## 3 Capacity Analysis of the IWMSNs
### The Linear MMSE Capacity
\begin{array}{l}\frac{d}{d_{7}}\left(\frac{d\mathbf{F}(u)}{du}\right)=\frac {d}{d_{7}}\left(\frac{1}{N}\frac{\prod_{i=1}^{N}\left(k-1\right)!}{V(-\gamma \mathbf{I}\right)\mathbf{\
abla}(\mathbf{0})}\frac{\det\mathbf{V}(u)}{\mathbf{ G}(u)}\cdot\left(\text{tr}\left(\mathbf{Z}(u)\cdot\left(\log(1+\gamma\theta_{i} \xi_{j})\cdot(1+\gamma\theta_{i}\xi_{j})^{u+N-1}\right)_{ij}\right)-\sum_{n=1}^ {N-1}\frac{n}{n+n}\right)\\ =\frac{1}{N}\frac{\prod_{i=1}^{N-1}\left(k-1\right)!}{V(\mathbf{\
abla}( \mathbf{0})\mathbf{G}(u)}\frac{d}{d_{7}}\left((-\gamma)^{-\frac{N(N-1)}{2}} \cdot\det\mathbf{Y}(u)\cdot\left(\text{tr}\left(\mathbf{Z}(u)\cdot\left( \log(1+\gamma\theta_{i}\xi_{j})\cdot(1+\gamma\theta_{i}\xi_{j}\right)^{u+N-1} \right)_{ij}\right)-\sum_{n=1}^{N-1}\frac{n}{n+n}\right)\right)\\ =\frac{1}{N}\frac{(-\gamma)^{-\frac{1}{2}N(N-1)}\prod_{i=1}^{N}\left(k-1\right)! }{V(\mathbf{\
abla}(\mathbf{0})\mathbf{G}(u)}\left(-\frac{1}{2\gamma}N(N-1) \cdot\det\mathbf{Y}(u)\cdot\text{tr}\left(\mathbf{Z}(u)\cdot\left(\log(1+ \gamma\theta_{i}\xi_{j})\cdot(1+\gamma\theta_{i}\xi_{j})^{u+N-1}\right)_{ij} \right)\\ +\frac{d}{d_{7}}\left(\det\mathbf{Y}(u)\cdot\text{tr}\left(\mathbf{Z}(u)\cdot \left(\log(1+\gamma\theta_{i}\xi_{j})\cdot(1+\gamma\theta_{i}\xi_{j})^{u+N-1 }\right)_{ij}\right)\\ +\det\mathbf{Y}(u)\cdot\frac{d}{d_{7}}\left(\text{tr}\left(\mathbf{Z}(u) \cdot\left(\log(1+\gamma\theta_{i}\xi_{j})\cdot(1+\gamma\theta_{i}\xi_{j} \right)^{u+N-1}\right)_{ij}\right)\\ +\frac{1}{2\gamma}N(N-1)\cdot\det\mathbf{Y}(u)\cdot\sum\limits_{n=1}^{N-1} \frac{n}{n+n}-\frac{d}{d_{7}}\left(\det\mathbf{Y}(u)\right)\cdot\sum\limits_{n= 1}^{N-1}\frac{n}{n+n}\end{array}  \tag{33}\]  
Being similar to the process of solving Equation (22), we get:  
\[\begin{array}{l}\frac{d}{d_{7}}\left(\det\mathbf{Y}(u)\right)\\ =\det\mathbf{Y}(u)\cdot\text{tr}\left(\mathbf{Y}(u)^{-1}\cdot\frac{d\mathbf{Y }(u)}{d_{7}}\right)\\ =\det\mathbf{Y}(u)\cdot\text{tr}\left(\left(\left((1+\gamma\theta_{i}\xi_{j})^ {u+N-1}\right)_{ij}\right)^{-1}\cdot\left(\frac{(u+N-1)\theta_{i}\xi_{j}}{1+ \gamma\theta_{i}\xi_{j}}\cdot(1+\gamma\theta_{i}\xi_{j})^{u+N-1}\right)_{ij} \right)\\ =\det\mathbf{Y}(u)\cdot\text{tr}\left(\mathbf{Z}(u)\cdot\left(\frac{(u+N-1) \theta_{i}\xi_{j}}{1+\gamma\theta_{i}\xi_{j}}\cdot(1+\gamma\theta_{i}\xi_{j} )^{u+From [55], the differential of trace \(d(\text{tr}(\mathbf{X}))=\text{tr}(d(\mathbf{X}))\); thus:  
\[ \begin{split}&\tfrac{d}{d\gamma}\text{tr}\left(\mathbf{Z}(u)\cdot \left(\log(1+\gamma\theta_{i}\zeta_{j})\cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N- 1}\right)_{ij}\right)\\ &=\text{tr}\left(\tfrac{d}{d\gamma}\left(\left(((1+\gamma\theta_{i }\zeta_{j})^{u+N-1})_{ij}\right)^{-1}\cdot\left(\log(1+\gamma\theta_{i}\zeta_{ j})\cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\right)_{i,j}\right)\right)\\ &=\text{tr}\left(\mathbf{Z}(u)\cdot\left(\log(1+\gamma\theta_{i} \zeta_{j})\cdot\tfrac{(u+N-1)\theta_{i}\zeta_{j}}{1+\gamma\theta_{i}\zeta_{j}} \cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\right)_{i,j}\right.\\ &\left.+\mathbf{Z}(u)\cdot\left(\tfrac{\theta_{i}\zeta_{j}}{1+ \gamma\theta_{i}\zeta_{j}}\cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\right)_{ i,j}\right.\\ &\left.-\tfrac{d}{d\gamma}\Big{(}\big{(}(1+\gamma\theta_{i}\zeta_ {j})^{u+N-1}\big{)}_{i,j}\Big{)}^{-1}\cdot\left(\log(1+\gamma\theta_{i}\zeta_{ j})\cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\right)_{i,j}\Big{)}\right.\end{split}  \tag{35}\]  
Furthermore, from [55], the differential of inverse matrix \(d(\mathbf{X}^{-1})=-\mathbf{X}^{-1}(d\mathbf{X})\mathbf{X}^{-1}\); then, we get:  
\[ \begin{split}&\tfrac{d(\mathbf{Z}(u))}{d\gamma}\\ &=-\bigg{(}\Big{(}(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\Big{)}_{i,j}\bigg{)}^{-1}\cdot\tfrac{d}{d\gamma}\Big{(}(1+\gamma\theta_{i}\zeta_{j})^{ u+N-1}\Big{)}_{i,j}\cdot\left(\Big{(}(1+\gamma\theta_{i}\zeta_{j})^{u+N-1} \Big{)}_{i,j}\right)^{-1}\\ &=-\mathbf{Z}(u)\cdot\Big{(}\tfrac{\theta_{i}\zeta_{j}}{1+\gamma \theta_{i}\zeta_{j}}\cdot(1+\gamma\theta_{i}\zeta_{j})^{u+N-1}\Big{)}_{i,j} \cdot\mathbf{Z}(u)\end{split}