[jason1990]

Let W be normal with mean w and standard deviation dw. Let H be normal with mean h and standard deviation dh. Suppose W and H are independent. Let X=WH. Define A = EX and let dA be the standard deviation of X.

By independence, A=wh. Also,

(dA)^2 = E(X^2) - (EX)^2
= E(W^2)E(H^2) - w^2 h^2.

Note that (dw)^2=Var(W)=E(W^2)-w^2. Hence, E(W^2)=w^2+(dw)^2. We therefore have

(dA)^2 = (w^2+(dw)^2)(h^2+(dh)^2) - w^2 h^2.

The (dw)^2(dh)^2 term is asymptotically negligible, so this yields the correct formula.

Edited to add:
Here's another way to look at it. The term dw is the standard deviation of the error in the measurement of the width. Let E_w be the actual error. Your derivation simply reproduces the standard linear approximation

E_A = wE_h + hE_w.

This is valid. But dA is the standard deviation of the left-hand side. By independence, the variance of the sum on the right is the sum of the variances. But the same is not true for standard deviations.