急 指數微分一題

f(x)=x^a^a+a^x^a+a^a^xd/dx[f(x)]=(a^a)[x^(a^a-1)] +(a^x^a)(ln a)[ax^(a-1)] +(a^a^x)(ln a)(a^x)(lna)d/dx[f(x)]=(a^a)x^(a^a-1) +a^(x^a+1)x^(a-1)ln a +a^(a^x+x)ln²a

2011-12-26 18:22:31 補充：
Mine is correct, please refer to:

http://www.wolframalpha.com/input/?i=d%2Fdx%28f%28x%29%3Dx%5Ea%5Ea%2Ba%5Ex%5Ea%2Ba%5Ea%5Ex%29

tcy's ans is wrong because a^a^x=a^(a^x)=\=(a^a)^x,
so d/dx(a^a^x)=(a^a^x)(ln a)(a^x)(ln a)=\=(a^a^x)(ln a^a)

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