# Differentiability of multivariable functions, and its relation to the chain rule.

I'm struggling with the conditions for the applicability of the chain rule.

$${df(C(t))\over dt} = \mathrm{grad}f(C(t))\cdot C'(t)$$

Where $C$ is in $\Bbb R^n$ and $f$ is differentiable in the following sense:

$f:\Bbb R^n\rightarrow \Bbb R$ is differentiable at $X$ if we have, for sufficiently small $H$ (in the sense of being close to the null vector) $$f(X+H)-f(X) = \mathrm{grad}f(x)\cdot H+||H||g(H)$$ for some real-valued $g$ which tends to $0$ as $H$ tends to the null vector.

$C$ also has to be differentiable in that all of its partial derivatives exist.

These requirements seem overly strong to me. What we really need is that $f\circ C:\Bbb R^n \rightarrow \Bbb R$ be differentiable (in the traditional sense of the word). For that, I don't see that either $f$ nor $C$ even need to be continuous. What if we have

$$C(t) = (0, t)$$ for $t$ irrational, and $$C(t) = (t, 0)$$ for $t$ rational. Then define $f$ on $C(\Bbb R)$ by extracting $f(0, t) = f(t, 0) = t$. Do we not have $f\circ C$ as the identity function, thus differentiable?

-
This is not really a question, and more of a blog. You don't list any of the definitions that you disagree with, nor any of their applications. How am I to know what is in the book, and how am I to help you? All I know is that your disagree quite strongly with what the author has written. –  Fly by Night Jun 21 '13 at 17:41
My specific question is: why do $f$ and $C$ need to be differentiable? That seems like a stronger condition than what we really need, which is that $f\circ C$ be differentiable. –  Jack M Jun 21 '13 at 17:53
@FlybyNight You were right, I completely rewrote the question. Thanks. I wanted people to see the context of my confusion, but I guess you don't need quite that much context. –  Jack M Jun 21 '13 at 18:09

"Looking at the proof, it almost seems like the only reason the chain rule formula is obtained is because the author's definition of differentiability is specifically set up for it to emerge."

The essential content of mathematics are examples, definitions and theorems. The definition of differentiability is not "the author's definition", but it is the essence that has been cristalized after 250 years of "calculus" out of various manifestations of differentiability in onedimensional and pluridimensional contexts. The chain rule is a prime feature of differentiability, having a high intuitive-geometrical power (I know that students see the chain rule mainly as a recipe to formally differentiate composed function terms). Therefore no general concept of differentiability could do without it.

It's no big deal when a crazy function composed with, say, a constant function is differentiable. But it is essential that the composition of any two differentiable functions is again differentiable, and that we have a formula for the derivative of this composition.

Now to your example: Assume (i) that the vector-valued function $$t\mapsto {\bf c}(t)=\bigl(c_1(t),\ldots,c_n(t)\bigr)\qquad (-h<t<h)$$ representing a curve in ${\mathbb R}^n$ is differentiable at $t=0$, i.e., that $${\bf c}'(0)=\bigl(c_1'(0),\ldots,c_n'(0)\bigr)$$ exists.

Assume (ii) that the scalar function $$f:\quad {\mathbb R}^n\to{\mathbb R},\qquad{\bf x}=(x_1,\ldots,x_n)\to f(x_1,\ldots, x_n)$$ representing, say, a variable temperature in space, is differentiable at the point ${\bf p}:={\bf c}(0)$. This means that we have an estimate of the form $$f({\bf p}+{\bf X})-f({\bf p})=A.{\bf X} + o(|{\bf X}|)\qquad({\bf X}\to{\bf 0})\ ,\tag{1}$$ where $$A=:df({\bf p}):\quad T_{\bf p}\to{\mathbb R}$$ is a certain linear map (a linear functional, to be exact). This implies (but is not implied by) that $f$ has partial derivatives $f_{.k}({\bf p})={\partial f\over\partial x_k}({\bf p})$ $\>(1\leq k\leq n)$. These partial derivatives can be collected into the gradient vector $\nabla f({\bf p})$, and $(1)$ can then be written in the form $$f({\bf p}+{\bf X})-f({\bf p})=\nabla f({\bf p})\cdot{\bf X} + o(|{\bf X}|)\qquad({\bf X}\to{\bf 0})\ ,$$ where the dot denotes the scalar product in ${\mathbb R}^n$.

Now we are interested in the composed function $$\phi:\quad ]-h,h[\ \to {\mathbb R},\qquad t\mapsto f\bigl({\bf c}(t)\bigr)$$ which represents the temperature measured along time by an experimenter sitting in the spaceship $t\mapsto{\bf c}(t)$. The chain rule tells us that $$d\phi(0)=df({\bf p})\circ d{\bf c}(0)\ ,$$ which when unpacked goes over into $$\phi'(0)=\nabla f({\bf p})\cdot {\bf c}'(0)\ .$$

-
Here's a second part to my question, then. Does the chain rule apply whenever 1) $f$ has a gradient (ie its partial derivatives exist) and 2) $C$ has a tangent vector (ie its components are differentiable functions)? –  Jack M Jun 21 '13 at 20:02
@JackM: See my edit. –  Christian Blatter Jun 22 '13 at 8:39