TITLE: Gradient of a Vector Valued function
QUESTION [18 upvotes]: I read somewhere, gradient vector is defined only for scalar valued functions and not for vector valued functions. When gradient of a vector is definitely defined(correct, right?), why is gradient vector of a vector valued function not defined? 
Is my understanding incorrect? Is there not a contradiction?
I would appreciate clear clarification.
Thank You.

REPLY [2 votes]: Sure enough a vector valued function ${\bf f}$ can have a derivative, but this derivative does not have the "type" of a vector, unless the domain or the range of ${\bf f}$ is one-dimensional. The general setup is the following: Given a function
$${\bf f}:\quad{\mathbb R}^n\to{\mathbb R}^m,\qquad {\bf x}\mapsto {\bf y}={\bf f}({\bf x})$$
and a point ${\bf p}$ in the domain of ${\bf f}$ the derivative of ${\bf f}$ at ${\bf p}$ is a linear map $d{\bf f}({\bf p})=:L$ that maps the tangent space $T_{\bf p}$ to the tangent space $T_{\bf q}$, where ${\bf q}:={{\bf f}({\bf p})}$. The matrix of $L$ with respect to the standard bases is the Jacobian of ${\bf f}$ at ${\bf p}$ and is given by
$$\bigl[L\bigr]=\left[{\partial y_i\over\partial x_k}\right]_{1\leq i\leq m,\ 1\leq k\leq n}\ .$$
If $m=1$, i.e., if ${\bf f}$ is in a fact a scalar function, then the matrix $\bigl[L\bigr]$ has just one row (of length $n$):
$$\bigl[L\bigr]=\bigl[{\partial f\over\partial x_1} \ {\partial f\over\partial x_2}\ \ldots\ {\partial f\over\partial x_n}\bigr]_{\bf p}\ .$$
The $n$ entries of this one-row matrix can be viewed as coordinates of a vector which is then  called the gradient of $f$ at ${\bf p}$.