求 y = [x + (x^2 + 24)^(1/2)]^(1/3) 的導數。
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求y=[x+(x^2+24)^(1/2)]^(1/3)的導
2011-08-27 9:27 am
回答 (1)
2011-08-27 4:56 pm
✔ 最佳答案
y = [x + (x^2 + 24)^(1/2)]^(1/3)y = t^1/3
t = x + (x^2 + 24)^(1/2)
dy/dt = (1/3)t^(-2/3)
dt/dx =1 +(1/2)(x^2 + 24)^(-1/2)d/dx(x^2 + 24)
dt/dx =1 +(1/2)(x^2 + 24)^(-1/2)(2x)
dt/dx =1 +x(x^2 + 24)^(-1/2)
dy/dx =( dy/dt)(dt/dx)
dy/dx = (1/3)t^(-2/3) [1 +x(x^2 + 24)^(-1/2)]
dy/dx = (1/3)[ x + (x^2 + 24)^(1/2)]^(-2/3) [1 +x(x^2 + 24)^(-1/2)]
收錄日期: 2021-04-22 00:10:31
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https://hk.answers.yahoo.com/question/index?qid=20110827000051KK00096