Derivatives as linear operators.

Because partial derivatives are co-ordinate dependent they are not a particularly useful way of thinking about derivatives if we want to move to a co-ordinate independent setting such as differentiable manifolds. It is more useful to think of the derivative of a function $f \colon U \to \mathbb{R}$ at $x$ as a linear map

$$df(x) \colon \mathbb{R}^n \to \mathbb{R}$$

defined by

$$df(x)(v) = \left. \frac{d \phantom{t} }{ dt} ( t \mapsto f(x + tv))\right\vert _{t=0}.$$

We think of this as the rate of change of $f$ at $x$ in the direction of $v$. For smooth functions $df(x)$ is linear. Note that $df(x)$ is akin to the notion of a directional derivative but we do not require that $v$ is of unit length. We can recover the partial derivatives from this definition by applying the linear operator $df(x)$ to the vector $e^i$. The result, $df(x)(e^i)$, is just the $i$th partial derivative of $f$ at $x$.

Similarly if $f\colon U \to \mathbb{R}^m$ then we define a linear map

$$df(x) \colon \mathbb{R}^n \to \mathbb{R}^m$$

by

$$df(x)(v) = \left. \frac{d \phantom{t} }{ dt} ( t \mapsto f(x + tv))\right\vert _{t=0}.$$

As a linear map we can expand $df(x)$ in a basis and we recover the Jacobian matrix

$$df(x)(e^i) = \sum_{j=1}^m \partial _i f^j e^j.$$