Multivariable Calculus – Are Derivatives Linear Maps?

derivativesmultivariable-calculus

I am reading Rudin and I am very confused what a derivative is now. I used to think a derivative was just the process of taking the limit like this
$$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}$$

But between Apostol and Rudin, I am confused in what sense total derivatives are derivatives.

Partial derivatives much more resemble the usual derivatives taught in high school

$$f(x,y) = xy$$

$$\frac{\partial f}{\partial x} = y$$

But the Jacobian doesn't resemble this at all. And according to my books it is a linear map.

If derivatives are linear maps, can someone help me see more clearly how my intuitions about simpler derivatives relate to the more complicated forms? I just don't understand where the limits have gone, why its more complex, and why the simpler forms aren't described as linear maps.

Best Answer

A derivative of a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ at $x \in \mathbb{R}^n$ is a linear map $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that

$$\lim_{v \rightarrow 0} \frac{f(x+v)-f(x)-L(v)}{\|v\|} = 0$$

or alternatively $f(x+v) = f(x) + L(v) + o(\|v\|)$, i.e. $f(x)$ is the constant part of $f$ at $x$, and $L$ is the linear part of $f$ at $x$, and everything else is sub-linear.

In the one-dimensional case, the linear map is just multiplication by a scalar, and we call that scalar $f'(x)$, with $L(v) = f'(x)\, v$ for $v \in \mathbb{R}$.

For the case $f: \mathbb{R}^n \rightarrow \mathbb{R}$, the partial derivative at $x$ in the direction $u$ (a unit vector in $\mathbb{R}^n$ would be $L(u) \in \mathbb{R}$.

This generalizes directly to functions $f: X \rightarrow Y$ between Banach spaces $X$ and $Y$ (vector spaces with a norm which are additionally complete [so that limits are well-behaved]).