Partial Derivatives and Implicit Function Thm.

[qspeechc]

Hi.

So I'm reading a physics book and I come across the following passage:

Let $p=(p_1,p_2,...,p_n)$ and $q=(q_1,q_2,...,q_n)$ and the system of n algebraic equations
$g_r(p,q)=0, \quad r\in \{1,2,...,n\}$
can be solved with respect to the $p_i$ in the form $p_i=\phi _i(q)$ for $i\in \{1,2,...,n\}$
Then we have
$\frac{d}{dq_h}g_r(\phi ,q)=0$
where $\phi =(\phi _1,...,\phi _n)$. If we denote $G_r(q)=g_r(\phi ,q)$ then the last relation gives

$0=\frac{\partial}{\partial q_h}G_s = \Big[\frac{\partial g_s}{\partial q_h} + \sum_{i=1}^{n}\frac{\partial g_s}{\partial p_i}\frac{\partial \phi _i}{\partial q_h} \Big]_{p=\phi (q)}$

Ok, up to this point I'm fairly confident I'm following along. But then they do the following:

Therefore the partial derivatives can be expressed in the form
$\frac{\partial g_s}{\partial q_h} = \sum_{i=1}^n\frac{\partial g_s}{\partial p_i}\frac{\partial (p_i-\phi _i)}{\partial q_h}$

and I have no idea where this comes from. I am guessing here that $p_i=\phi _i(q)$ is only in some sufficiently small region and not the entire space, since they do not explicitly say anything in this regard, so that $p_i-\phi _i$ is not identically zero on the whole space, so I do not know where the above comes from. This is probably something really simple and I'm just being stupid, but any help is greatly appreciated.

 

[qspeechc]

Anyone
Please?

 

[Astronuc]

What is the implication of

$$\frac{\partial p_i}{\partial q_h}\,=\,0$$

 

[qspeechc]

If
$$\frac{\partial p_i}{\partial q_h} = 0$$
then
$$\frac{\partial \phi _i}{\partial q_h} = 0$$
in the region where $$p_i=\phi _i$$, and outside this region, well, it's identically zero since $$p_i$$ is not explicitly a function of the $$q_i$$'s. Am I on the right track? I don't see where these train-tracks are leading.

 

[qspeechc]

Aha! Does this mean that, regardless of where we are in the space, if we have the $$p_i$$ as a function of the $$q_i$$ or not, the adding of the term $$p_i$$ has no effect of the partial derivative because:
If we are in the region where $$p_i-\phi _i=0$$ the partial derivatives
$$\frac{\partial g_s}{\partial q_h} = 0$$
anyway, and if we are outside this region then
$$\frac{\partial p_i}{\partial q_h} = 0$$
Is this right?

 

[qspeechc]

Ok, anyone please? Even if it's just a " yea I don't get that either"...