Suppose that you have a function $f:{\mathbb R}^n\rightarrow{\mathbb R}^m.$ Then we
say that it is differentiable at some $\v a\in{\mathbb R}^n$ if
locally at $\v a$ the function may be "well approximated" by an
(affine) linear function passing through $(\v a, f(\v a)).$ Such a
function has the form
$$
g(\v x) = f(\v a) + D (\v x -\v a),
$$
where $D$ is a $m\times n$ matrix.
This
linear function is what we call the tangent if $m=n=1$ and the
tangent plane if $m=1,~n=2.$
This $D$ is
called the derivative of $f$ at $\v a.$ More commonly,
it is also called the Jacobian of $f$ at $\v a.$
If $m=1$ and $n>1$ then it is often denoted
by $\nabla f$ and called "grad $f$".
If you are curious about what is meant by "well approximated",
then here it is:
$$
\frac{f(\v x)-g(\v x)}{\|\v x - \v a\|}\rightarrow \v 0\mbox{ as } \v x\rightarrow\v a.
$$
The $\|\cdots\|$ is the length in ${\mathbb R}^n.$
Check that
for $n=1$ this is same as our familiar concept of differentiation.
The derivative of $f:{\mathbb R}^n\rightarrow{\mathbb R}^n$ at any given $\v
a\in{\mathbb R}^n$ is an $m\times n$ matrix $D.$
Its $(i,j)$-th entry is given by
$$
\frac{\partial f_i}{\partial x_j} (\v a).
$$
EXAMPLE:
If $f(x_1,x_2,x_3) = 4x_1x_2 - x_3^2,$ then direct
computation shows
$$
\nabla f = (4x_2, 4x_1, -2x_3).
$$
If
$f:{\mathbb R}^n\rightarrow{\mathbb R}$ is of the form
$$
f(\v x) = \v x' M \v x
$$
for some symmetric matrix $M,$ then
$$
\nabla f = 2\v x' M.
$$
This is the formula we have used in the soap film example.
Just like the familiar 1-dim derivative, the higher dimensional
derivatve also vanishes if $f$ attains a local max/min (for $m=1$). We
also have an analogous second order derivative check, but we
shall not go into that here.
Let $f:{\mathbb R}^m\rightarrow{\mathbb R}^n$ and $g:{\mathbb R}^n\rightarrow{\mathbb R}^p$ be both
differentiable with derivatives $D_f$ and $D_g,$
respectively. Let $h:{\mathbb R}^n\rightarrow{\mathbb R}^p$ is defined as $h =
g\circ f,$ then $h$ must be differentiable with
derivative at any $\va\in{\mathbb R}^n$ given by
$$
D_h = D_g( f(\v a)) D_f(\v a).
$$
For a function $f:{\mathbb R}^m\rightarrow{\mathbb R}^n$ that is differentiable at
some $\v a\in{\mathbb R}^n, $ we know that $f$ is well
approximated by the affine linear function
$$
g(\v x) = f(\v a) + D (\v x -\v a),
$$
where $D$ is the derivative of $f$ at $\v a.$ This
is called the 1st order Taylor approximation of $f.$ Under
suitable assumptions (related to existence and continuity of the second
partial derivatives of $f$), the error involved in this
approximation is of quadratic order.
If $f:{\mathbb R}^2\rightarrow{\mathbb R}$ is differentiable at
some $(x_0,y_0)$ then the tangent plane is given by
$$
z = f(x_0,y_0) + f_1(x_0,y_0)(x-x_0) + f_2(x_0,y_0)(y-y_0),
$$
where $f_i$ is the partial derivative of $f$
w.r.t. the $i$-th argument.
In particular, the partial derivatives must exist if $f$ is differentiable.
However, the converse is not true. It is possible that the
partial derivatives exist at some $(x_0,y_0)$ (and so the
above plane is well defined), but still $f$ is not
differentiable there. Thus, even if the plane exists, it need not
be the tangent plane. One such example is given by the function
$$
f(x,y) = \left\{\begin{array}{ll}-y&\text{if }x\neq 0\\0&\text{if }x=0\\\end{array}\right..
$$
The graph is shown below:
Not differentiable at $(0,0)$
Here $f_1(0,0) = f_2(0,0) = 0,$ and so the plane is just
the $xy$-plane, but clearly it is not tangent to the
surface.
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