$$y=\frac{e^x-1}{e^x+1}$$

(正答率 75%)

$$\begin{eqnarray*} y'&=&\frac{(e^x+1)'(e^x-1)-(e^x+1)(e^x-1)'}{(e^x+1)^2}\\ &=&... ...e^x(e^x-1)-(e^x+1)e^x}{(e^x+1)^2}\\ &=&-\frac{e^x}{(e^x+1)^2} \end{eqnarray*}$$