用定义法求3次根号下X的平方的导数，谢谢啦！！！

这个吧,要用因式分解的平方差,立方和和立方差公式,挺麻烦的
平方差:x�0�5-h�0�5=(x+h)(x-h)
立方和:x�0�6+h�0�6=(x+h)(x�0�5-xh+h�0�5)而(x+h)=(x^1/3+h^1/3)(x^2/3-x^1/3*h^1/3+h^2/3)
立方差:x�0�6-h�0�6=(x-h)(x�0�5+xh+h�0�5)而(x-h)=(x^1/3-h^1/3)(x^2/3+x^1/3*h^1/3+h^2/3)
过程中是乘以(x^2/3-x^1/3*h^1/3+h^2/3)和(x^1/3-h^1/3)(x^2/3+x^1/3*h^1/3+h^2/3)凑出(x+h)和(x-h)

过程繁复,这样好看点吧:
y=x^(2/3),根据导数基本定义,f'(x)=lim(h→0) [f(x+h)-f(x)]/h

导数y'=lim(h→0) 1/h*[(x+h)^(2/3)-x^(2/3)]
=lim(h→0) 1/h*{[(x+h)^(1/3)]�0�5-[x^(1/3)]�0�5}

=lim(h→0)
分子:[(x+h)^(1/3)+x^(1/3)]*[(x+h)^(1/3)-x^(1/3)],这里是平方差公式
分母:h

=lim(h→0)
分子:[(x+h)^(1/3)+x^(1/3)][(x+h)^(2/3)-(x+h)^(1/3)*x^(1/3)+x^(2/3)],这里是立方和公式
*[(x+h)^(1/3)-x^(1/3)][(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)],这里是立方差公式
分母:h*[(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)]

=lim(h→0)
分子:[(x+h)+x)][(x+h)-x]
分母:h*[(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)]

=lim(h→0)
分子:2x+h
分母:[(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)],约去h

=1/[x^(2/3)+x^(2/3)+x^(2/3)][x^(2/3)-x^(2/3)+x^(2/3)]*(2x)
=1/[3x^(2/3)*x^(2/3)]*2x
=2/3*x/x^(4/3)
=2/3*1/x^(1/3)
=2/[3x^(1/3)],即导数为(三乘以x的立方根)分之(二)

若用导数公式(x^n)'=nx^(n-1)过程简单多了,就是[x^(2/3)]'=(2/3)*x^(2/3-1)=(2/3)*x^(-1/3)=2/[3x^(1/3)]

数学不好啊

这个啊 ，，，，不是很会啊