I'm trying to find the tangent plane at a point $p = (x,y,u,v)$ to a given submanifold of $\mathbb{R}^4$. I have found that this submanifold is the solution set to $$f(x,y,g_1(x,y), g_2(x,y)) = (0,0)$$where $g_i:\mathbb{R}^2 \rightarrow \mathbb{R}$.
Typically, I would solve for $\nabla f(p)\cdot(x-p)=0$, but in this case, how do I proceed? What is the analogue to the gradient here?
If you need additional context for the question, please let me know. I'm looking for more of a general approach to solving problems like these, than any particular problem.
"The gradient is the special case of the derivative for a function $f: \mathbb{R}^n \to \mathbb{R}^m$" i.e when $m =1$ we have $f'=\nabla f$.
In $\mathbb{R}^4$, your submanifold $S$ is the graph of $\phi = (g_1,g_2)$ i.e $S= \{(x,y, \phi(x,y))\}$. We will have to use the derivative of the map $\Phi(x,y) = (x,y,g_1(x,y),g_2(x,y))$ i.e $D_{(x,y)}\Phi: T_{(x,y)}\mathbb{R}^2 \to T_pS$.
Now just recall the definition and so in this case,
$$ D_q\Phi = \begin{pmatrix} \nabla(x)\\ \nabla (y) \\ \nabla g_1 \\ \nabla g_2 \end{pmatrix}$$
where $q = \Phi^{-1}(p)$. For instance, $\nabla (x) = \begin{pmatrix} \frac{\partial (x)}{\partial x} \\ \frac{\partial (x)}{\partial y} \end{pmatrix} = (1,0)^T$. Here the tangent plane is the image of this map and translated by $q$ i.e,
$$ D_q\Phi \begin{pmatrix} x \\ y \end{pmatrix} + q$$