Prove of $\frac{\partial F}{\partial n^m}=\left(\frac{\partial n^{m+1}}{\partial n^m}\right)^T\frac{\partial F}{\partial n^{m+1}}$ How to prove this
$\dfrac{\partial F}{\partial n^m}=\left(\dfrac{\partial n^{m+1}}{\partial n^m}\right)^T\dfrac{\partial F}{\partial n^{m+1}}$ where $n$ is a vector ($\mathbf n$), $\left(\dfrac{\partial n^{m+1}}{\partial n^m}\right)$ is of size $S^{m+1}\times S^m$, and $\dfrac{\partial F}{\partial n^{m+1}}$ is of size $S^m\times1$?
Attempt: To simplify, assume $S^{m+1}=S^m=2.$
Thus
$\dfrac{\partial n^{m+1}}{\partial n^{m}}=\begin{pmatrix} \dfrac{\partial n_1^{m+1}}{\partial n_1^{m}}&\dfrac{\partial n_1^{m+1}}{\partial n_2^{m}}\\ \dfrac{\partial n_2^{m+1}}{\partial n_1^{m}} & \dfrac{\partial n_2^{m+1}}{\partial n_2^{m}}
\end{pmatrix},$ its transpose is $\left(\dfrac{\partial n^{m+1}}{\partial n^{m}}\right)^T=\begin{pmatrix} \dfrac{\partial n_1^{m+1}}{\partial n_1^{m}}&\dfrac{\partial n_2^{m+1}}{\partial n_1^{m}}\\ \dfrac{\partial n_1^{m+1}}{\partial n_2^{m}} & \dfrac{\partial n_2^{m+1}}{\partial n_2^{m}}
\end{pmatrix},$
and $\dfrac{\partial F}{\partial n^{m+1}}=\begin{pmatrix} \dfrac{\partial F}{\partial n_1^{m+1}}\\ \dfrac{\partial F}{\partial n_2^{m+1}}
\end{pmatrix}$
Thus $\left(\dfrac{\partial n^{m+1}}{\partial n^m}\right)^T\dfrac{\partial F}{\partial n^{m+1}}=\begin{pmatrix} \dfrac{\partial F}{\partial n_1^{m}}+ \dfrac{\partial F}{\partial n_1^{m}}\\ \dfrac{\partial F}{\partial n_2^{m}}+\dfrac{\partial F}{\partial n_2^{m}}
\end{pmatrix}.$
Which is not correct because $\dfrac{\partial F}{\partial n^m}=\begin{pmatrix} \dfrac{\partial F}{\partial n_1^{m}}\\ \dfrac{\partial F}{\partial n_2^{m}}
\end{pmatrix}.$
Why is that?
Thank you in advance.

This is from the book Neural Network Design by Hagan, Demuth, Beale, De Jesús, and it's a step of an equality from page [368] in the pdf book.
 A: In your simplification you are assuming that
$$\frac{\partial f}{\partial y} \frac{\partial y}{\partial x} = \frac{\partial f}{\partial x}$$
as would be the case with single variable chain rule. But the problem is that partial derivatives alone cannot account for the "whole" variation. If I have a change of variables from $(x,y)$ to $(s,t)$ the correct partials by chain rule would be as follows
$$\frac{\partial f}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$
This is because when we want to account for a variation in $s$ we have to hunt down and add all of its contributions since a change of variable could also be written as $f(s,t)=f(x(s,t),y(s,t))$. In your case this means the correct chain rule would give you
$$\frac{\partial F}{\partial n_1^{m+1}}\frac{\partial n_1^{m+1}}{\partial n_1^m} + \frac{\partial F}{\partial n_2^{m+1}}\frac{\partial n_2^{m+1}}{\partial n_1^m} = \frac{\partial F}{\partial n_1^m}$$
and so on for the second term.
