How to take the frobenius norm of Jacobian of inner product of these two q dimensional vectors? $$
\begin{array}{l}
B_1 \cdot s=\left[\begin{array}{cccc}
b_{11} & b_{12} & \cdots & b_{1d_1} \\
\vdots & \cdots & & \vdots \\
b_{q_1 1} & \cdots    & b_{q_1 d_1}
\end{array}\right]_{q_1 \times d_1} \cdot \left[\begin{array}{c}
s_1 \\
s_2 \\
s_3 \\
s_{d_1}
\end{array}\right]_{d_1 \times 1} \\
=\left[\begin{array}{ccc}
b_{11} s_1+b_{12} s_2+\cdots b_{1 d_1} s_{d_1} \\
\vdots \\
b_{q_1 1} s_1 & \cdots & b_{q_1 d_1} s_{d_1}
\end{array}\right]_{q_1 \times 1} \\
\sigma\left(B_1 \cdot s\right)=\left[\begin{array}{c}
\sigma\left(b_{11} s_1+\cdots b_{1 d_1} s_{d_1}\right) \\
\vdots \\
\sigma\left(b_{q_1} s_1+\cdots b_{q_1 d_1} s_{d_1}\right)
\end{array}\right]_{q_1 \times 1} \\
=\left[\begin{array}{c}
u_1 \\
u_2 \\
\vdots \\
u_{q_1}
\end{array}\right]_{q_1 \times 1} \\
\end{array}
$$
Following the same calculation for $\sigma\left(T_1 p\right)$ $=\left[\begin{array}{cc}\sigma\left(t_{11} p_1+\cdots \cdot t_{1 d_2} p_{d_2}\right) \\ \vdots \\ \sigma\left(t_{q_2 1} p_1+\cdots \cdot t_{q_2 d_2} p_{d_2}\right)\end{array}\right]$ $=\left[\begin{array}{c}v_1 \\ v_2 \\ v_{q_2}\end{array}\right]_{q_2 \times 1}$
$h=\left\langle\sigma\left(B_1 \cdot s\right), \sigma\left(T_1 \cdot p\right)\right\rangle$
How to get $h$ ? Here $\sigma$ = sigmoid
Edit
Here $q_1 = q_2 = q$ and ,$d$ = 2
I particularly interested to implement this calculation for my $h(s,p)$ . But getting no clue :(


 A: For typing convenience, define the variables
$$\eqalign{
\def\s{\sigma}
\def\D{{\rm Diag}}
\def\qiq{\quad\implies\quad}
\def\p{{\partial}}
\def\g#1#2{\frac{\p #1}{\p #2}}
z &= B_1s,\qquad e=\s(z),\qquad E=\D(e) \\
y &= T_1p,\qquad f=\s(y),\qquad F=\D(f) \\
}$$
The derivative of the logistic sigmoid is well known
$$\eqalign{
de &= \left(E-E^2\right)dz \\
df &= \left(F-F^2\right)dy 
\qquad\qquad\qquad\qquad\qquad \\
}$$
Now calculate the derivative of the $h$-function wrt some independent $x$ variable
$$\eqalign{
h &= e^Tf \\
dh &= e^Tdf \;+\; f^Tde \\
 &= e^T\left(F-F^2\right)dy \;+\; f^T\left(E-E^2\right)dz \\
 &= e^T\left(F-F^2\right)\g{y}{x}\:dx \;+\; f^T\left(E-E^2\right)\g{z}{x}\:dx \\
J = \g{h}{x}
 &= e^T\left(F-F^2\right)\g{y}{x} \;+\; f^T\left(E-E^2\right)\g{z}{x} \\
}$$
The norm of $J$ can be calculated as
$$\|J\|_F^2 = {\rm Trace}\left(J^TJ\right)$$
Update
The independent variable $x$ was not specified in the question, but assuming $\,x=s$
$$\eqalign{
\g{z}{x} = \g{(B_1s)}{s} = B_1,\qquad
 \g{y}{x} = \g{(T_1p)}{s} = 0\; \\
}$$
Conversely, assuming $\,x=p\;$ yields
$$\eqalign{
\g{z}{x} = \g{(B_1s)}{p} = 0,\qquad
\;\; \g{y}{x} = \g{(T_1p)}{p} = T_1 \\
}$$
