2
$\begingroup$

I was finishing up Chapter 10 of Spivak's Calculus when I stumbled upon the following remark which I have trouble wrapping my head around. The paragraph in question went something like this:

lt is often tempting, and seems more elegant, to write some of the theorems in this chapter as equations about functions, rather than about their values. Thus Theorem 3 might be written

$$ (f+g)' = f' + g' $$

For which he follows up with: Strictly speaking, these equations may be false, because the functions on the left- hand side might have a larger domain than those on the right.

I feel like I am missing the main message here, and do not quite understand what he means by thinking about the theorems (mainly, the rules for differentiation) in terms of values instead of functions. Does he mean that such an 'expansion' only makes sense if a specific value is attached to a function? e.g. $(f(a) + g(a))' = f'(a) + g'(a)$ as illustrated earlier in the chapter. I have also been unsuccessful in generating some examples to help me break down this point better.

Any pointers would be greatly appreciated!

$\endgroup$
1
  • $\begingroup$ Be careful! In the expression $f(a)+g(a)$, the functions are already evaluated, so strictly speaking you have $(f(a)+g(a))'=0$. $\endgroup$ Aug 30, 2020 at 0:26

3 Answers 3

3
$\begingroup$

For example, take $f(x)=|x|$ and $g(x)=-|x|.$ Then, $\mathcal D_f=\mathcal D_g=\mathbb R$ and $f+g=0$ so $(f+g)'(x)=0$ but neither $f$ nor $g$ is differentiable at $x=0.$

$\endgroup$
2
  • $\begingroup$ That is a clear example. Thank you very much! $\endgroup$
    – iobtl
    Aug 30, 2020 at 1:46
  • $\begingroup$ You are welcome Glad to help! $\endgroup$ Aug 30, 2020 at 2:58
1
$\begingroup$

He means that $f+g$ can be differentiable even when $f$ and $g$ are not differentiable. In other words this theorem is not always applicable when you want to see what is going on with the derivative of a sum (although it is not very common to find natural examples in which $f+g$ is differentiable but $f$ and $g$ are not )

$\endgroup$
1
  • $\begingroup$ Could not have been put any clearer. Thank you! $\endgroup$
    – iobtl
    Aug 30, 2020 at 1:48
1
$\begingroup$

Jorge already said it all. I just want to add the following example: take the functions $f,g:\mathbb R\to \mathbb R$ given by $$f(x)=\begin{cases}0 & x\leq 0 \\ x & x\geq 0\end{cases}$$ $$g(x)=\begin{cases}x & x\leq 0 \\ 0 & x\geq 0\end{cases}$$

Then the domain of both $f'$ and $g'$ is $\mathbb R\setminus \{0\}$, but the domain of $(f+g)'$ is all of $\mathbb R$.

Hence the identity $$(f+g)'=f'+g'$$ is not strictly correct because you are equating two functions with different domains. In this case, one ought to write $$(f+g)'|_{\mathbb R\setminus \{0\}}=f'+g'$$ or $$(f+g)'(x)=f'(x)+g'(x), \hspace{5mm} \text{for } x\in\mathbb R, x\neq 0.$$

$\endgroup$
1
  • $\begingroup$ Thank you for the clear example and recommended workings as well! $\endgroup$
    – iobtl
    Aug 30, 2020 at 1:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .