Question

用定义法求三次根号下x平方的导数

用定义法求三次根号下x平方的导数

Answer 1

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) [(x+h)^(2/3)-x^(2/3)]/h
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)]*[(x+h)^(1/3)-x^(1/3)][(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)],
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)]
因此y=x^(2/3)的导数为2/3*x^(-1/3)