f'(x)=(x/3)'(x-2)^2+(x/3)[(x-2)^2]'
=(x-2)^2/3+(x/3)(2x-4)
=(x^2-4x+4)/3+(2x^2-4x)/3
=(3x^2-8x+4)/3
=(3x-2)(x-2)/3不是的，2/3是平方就是x(x-2) 的2/3次方- -括號寫清楚啊不好意思阿幫我化簡一下吧f(x)=(x^2-2x)^(2/3) f'(x)={2/[3(x^2-2x)^(1/3)]}*(x^2-2x)' = 4(x-1)/{3[x(x-2)]^(1/3)}謝謝你，但是恐怕不對吧哪里不對？

我覺得應該這樣繼續(xù)化簡題目要求單調(diào)區(qū)間和極值- -早點發(fā)題目圖片上來嘛因為我求導化簡有困難f(x)=x(x-2)^(2/3) f'(x)=1*(x-2)^(2/3)+x[(x-2)^(2/3) ]' =(x-2)^(2/3)+x[(x-2)^(2/3) ]' =(x-2)^(2/3)+x*1/[(x-2)^(1/3) ] =(x-2)^(2/3)+x(x-2)^(2/3) /(x-2) =[(x-2)^(2/3)]*[1+x/(x-2)] =[(x-2)^(2/3)]*[(2x-2)/(x-2)] =2(x-1)/(x-2)^(1/3) 注意括號還有先次方后乘除，就幫到這了。好的謝謝你了