Question

y=x^(2/3)的导数，用定义求导，谢谢

Answer 1

令x=t³，x0=(t0)³则
f'(x0)=lim【x→x0】
[f(x)-f(x0)]/(x-x0)
=lim【x→x0】
[x^(2/3)-(x0)^(2/3)]/(x-x0)
=lim【t→t0】
[t²-(t0)²]/[t³-(t0)³]
=lim【t→t0】(t+t0)/(t²+t*t0+(t0)²)
=2t0/3(t0)²
=(2/3)*(t0)^(-1)
=(2/3)*(x0)^(-1/3)
所以可知y=x^(2/3)的导数为y'=(2/3)*x^(-1/3)
不明白可以追问，如果有帮助，请选为满意回答！

Answer 2

令x=t³,x0=(t0)³则
f'(x0)=lim【x→x0】 [f(x)-f(x0)]/(x-x0)
=lim【x→x0】 [x^(2/3)-(x0)^(2/3)]/(x-x0)
=lim【t→t0】 [t²-(t0)²]/[t³-(t0)³]
=lim【t→t0】(t+t0)/(t²+t*t0+(t0)²)
=2t0/3(t0)²
=(2/3)*(t0)^(-1)
=(2/3)*(x0)^(-1/3)
所以可知y=x^(2/3)的导数为y'=(2/3)*x^(-1/3)
或者 由立方差公式,
△y＝(x+h)^(2/3)－x^(2/3)

＝[(x+h)^2-x^2]/[(x+h)^(4/3)＋(x^2+hx)^(2/3)＋x^(4/3)]

＝h(2x＋h)/[(x+h)^(4/3)＋(x^2+hx)^(2/3)＋x^(4/3)]

所以,

△y/h＝(2x＋h)/[(x+h)^(4/3)＋(x^2+hx)^(2/3)＋x^(4/3)],

当h→0时,极限是 2x/[x^(4/3)＋x^(4/3)＋x^(4/3)]

＝2/3×x^(-1/3)

所以,x^(2/3)的导数是2/3×x^(-1/3)