Question

导数与微分

Answer 1

因为lim (h->0) h/[f(x0-2h)-f(x0)]=1/4,
所以lim (h->0) [f(x0-2h)-f(x0)]/h=4
从而
f'(x0)=lim(-2h->0) [f(x0+(-2h))-f(x0)]/(-2h)=4/(-2)=-2

Answer 2

lim[f(x0-2h)-f(x0)]/h=4
lim[f(x0)-f(x0-2h)]/h=-4
lim[f(x0)-f(x0-2h)]/(2h)=-4/2=-2
所以f'(x0)=lim[f(x0)-f(x0-2h)]/(2h)=-4/2=-2