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f(x)=x(x-2)^2/3 求导化简求具体过程!
人气:140 ℃ 时间:2020-05-14 12:39:06
解答
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- -括号写清楚啊不好意思阿帮我化简一下吧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)}谢谢你,但是恐怕不对吧哪里不对?

我觉得应该这样继续化简题目要求单调区间和极值- -早点发题目图片上来嘛因为我求导化简有困难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) 注意括号还有先次方后乘除,就帮到这了。好的谢谢你了
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