题目

用定义法求3次根号下X的平方的导数,

提问时间：2020-08-25

答案

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-h=(x+h)(x-h)
立方和:x+h=(x+h)(x-xh+h)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-h=(x-h)(x+xh+h)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)
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)]-[x^(1/3)]}=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)