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):

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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?


1 Answer 1


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.


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