計First Principle應該唔可以用呢招……
2009-01-30 20:52:29 補充：
First Principle:
f(x) = e^{-x}
f'(x) = lim(h→0) [e^{-(x+h)} - e^{-x}]/h
= lim(h→0) e^{-x}*[e^{-h} - 1]/h
= e^{-x}*lim(h→0) [e^{-h} - 1]/h
= ...
Now let k = e^{-h} - 1, then h = - ln(1+k),
also, when h→0, we have k→0,
and so
...
= e^{-x}*lim(k→0) k/[- ln(1+k)]
= - e^{-x}*[lim(k→0) ln(1+k) / k]^{-1}
= - e^{-x}*[ ln [lim(k→0) (1+k)^{1/k}]]^{-1}
= - e^{-x}*[ln e]^{-1}
= - e^{-x}
where lim(k→0) (1+k)^{1/k}= e is the definition of e.
參考: ME
2009-01-31 12:27 am
f(x)=e^(-x)
f'(x)
=lim (h->0) f(x+h)-f(x)/h
=lim (h->0) [e^(-x-h)-e(-x)]/h
=lim (h->0) e^(-x)[e^(-h)-1]/h
=e^(-x)lim (h->0) [e^(-h)-1]/h
=e^(-x)lim (h->0) [-e^(-h)]
=-e^(-x)
f'(x)
=lim (h->0) f(x+h)-f(x)/h
=lim (h->0) [e^(-x-h)-e(-x)]/h
=lim (h->0) e^(-x)[e^(-h)-1]/h
=e^(-x)lim (h->0) [e^(-h)-1]/h
=e^(-x)lim (h->0) [-e^(-h)]
=-e^(-x)
收錄日期: 2021-04-26 13:04:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090130000051KK01304