怎样证明乘法求导法则

2025-05-02 09:37:21
推荐回答(1个)
回答1:

(f(x)g(x))'
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x)]/h
=lim(h→0)[f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)]/h
=lim(h→0)g(x+h)*[f(x+h)-f(x)]/h+f(x)*[g(x+h)-g(x)]/h
=g(x)f'(x)+f(x)g'(x)

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