sllyyy
幼苗
共回答了18个问题采纳率:83.3% 举报
服你了.
△y=y(x+h)-y(x) (h=△x,为阅读方便)
=1/√(x+h)-1/√x
=(√x-√(x+h)) / √(x(x+h)) (通分)
=-h/[√(x(x+h))*(√x+√(x+h))](上下都乘(√x+√(x+h)))
△y/h=-1/[√(x(x+h))*(√x+√(x+h))]
lim△y/h=-1/[√(x(x+0))*(√x+√(x+0))] h->0
==-1/(2x√x)
明白了,你是想用定义来求
y'=lim[y(x+h)-y(x)]/h h->0
=lim[1/√(x+h)-1/√x]/h
=lim[(√x-√(x+h)) / √(x(x+h))]/h
=lim{-h/[√(x(x+h))*(√x+√(x+h))]}/h
=lim-1/[√(x(x+h))*(√x+√(x+h))]
=-1/[√(x(x+0))*(√x+√(x+0))]
=-1/(2x√x)
下面是原来的复合函数求导
dy=d(1/√x)
=d((√x)^-1)
=-1*(√x)^(-2)*d(√x)
=(-1/x)* [(1/2)*(x)^(-1/2)*dx]
=(-1/2x)*(1/√x)dx
=(-1/2x√x) dx
1年前
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