公式[v(x)yx)=l(x)y(x)+l(x)v(x)的证明: Lu(x) v(x]=im u(x+h)v(x +hu(v(x) h>0 im;[v(x+h)v(x+h)-l4(x)v(x+h)+(x)w(x+h)-l(x)v(x) h->0 h im l(x+h)-(x) v(x+h+u(x) v(x +h)-v(x) h→>0 h h Elim u(x+h)u( Hm v(xth)+u(xlm v(x+h)-v(x) h->0 h h→>0 h->0 =(x)v(x)+(x)v(x), 其中imnu(x+h)=(x)是由于v(x)存在 h→>0 反囗
上页 返回 下页 公式 [u(x)v(x)] =u(x)v(x)+u(x)v(x)的证明: = 0 lim h→ h u(x + h)v(x + h) − u(x)v(x) = 0 lim h→ h 1 [u(x+h)v(x+h)−u(x)v(x+h)+u(x)v(x+h)−u(x)v(x)] = 0 lim h→ + − h u(x h) u(x) v(x+h)+u(x) + − h v(x h) v(x) = 0 lim h→ h u(x + h) − u(x) 0 lim h→ v(x+h)+u(x) 0 lim h→ h v(x + h) − v(x) =u(x)v(x)+u(x)v(x), [u(x)v(x)] 其中 0 lim h→ v(x+h)=v(x)是由于 v(x)存在。 返回