Question

有关导数的定义的题，两道，求解

Answer 1

解：
1.
原式=lim(k->0)(-1/2)*{f[x0+(-k)]-f(x0)}/(-k)
由导数的定义lim(k->0){f[x0+(-k)]-f(x0)}/(-k)=f'(x0)
原式=-1/2*f'(x0)
=-1
2.
原式=lim(h->0)[f(x0+h)-f(x0)+f(x0)-f(x0-3h)]/h
=lim(h->0)[f(x0+h)-f(x0)]/h+lim(h->0)[f(x0)-f(x0-3h)]/h
=lim(h->0)[f(x0+h)-f(x0)]/h+3lim(h->0)[f(x0)-f(x0-3h)]/3h
=f'(x0)+3f'(x0)
=-3+3*(-3)
=-12

Answer 2

lim(k->0) [f(x0+(-k) -f(x0)]/(-k) * 1/(-2)
=-1/2 *f'(x0)
= -1

lim(h->0) [(f(x0+h) -f(x0)) + (f(x0)- f(x0+(-3h))]/h
= lim(h->0) [f(x0+h) -f(x0)]/h - lim(h->0) [(f(x0+(-3)- f(x0)]/(-3h) *(-3)
= f'(x0) - (-3)*f'(x0)
= -12