[微積] x^(x)作微分
題目:f(x)= x^(x),請求解f'(x)
答案:[x^(x)]*[1+lnx]
小弟的想法:
(解法一)
設 y=x^(x) 採對數解
lny = ln[x^(x)]
(d/dx)lny=(d/dx)ln[x^(x)]
y'/y = (d/dx)x*lnx
y'/y = 1*lnx + x*(1/x)
y'/y=(1+lnx)
y'=(1+lnx)*y
y'=f'(x)=[x^(x)]*(1+lnx)
------------------------------------------------------------------------------
(解法二)
e^(lnx) = x
x^(x) = [e^(lnx)]^(x)
(d/dx)*[e^(lnx)]^(x)
= (d/dx)e^(xlnx)
= [e^(xlnx)]*(d/dx)(xlnx)
= [e^(xlnx)]*[1*lnx+x*(1/x)]
= [e^(xlnx)]*(lnx+1)
= [e^(xlnx)]*(1+lnx)
= [x^(x)]*(1+lnx)
以上為小弟的解法,雖然答案一樣,但是不太清楚自己的計算過程是否有錯誤,
麻煩版上前輩們不吝嗇指導,謝謝!
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