搵 f(x)=x^(1/3) 的derivative, 用first principles
請列出full step!!! PLEASE!!!
differentiation 問題
2011-06-11 8:16 am
回答 (3)
2011-06-11 5:03 pm
✔ 最佳答案
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EIther one of the last answers is correct.
2011-06-11 9:10 am
f(x+h)-f(x)
=(x+h)^(1/3)-x^(1/3)
=[(x+h)^(1/3)-x^(1/3)][(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]/[(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
=(x+h-x) / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
= h / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
(f(x+h)-f(x))/h
= 1 / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
take lim h to 0, the derivative is
f'(x) = 1 / (3x^(2/3))
=(x+h)^(1/3)-x^(1/3)
=[(x+h)^(1/3)-x^(1/3)][(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]/[(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
=(x+h-x) / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
= h / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
(f(x+h)-f(x))/h
= 1 / [(x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)]
take lim h to 0, the derivative is
f'(x) = 1 / (3x^(2/3))
2011-06-11 8:25 am
f(x) = e^(x+1)^(1/3)
ln f(x)=(x+1)^(1/3)
f'(x)/f(x)=(1/3)(x+1)^(-2/3)
f'(x)=(1/3)(x+1)^(-2/3)*e^(x+1)^(1/3)
吾sure嫁...
ln f(x)=(x+1)^(1/3)
f'(x)/f(x)=(1/3)(x+1)^(-2/3)
f'(x)=(1/3)(x+1)^(-2/3)*e^(x+1)^(1/3)
吾sure嫁...
收錄日期: 2021-04-24 23:23:01
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