A gradient by hand
A single sigmoid unit has z=w_0+w_1 x, w_0=0, w_1=1, and a single pattern x=2, y=0. Compute E=(y-o)^2, \partial E/\partial w_1 using the single-layer formula, and the same derivative by a central difference with \delta=0.001. Then show that the formula without the factor e^{-z} would give a different answer.
Here z=1\cdot 2=2, o=g(2)=0.8808 and E=(0-0.8808)^2=0.7758. The single-layer formula gives
\frac{\partial E}{\partial w_1} =-2(y-o)\,e^{-z} o^2 x =-2(-0.8808)(0.1353)(0.7758)(2) =0.3699.
Equivalently, 2\,o\cdot o(1-o)\,x=2(0.8808)(0.1050)(2)=0.3699.
Central difference with \delta=0.001: E(w_1=1.001)=g(2.002)^2 and E(0.999)=g(1.998)^2. Now g(2.002)=0.88101 and g(1.998)=0.88059, with squares 0.77617 and 0.77543. The difference divided by 0.002 is about 0.370, agreeing with 0.3699 to rounding.
Without the factor e^{-z} the formula would give 0.3699/0.1353=2.734, wrong by a factor e^2=7.39.