y=(x-1)乘三次根号下x的平方,令Y的导数等于0，那么如何求X的值？
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解决时间 2021-03-03 14:11
- 提问者网友：伴风望海
- 2021-03-03 04:02
y=(x-1)乘三次根号下x的平方,令Y的导数等于0，那么如何求X的值？
最佳答案
- 五星知识达人网友：玩世
- 2021-03-03 05:27
= (x-1)乘三次根号下x² = (x-1) * x^(2/3)
y' = (x-1) * [x^(2/ + (x-1)' * x^(2/3)
= (x-1) * 2/3* x^(-1/ [3x^(1/ x^(1/3 (x-1) + x ] / x^(1/3)
= [2(x-1) + 3x ] / [3x^(1/3)]
= [2x-2 + 3x ] /3)]
= 5(x-2/5) / [3x^(1/3)] = 0
x = 2/3)
= 2/3) + x^(2/3)
= 2/3 (x-1) /3 (x-1) * x^(-1/3) + x^(2/3)
= [2/3)]'3) + 1 * x^(2/
y' = (x-1) * [x^(2/ + (x-1)' * x^(2/3)
= (x-1) * 2/3* x^(-1/ [3x^(1/ x^(1/3 (x-1) + x ] / x^(1/3)
= [2(x-1) + 3x ] / [3x^(1/3)]
= [2x-2 + 3x ] /3)]
= 5(x-2/5) / [3x^(1/3)] = 0
x = 2/3)
= 2/3) + x^(2/3)
= 2/3 (x-1) /3 (x-1) * x^(-1/3) + x^(2/3)
= [2/3)]'3) + 1 * x^(2/
全部回答
- 1楼网友：举杯邀酒敬孤独
- 2021-03-03 05:50
y=(x-1)*[(3x+1)^(2/3)]*(x-2)
y'=(x-1)'*[(3x+1)^(2/3)]*(x-2)+(x-1)*[(3x+1)^(2/3)]'*(x-2)+(x-1)*[(3x+1)^(2/3)]*(x-2)'
=[(3x+1)^(2/3)]*(x-2)+(x-1)*2[(3x+1)^(-1/3)*(x-2)+(x-1)*[(3x+1)^(2/3)]
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