Finding the tangent hyperplane to a function at a given point Given a function $f : \mathbb{R}^n \to \mathbb{R}$ and a point $p$ in the domain, how would you find the equation of the tangent hyperplane to the surface at point $p$?
We know that if $h \to 0$, then the Jacobian matrix $J_f(p)h = 0$. Can we use this to find the equation of the hyperplane? Maybe something along the lines of $J_f(p) \cdot [x_i - p_i]^{\text{T}} = c$?
 A: The equation of a tangent hyperplane (tangent plane in $\Bbb R^3$) to a function $f$ in a point $P_0\in\mathbb{R}^n$, (let be $n=3$) assuming that all conditions of differentiability are satisfied, is:
$$\left(\frac{\partial f}{\partial x}\right)_{P_0}(x-x_0)+\left(\frac{\partial f}{\partial y}\right)_{P_0}(y-y_0)+\left(\frac{\partial f}{\partial z}\right)_{P_0}(z-z_0)=0$$
$$\iff (\boldsymbol\nabla f)_{P_0}\cdot (\mathbf x-\mathbf x_0)=0$$
Considering
$$\left(\frac{\partial f}{\partial x}\right)_{P_0}(x-x_0)+\left(\frac{\partial f}{\partial y}\right)_{P_0}(y-y_0)+\left(\frac{\partial f}{\partial z}\right)_{P_0}(z-z_0)=0$$
if $\left(\frac{\partial f}{\partial x}\right)_{P_0}=\color{red}{a}\in \Bbb R$, $\left(\frac{\partial f}{\partial y}\right)_{P_0}=\color{blue}{b}\in \Bbb R$ and
$\left(\frac{\partial f}{\partial z}\right)_{P_0}=\color{magenta}{c}\in \Bbb R$ we have
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \iff $$
$$ax+by+cz-(ax_0+by_0+cz_0)=0 \tag{*}$$
If I denote with $C=ax_0+by_0+cz_0$ the $(*)$ become:
$$\bbox[9px,border:2px solid #F5B041]{\underbrace{F(x,y,z)}_{ax+by+cz}=C}\quad \text{tangent plane in } \Bbb R^3$$

If $f \colon A\subseteq \Bbb R^3 → \Bbb R$ is differentiable in
$\mathbf{x}_0\in A$, then the graph of the equation
$$z=f(\mathbf{x}_0)+J_f(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0)\equiv f(\mathbf{x}_0)+(\boldsymbol\nabla f)_{P_0}(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0)$$
is the tangent hyperplane. Consider that the Jacobian has as matrix elements precisely the components of the gradient:
$$J_f(\mathbf{x}_0)=\begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{bmatrix}_{\mathbf{x}_0}$$
Definitively

You use the transpose symbol, in your question, as you do the row-by-column product.

