How to calculate partial derivative using chain rule?

Question

I have a function:

$${{\mathop{\rm F}\nolimits} _i}\left( {\bf{\xi }} \right) = \sum\limits_k^N {{\mathop{\rm D}\nolimits} \left( {\frac{1}{N}\sum\limits_j^N {{\mathop{\rm G}\nolimits} \left( {j,{\mathop{\rm I}\nolimits} \left( {j,{\bf{\xi }}} \right)} \right)} - {\mathop{\rm G}\nolimits} \left( {k,{\mathop{\rm I}\nolimits} \left( {k,{\bf{\xi }}} \right)} \right)} \right)}$$

$${\rm F}_i(\xi)=\sum_k^N {\rm D}_k\left(\frac1N\sum_j^N{\rm G}_j({\rm I}_j(\xi))-{\rm G}_k({\rm I}_k(\xi))\right).$$

$\xi$ is a vector.

How do I calculate the partial derivative using the chain rule?

$$\frac{\partial{\rm F}_i}{\partial\xi}=? $$

I guess...

$$\frac{\partial{\rm F}_i}{\partial\xi}=\sum_k^N\frac{\partial{\rm F}_i}{\partial{\rm D}_k}\left(\frac1N\sum_j^N\frac{\partial {\rm D}_k}{\partial {\rm G}_j}\frac{\partial {\rm G}_j}{\partial {\rm I}_j}\frac{\partial {\rm I}_j}{\partial \xi}-\frac{\partial {\rm D}_k}{\partial {\rm G}_k}\frac{\partial {\rm G}_k}{\partial {\rm I}_k}\frac{\partial {\rm I}_k}{\partial \xi}\right). $$

• full version

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Answer 1

Personally, I think the total derivative chain rule is easiest to remember:

$$D_{f\circ g}(x) = D_f(g(x)) \cdot D_g(x)$$

What the $D$'s look like $($as $m\times n$ matrices$)$ of course depends on the domain and codomain of each function.

Answer 2

$${\rm F}^\prime_i(\xi)=\sum_k^N {\rm D}^\prime_k\left(\frac1N\sum_j^N\left({\rm G}_j({\rm I}_j(\xi))-{\rm G}_k({\rm I}_k(\xi))\right)\right)\cdot \frac1N\sum_j^N\left({\rm G}^\prime_j({\rm I}_j(\xi)){\rm I}^\prime_j(\xi)-{\rm G}^\prime_k({\rm I}_k(\xi)){\rm I}^\prime_k(\xi)\right)$$

Edit: This is the derivative of what was originally asked about but is certainly not the same as the derivative of the function contained in the image of the updated question.