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If I make the following argument Since rhd of $f$ is same as rhl of $f'$, similarly for left side, and we know rhd =lhd= $f'(a)$, as it is given that the function is differentiable, then have we not proven that $f'(x)$ is continuous at x=a?

Since rhl of f '(x) at x=a (=rhd of f(x)) = lhl of f '(x) at x=a (=lhd of f(x))= f'(a) (by definition of the derivative at x=a)

If you agree, then consider

$ f(x) = \left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & \mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0 \end{array} \right.$

This function is differentiable at $0$, but its derivative is discontinuous.

If my premise is wrong, please prove so....thanks I know that a proof by counter example is valid in maths, but I would appreciate an algebraic proof as to why my statement is wrong.

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  • $\begingroup$ You consider only one-sided derivatives at a particular point. They are not limits of derivatives, but of difference quotients. You don't say anything about behavior of the derivative in other points. How could that prove continuity? $\endgroup$ – Marcin Łoś May 9 '14 at 17:18
  • $\begingroup$ Firstly, I did not clearly get the first 3 lines. I am in the last yr of high school, so if u restrict ur arguments to this domain, it will be easier for me to understand. $\endgroup$ – Saurabh Raje May 10 '14 at 1:10
  • $\begingroup$ Moreover, I am talking abt continuity at THAT point. So why does the derivative need to be defined over an interval containing the same point? Maybe you could take another look at the function I talked about. Its derivative exists for all real nos. Then why isnt that derivative continuous at 0? $\endgroup$ – Saurabh Raje May 10 '14 at 1:15
  • $\begingroup$ Upvote because it's a good question but it is one of the text-book classics, my fav analysis book has the graph on the front cover. $\endgroup$ – Alec Teal May 12 '14 at 9:02
  • $\begingroup$ Alec, it is a typical eg of the invalid claim I made. I know the graph, but I dont see where my basic argument has gone wrong $\endgroup$ – Saurabh Raje May 12 '14 at 10:17
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The RHD of $f(x)$ at $x=a$ is $\lim_{x \to a^+} \frac{f(x)-f(a)}{x-a}$.

The RHL of $f'(x)$ at $x=a$ is $\lim_{x \to a^+} f'(x)$ = $\lim_{x \to a^+, h \to 0} \frac{f(x+h)-f(x)}{h}$.

Your mistake is stating these are the same thing.

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  • $\begingroup$ I do see that there is a problem in my assumption, however, on a cartesian plane, for an arbitrary curve, the assumption makes perfect sense. $\endgroup$ – Saurabh Raje May 12 '14 at 10:19
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    $\begingroup$ It only makes sense for precisely those 'curves' for which $f'(x)$ is continuous, which is what you are trying to prove. Clearly there are other graphs for which this does not apply, such as the one in your post. $\endgroup$ – user11977 May 12 '14 at 10:22

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