I'm currently trying to understand the following equation.
For $h(x) = n^{-1}\sum_{i=1}^pa_i/(a_i+x), a_i > 0, x > 0,\ n,p \in \mathbb N$, consider a equation: $$1 = \lambda/x + h(x)$$ and let the unique solution of the equation be $x(\lambda)$. Then the implicit function theorem leads to $$\partial_{\lambda}x(\lambda) = \Bigl[1 - \frac{1}{n}\sum_{i=1}^p\frac{a_i^2}{(a_i + x(\lambda))^2} \Bigr]^{-1}. \tag1$$
I tried to induce the result (1) by following: (regarding $\lambda$ as a function of $x$) $$ \lambda = x - \frac{1}{n}\sum_{i=1}^p\frac{a_ix}{a_i + x} \\ \frac{\partial \lambda}{\partial x} = 1 - \frac{1}{n}\sum_{i = 1}^p\frac{a_i^2}{(a_i+x)^2} $$ Taking an inverse leads to (1).
Now, I'm curious about whether the above approach is correct.
If that's the case, I don't think I used the implicit function theorem. What is the relationship between the above approach and the implicit function theorem? If there is no such relation, how can I use the theorem to induce the result (1)?
Any help would be appreciated.
The implicit function applies here to say that (local) solutions $x(\lambda)$ exist and are $C^1$ for any fixed $\lambda>0$, $x>0$. There are no obstructions and I believe you could glue these to a global solution on $(0,\infty)^2$. But the implicit function theorem normally comes packaged with an additional statement:
Describing the derivative of this local solution. Here $F$ would be $F(x,\lambda)=\frac{\lambda}{x}+h(x)$. So the theorem claims: $$\frac{\mathrm{d}x}{\mathrm{d}\lambda}=-\frac{x^{-1}}{-\lambda x^{-2}+h'(x)}=-\frac{x^{-1}}{-x^{-1}+x^{-1}h(x)+h'(x)}=(1-h(x)-xh'(x))^{-1}$$Which works out to equal the given expression.
So you could have used the implicit function theorem to do the thinking for you here, but you didn't have to; your work looks alright to me. It gets more complicated in higher dimensions though, in which case the derivative given by the implicit function theorem is useful.