lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
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![lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂](/uploads/image/z/5526062-62-2.jpg?t=lim+%5Bf%28x0%2Bh%29-f%28x0-h%29%5D%2Fh+%3D+2+lim+%5Bf%28x0%2Bh%29-f%28x0-h%29%5D%2F2h+%3D+2+f%27%28x0%29%E8%BF%99%E4%B8%AA%E6%88%91%E8%BF%98%E6%98%AF%E4%B8%8D%E6%87%82)
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lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
lim [f(x0+h)-f(x0-h)]/2h
凑成导数定义就是
= lim [f((x0-h)+2h)-f((x0-h)-0)]/2h
把2h看成delta-x
= lim [f'(x0-h)]
h-->0
=f'(x0)
lim [f(x0+h)-f(x0-h)]/h = 2 lim [f(x0+h)-f(x0-h)]/2h = 2 f'(x0)这个我还是不懂
若f′(x0)=-2,则lim[f(x0+h)-f(x0-h)]/h=
若f′(x0)=-3,则lim[f(x0+h)-f(x0-h)]/h=
若f'(x0)=-2,则lim[(f(x0-h/2)-f(x0))/h]=
为什么lim f(x0-h)-f(x0)/-h为什么是f'(x0)
lim h趋于0时,(f(x0+h)-f(x0-h))/2h=f`(x0) 看不懂
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0) 为什么不能这么做?原式=lim(h->0)[f(x0+h)-f(x0)]/h^2 — lim(h->0)[f(x0)-f(x0-h)]/h^2=f'(x)/h-f'(x)/h=0
设函数f(x)在x=x0处可导,则lim(h>0)[f(x0)-f(x0-2h)]/h
若f′(x0)=-3,则lim[f(x0+h)-f (x0-3h)]/h=
若f′(x0)=-3,则lim[f(x0+h)-f (x0-3h)]/h=过程
证明题:如果y=f(x)在x0处可导,那么lim(h->0)[f(x0+h)-f(x0-h)]/2h=f'(x0).证明逆定理全题:如果y=f(x)在x0处可导,那么lim(h->0)[f(x0+h)-f(x0-h)]/2h=f'(x0).反之,如果lim(h->0)[f(x0+h)-f(x0-h)]/2h存在,那么f'(x0)
设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h
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已知函数f(x)在x0可导,且lim(h→0)h/[f(x0-2h)-f(x0)]=1/4,则f‘(x0)=?
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其请问 lim(h→0) [ f(x0+3h)-f(x0-2h) ] / h 为什么是5f’(x0) 是怎么化的呀