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I am not studying maths, hence I need a little help with the algebra performed in the following image:

enter image description here

I can follow every step until equation 20. Basically my question is how to come up with 20, starting at 19.

I am not sure about that derivative. It confuses me that apparently this is the derivative wrt. a row vector. Furthermore I do not know how the individual parts come into the equation. Would someone be so kind and explain to me what is happening here? What is the role of the transpose p and why are the parts the way they are stated?

I need to obtain a similar sensitivity matrix for a project, so understanding how this equation arises is important for me.

Edit: Since I managed to come up with the left part, given a certain way to write a product rule, I asked this more general question here: Product-rule for Jacobian calculation, i.e. $\frac{d}{dx}(Ay)$ where A is a matrix and y a vector and both depend on x . See my third comment. I would really appreciate it if someone can tell me if it is okay to write the product rule like this.

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  • $\begingroup$ Hello again,I managed to get the right side of the equation. There is no product rule involved. It is really a pure Jacobian. By solely looking at the first row and splitting eq. 19 into three parts: $(I-\Lambda-M)^{-1} \omega^m$, $(I-\Lambda-M)^{-1} b$ and $(I-\Lambda-M)^{-1} \eta$ it worked. In the process of that, I had to use eq. 13 and solve it for $\omega_i^m+b_i+\eta_i$. $\endgroup$
    – Elarion
    Mar 26, 2015 at 8:43
  • $\begingroup$ This yields the second and third diagonal matrix of $\omega_g$ and its abs(). The derivative needed is $\frac{\partial}{\partial x}\frac{1}{1-a-x}$, which is easy. Can you help me with the left part of the equation? That looks like a product rule to me, however the way I know it, the term $\frac{\partial \overline{G}}{\partial p}$ would come up. This is the derivative of a matrix wrt. a vector. From what I've read this is a third order tensor, so I do not think that this is how they came up with the left part. Any ideas? $\endgroup$
    – Elarion
    Mar 26, 2015 at 8:51

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