# Implicit function theorem: The result about equivalence of partial derivatives

I am trying to understand how one obtains the result of the Implicit Function Theorem which involves the equivalence of the derivatives as stated in the related Wikipedia page (https://en.wikipedia.org/wiki/Implicit_function_theorem): Here, $$f$$ is a continuous differentiable function $$f: \mathbb{R}^{n+m} \to \mathbb{R}^{m}$$. At a point $$(a,b)$$, we have $$f(a,b) = 0 \in \mathbb{R}^{m}$$. Then in a neighborhood $$U \in \mathbb{R}^{n}$$ around $$a$$, we have a $$C^1$$ function $$g: \mathbb{R}^{n} \to \mathbb{R}^{m}$$ such that $$f(x,g(x))=0$$ and $$g(a) = b$$ in this neighborhood. The above equation of derivatives holds in this neighborhood as well.

I tried to replicate the equation above by applying the chain rule straightforwardly to $$f(x,g(x))$$. Considering the total derivative of a single component $$f_i$$ of $$f$$ with respect to $$x_j$$ in $$U$$, we should have:

$$\nabla_{x_j} f_i = \sum_{t=1}^{m}\dfrac{\partial f_t}{\partial g_t}(x,g(x))\dfrac{\partial g_t}{\partial x_j}(x) + \dfrac{\partial f_i}{\partial x_j}(x,g(x))$$

This is simply the sum of all $$f_i$$'s components' derivatives with respect to $$x_j$$. Generalizing the above to all $$f_i$$ $$(1 \leq i \leq m)$$:

$$\left[\nabla_{x_j} f_1, \dots, \nabla_{x_j} f_m\right]^T_{m \times 1} = [J_{f,y}(x,g(x))]_{m \times m}\left[\dfrac{\partial g_1}{\partial x_j}(x), \dots, \dfrac{\partial g_m}{\partial x_j}(x)\right]^T_{m \times 1} + \left[\dfrac{\partial f_1}{\partial x_j}(x,g(x)), \dots, \dfrac{\partial f_m}{\partial x_j}(x,g(x))\right]^T_{m \times 1}$$

Here $$J_{f,y}$$ is the Jacobian of $$f$$ with respect to all $$g_t$$ components. Now, rearrenging I obtain:

$$\left[\dfrac{\partial g_1}{\partial x_j}(x), \dots, \dfrac{\partial g_m}{\partial x_j}(x)\right]^T_{m \times 1} = [J_{f,y}(x,g(x))]_{m \times m}^{-1}\left[\nabla_{x_j} f_1 - \dfrac{\partial f_1}{\partial x_j}(x,g(x)), \dots, \nabla_{x_j} f_m - \dfrac{\partial f_m}{\partial x_j}(x,g(x))\right]^T_{m \times 1}$$

This is not quite the result shown on the Wikipedia page, as I have a subtraction of the partial derivatives with respect to $$x_j$$ from the total derivatives. What am I missing here?

The only thing that you're missing is that all the total derivatives are zero, since $$f(x,g(x))$$ is constant (that's how $$g(x)$$ is defined to begin with). So in your last line, you have zero minus the partial derivatives, where you can factor out the minus sign to get the formula from Wikipedia.