Geometric interpretation of derivative of a function of more than one variable

A function $f$ is defined on an open set $D$ of $\mathbb R^{2}$ is called a differentiable at a point $x\in D$ if there is a vector $m \in \mathbb R^{2} $ such that $$\lim_{h\to 0} \frac{f(x+h)-f(x)-m\cdot h}{|h|}=0.$$

My questions are: (1) What is a geometric interpretation of $f:\mathbb R^{2} \to \mathbb R$ is a differentiable at a point in $D$ ?

( Let $f:\mathbb R^{2} \to \mathbb R $such that $f(x, y)= \frac{x^{3}y}{x^{4}+y^{2}}$ for $(x,y)\not = (0,0)$ and $f(0,0)= 0$. Notice that all the directional derivatives of $f$ exists at $(0, 0)$ and they are equal at $(0, 0)$ but although $f$ is fails to be differentiable at $(0,0)$. )

(2) What is a geometric interpretation of $f:D\subset \mathbb R^{n}\to \mathbb R^{m}$ is differentiable at point in $D$ ?

3 Answers

Given a function ${\bf f}:\>{\mathbb R}^n\to{\mathbb R}^m$ and a "working point" ${\bf p}$ in the domain of ${\bf f}$ one may ask how the value of ${\bf f}$ changes when one moves from ${\bf p}$ to a nearby point ${\bf p}+{\bf X}$, $\>|{\bf X}|\ll1$. This means that we are interested in the auxiliary function $$\Delta_{\bf p}{\bf f}({\bf X}):={\bf f}({\bf p}+{\bf X})-{\bf f}({\bf p})$$ as a function of the increment vector ${\bf X}$. When this function is "in first approximation" linear in the increment variable ${\bf X}$ the given function ${\bf f}$ is called differentiable at ${\bf p}$.

What does that mean precisely? It means that there is a certain linear map $A:\>{\mathbb R}^n\to{\mathbb R}^m$ such that $${\bf f}({\bf p}+{\bf X})-{\bf f}({\bf p})=A\>{\bf X}\ +r({\bf X})\ ,\tag{1}$$ where the remainder term $r({\bf X})$ is essentially smaller than linear in ${\bf X}$, i.e., one has $${r({\bf X})\over |{\bf X}|}\to{\bf 0}\qquad({\bf X}\to{\bf 0})\ .\tag{2}$$ The relations $(1)$ and $(2)$ can then be condensed in the intuitive formula $${\bf f}({\bf p}+{\bf X})-{\bf f}({\bf p})=A\>{\bf X}\ +o(|{\bf X}|)\qquad({\bf X}\to{\bf 0})\ .\tag{3}$$ The linear map $A$ appearing in $(1)$ and $(3)$ is called the derivative (or differential) of ${\bf f}$ at ${\bf p}$ and is denoted by ${\bf f}'({\bf p})$ or $d{\bf f}({\bf p})$. The $m\, n$ matrix elements of $A$ are the partial derivatives of ${\bf f}$ at ${\bf p}$.

One of the difficult to explain miracles of analysis is that most functions ${\bf f}$ occurring in practice are differentiable at most points in their domain.

The geometric interpretation of $f : \mathbb{R}^2 \to \mathbb{R}$ is a surface in $\mathbb{R}^3$. The geometric interpretation of the differentiability of $f$ at a point $x$ is that the surface has a tangent plane at $x$. The components of the vector $m$ are the partial derivatives of $f$.

In the multidimensional case $f : \mathbb{R}^n \to \mathbb{R}$ the map is differentiable at a given point when, geometrically speaking, it has a tangent hyper-plane at that point.

(1) A function $f: D \rightarrow \mathbb{R}$ is differentiable at an interior point $x_0$ of $D$ if $\nabla f(x_0)$ exists and $$\lim_{x \rightarrow x_0} (f(x_0)+ \nabla f(x_0)(x-x_0) + o\left( \| x-x_0 \right \|)=0$$ holds (here $x,x_0$ are vectors of two or more dimensions). Geometrically, for one variable functions, differentiability at $x_0$ implies the existence of the tangent line to the graph of $f$ at the point $P_0=(x_0,f(x_0))$ $$t(x)=f(x_0) + f'(x_0)(x-x_0)$$. The multivariate case is more complex and the existence of $\nabla f(x_0)$, or the existence of the tangent plane to $f$ at $P_0$ does not guarantee the validity of the definition above. Like in your exercise, not even the existence of every partial derivative at $x_0$ guarantees differentiability. For $n=2$ the tangent plane to the graph of $f$ at $P_0=(x_0,y_0,f(x_0,y_0)$ is defined by $$z=f(x_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial x}(x_0,y_0)(y-y_0)$$. This plane best approximates the graph of $f$ on a neighborhood of $P_0$. The differentiability of $f$ at $x_0$ means the existence of the tangent plane. In some cases this property fail to hold.

(2) This other function is a vector-valued map. We have differentiability if each component of $f$ is differentiable at $x_0 \in D$, i.e. $$f_{i}(x)=f_i(x_0)+\nabla f_i(x_0) \cdot (x-x_0) +o\left( \| x-x_0 \right \|), \quad x \rightarrow x_0$$, where the scalar product $\nabla f_i(x_0) \cdot (x-x_0)$ is the matrix product between the row vector $\nabla f_i(x_0)$ and the column vector $(x-x_0)$. The $n \times n$ matrix $Jf(x_0)$ formed by the row vectors $\nabla f_i(x_0)$ is also called Jacobian matrix. Putting $\Delta (x)=x-x_0$ the latter equation is $$f_{i}(x)=f_i(x_0)+J f(x_0)\Delta(x) +o\left( \| \Delta(x) \right \|), \quad x \rightarrow x_0$$. Up to infinitesimal of order greater than one the formula says that the increment $\Delta f= f(x_0 + \Delta x)-f(x_0)$ is approximated by the value of the differential $J f(x_0)\Delta(x)$. In other words the increment $\Delta f$ goes to zero quickly than $\Delta x$ as $x \rightarrow x_0$.