How Do You Handle a Discontinuous Derivative in Calculus?

[Arnoldjavs3]

Homework Statement

http://prntscr.com/czcn8h

Homework Equations

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The Attempt at a Solution

I know that if you derive x^2sin(1/x)
you get
-cos(1/x) + sin(1/x)(2x).
But what do I do from here? If I use the limit definition, i'll end up getting something like h(sin(1/h)) after evaluating. I still don't understand how the limit definition will show that this exists.

 

[Mark44]

Homework Statement

http://prntscr.com/czcn8h

Homework Equations

n/a

The Attempt at a Solution

I know that if you derive x^2sin(1/x)
you get
-cos(1/x) + sin(1/x)(2x).
But what do I do from here? If I use the limit definition, i'll end up getting something like h(sin(1/h)) after evaluating. I still don't understand how the limit definition will show that this exists.

Please show us what you did in using the definition of the derivative.

BTW, you don't "derive" x^2 sin(1/x) -- you differentiate it. If you start from a quadratic equation, you can use completing the square to derive the quadratic formula.

 

[andrewkirk]

You need to do two things

First show that $\lim_{x\to 0}g'(x)$ does not exist. That should be easy using the derivative you have calculated above.

Second, try to calculate $g'(0)$ which is defined as
$$\lim_{h\to 0}\frac{g(h)-g(0)}{h}$$
If that limit exists then you are finished.

The reference to the 'limit definition' in the question is a bit confusing as there are two different limits involved in this question. They are referring to the definition of the derivative as a limit (the second formula I wrote above), not to the limit of $g'(x)$ as $x\to 0$