以下を考えます．
$$\displaystyle\frac{d}{dx}\biggl(\displaystyle\frac{b_1(x)b_2(x)\cdots}{a_1(x)a_2(x)\cdots}\biggr)$$
まず規則を見たいので，以下の微分を先に見てみます．
$$\displaystyle\frac{d}{dx}\biggl\{\displaystyle\frac{b_1(x)b_2(x)}{a_1(x)a_2(x)}\biggr\}$$
$$=\displaystyle\frac{\{b_1(x)b_2(x)\}’a_1(x)a_2(x)-b_1(x)b_2(x)\{a_1(x)a_2(x)\}’}{\{a_1(x)a_2(x)\}^2}$$
$$=\displaystyle\frac{b_1(x)b_2(x)}{a_1(x)a_2(x)}\cdot\biggl\{\biggl(\displaystyle\frac{{b_1}'(x)}{b_1(x)}+\displaystyle\frac{{b_2}'(x)}{b_2(x)}\biggr)-\biggl(\displaystyle\frac{{a_1}'(x)}{a_1(x)}+\displaystyle\frac{{a_2}'(x)}{a_2(x)}\biggr)\biggr\}$$
同様の計算で$\displaystyle\frac{d}{dx}\biggl\{\displaystyle\frac{b_1(x)b_2(x)b_3(x)}{a_1(x)a_2(x)a_3(x)}\biggr\}$を計算すると
$$\displaystyle\frac{d}{dx}\biggl\{\displaystyle\frac{b_1(x)b_2(x)b_3(x)}{a_1(x)a_2(x)a_3(x)}\biggr\}$$
$$=\displaystyle\frac{b_1(x)b_2(x)b_3(x)}{a_1(x)a_2(x)a_3(x)}\cdot\biggl\{\biggl(\displaystyle\frac{{b_1}'(x)}{b_1(x)}+\displaystyle\frac{{b_2}'(x)}{b_2(x)}+\displaystyle\frac{{b_3}'(x)}{b_3(x)}\biggr)-\biggl(\displaystyle\frac{{a_1}'(x)}{a_1(x)}+\displaystyle\frac{{a_2}'(x)}{a_2(x)}+\displaystyle\frac{{a_3}'(x)}{a_3(x)}\biggr)\biggr\}$$
となるので，以下が推測できそうです．
$$\displaystyle\frac{d}{dx}\biggl\{\displaystyle\prod_i\displaystyle\frac{b_i(x)}{a_i(x)}\biggr\}=\displaystyle\prod_i \displaystyle\frac{b_i(x)}{a_i(x)}\biggl(\displaystyle\sum_j \displaystyle\frac{{b_j}'(x)}{b_j(x)}-\displaystyle\sum_k \displaystyle\frac{{a_k}'(x)}{a_k(x)}\biggr)$$
また，対数微分を利用すればスマートに導出できそうです．
$$y=\displaystyle\frac{b_1(x)b_2(x)\cdots b_n(x)}{a_1(x)a_2(x)\cdots a_n(x)}$$
とおき，両辺対数をとると
$$\log y=\displaystyle\sum_i \log b_i(x)-\displaystyle\sum_j \log a_j(x)$$
両辺微分して
$$\displaystyle\frac{y’}{y}=\displaystyle\sum_i\displaystyle\frac{{b_i}'(x)}{b_i(x)}- \displaystyle\sum_j \displaystyle\frac{{a_j}'(x)}{a_j(x)}$$
以上から
$$\displaystyle\frac{dy}{dx}=\displaystyle\prod_i
\displaystyle\frac{b_i(x)}{a_i(x)}\biggl(\displaystyle\sum_j\displaystyle\frac{{b_j}'(x)}{b_j(x)}- \displaystyle\sum_k \displaystyle\frac{{a_k}'(x)}{a_k(x)}\biggr)$$
となり同様の結果となりますね．