6#
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小 發表於 2011-5-5 09:59 PM (第 4740 天)
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u = x^(2x) + 1
u-1 = x^(2x)
ln(u-1) = 2x ln x
Diff. both sides w.r.t. x,
1/(u-1) *(du/dx)= 2(1+ln x)
du/dx = 2(u-1)(1+ln x)
h'(x)
=dh/du * du/dx
=(1/u) * 2(u-1)(1+ln x)
=2(u-1)(1+ln x) / u
When x=2, u=2^4 + 1=17
h'(2) = 2*16 (1+ln 2) / 17
=(32/17) (1+ln 2)
I hope that I didn't make careless mistakes.
[ 本帖最後由 桃子 於 2011-5-5 10:02 PM 編輯 ]