导数的乘法法则推倒uv)'=lim(h→0)[u(x+h)v(x+h)-uv]/h=lim(h→0)[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h=lim(h→0)[u(x+h)]×[v(x+h)-v(x)]/h+lim(h→0)[v(x)]×[u(x+h)-u(x)]/h=u(x)v'(x)+u'(x)v(x)=u'v+uv'请问这个第一步lim(h→0)[u(x+h)v(x+h)-

　　导数的乘法法则推倒

　　uv)'=lim(h→0)[u(x+h)v(x+h)-uv]/h

　　=lim(h→0)[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h

　　=lim(h→0)[u(x+h)]×[v(x+h)-v(x)]/h+lim(h→0)[v(x)]×[u(x+h)-u(x)]/h

　　=u(x)v'(x)+u'(x)v(x)

　　=u'v+uv'请问这个第一步lim(h→0)[u(x+h)v(x+h)-uv]/h是怎么来的

　　我算出来的的第一步是

　　[u(x+h)-u(x)]/h*[v(x+h)-v(x)]/h

　　=u(x+h)v(x+h)-u(x+h)v(x)-u(x)v(x+h)+u(x)v(x)/h*h

　　然后呢因为h->0就消掉么?