Darboux's theorem says.
If x=x(t) has a derivative at every point of [a,b], isn't that derivative automatically continuous? If not, please give an example of a derivative which exists at every point but which is not continuous. All I can think of is a spike in the curve which has different (one sided) derivatives on either side, but the derivative doesn't exist at the spike.
This question has its origin in the definition of a smooth curve, which requires that the derivatives of x=x(t), y=y(t), and z=z(t) exist AND be continuous.
Darboux's theorem says.
Let f(x) = (x^2)sin(1/x). f'(0) doesn't exist because 1/0 doesn't exist. Defining f'(0) = 0* makes no sense because f'(x) = 1/cosx doesn't exist.
*f'(0) = (f(x) - f(0))/x = (x^2sin(1/x)-0)/x = 0
If f(x) = (x^3)sin(1/x), defining f'(0) = 0 makes sense because f'(x) = 0. In this case existence of f'(0) implies continuity at 0.
In fact, a more elaborate example gives a function everywhere differentiable with the deerivative having a set of discontinuities of positive measure, see here.
I'll pass on the more elaborate example. Not my cup of tea. I'm really just trying to get a working feeling for what a curve looks like with derivatives existing everywhere on a continuous curve but not continuous. I don't have the ability to remember abstract derivations.
For example, sinx/x is undefined at x=0. But since lim sinx/x -> 0 exists, it makes sense to define it as 0. If limit didn't exist, defining it as 0 would be meaningless.
Suppose f'(x) = x^2 except at x=1 where I define it to be 10. Then clearly existence of derivative everywhere does not imply continuity of f'(x). But the example is contrived in a manner that doesn't make sense. I'm not saying existence of f'(x) implies continuity, just trying to get a feel for why it isn't true. f(x) = (x^3) sin(1/x) made sense to me because definition of f(0) and limit f'(x) were the same.
Perhaps it would help to elaborate on my original post. I didn't understand the requirement of a smooth curve that f'(x) exist everywhere AND be continuous. I have made some progress. I believe it relates to the definition of line integral. If you set up partitions to define a line integral by summation and use the derivative and intermediate value theorem along the way, then intuitiveley I see that in order for the limit to exist independent of definition of partition just as long as all partitions go to zero, f'(t) has to be continuous.
But out of curiousity I wondered what a continuous (part of the requirement) curve with existing derivatives that weren't continuous would look like. It's easy enough to imagine what a continuous curve looks like though I vaqueley recollect some non-intuitive examples.
The derivative of the given example function does exist. Remember that the derivative of a function is given by
Given the function
we calculate the derivative at by computing
Compute this and see what you find. Then compute (using standard differentiation techniques) the for . Note that is continuous if and only if exists and is equal to . So does the limit exist ?
EDIT: However I would be interested in knowing if it's possible to find an example of a function which is differentiable on some domain such that left- and right-sided limits of exist on all of , but such that is discontinuous in the particular sense of having the left- and right-sided limits unequal at some . Perhaps that is impossible.
Still looking for a non-contrived example of a continuous function on a closed interval which has a derivative defined at each point but which (the derivative) isn't continuous. Kaplan mentions that a curve can have jump discontinuities in the derivative but doesn't elaborate. Yes, yes, I know- Drexel28's example in first post. Oh well, i beat my objection to death and since nobody is buying it or maybe doesn't understand it, I'll let it ride without a feel for what a curve with discontinuous derivatives looks like.
Thanks for the example Drexel28.
PS I missed edit on a previous post so let me correct a typo. sinx/x, x=0 isn't defined. But BECAUSE lim sinx/x as x->0 =1 (not 0), it makes sense to define it as 1. So-so analogy.
What I am saying may make more sense if you read my last post in "show that f'(c) exists" nearby.
let's try to compute the derivative of , at x = 0 (remember we are defining f(0) = 0, this is ok, since this defintion makes f continuous), since its existence seems to be in doubt:
now, here, we have a problem, because we can't evaluate sin(1/0).
but...let's try doing the left-hand and right-hand limits separately:
for small enough h > 0, we can ensure that sin(1/h) > 0, thus for these same h,
hsin(1/h) > 0, and we know that sin(1/h) ≤ 1, so by the squeeze theorem:
similarly, for h < 0 and |h| sufficiently small,
-1 ≤ sin(1/h) < 0, so
so we conclude that:
since the left-hand and right-hand limits are equal,
, so f'(0) exists.