How Do You Handle a Discontinuous Derivative in Calculus?

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Homework Statement



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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.
 
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Arnoldjavs3 said:

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.
 
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##
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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