y'=lim[h→0] [f(x+h)-f(x)]/h
=lim[h→0] [(x+h)^(1/6)-x^(1/6)]/h
=lim[h→0] x^(1/6)[(1+h/x)^(1/6)-1]/h
等價(jià)無窮小代換：u→0時(shí),(1+u)^a-1等價(jià)于au
=lim[h→0] x^(1/6)[(1/6)(h/x)]/h
=[(1/6)x^(1/6)]/x
=(1/6)x^(-5/6)
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