Applying the Product Rule to a Non-Differentiable Function

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Homework Help Overview

The discussion revolves around the application of the product rule in calculus, specifically concerning a function that is not differentiable at a certain point. The original poster presents a function defined piecewise, questioning the validity of applying the product rule when one of the components is not differentiable.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of applying the product rule to a function involving a non-differentiable component. There are inquiries about the existence of limits and the conditions under which the product rule can be applied. Some participants question the approach of using the limit definition of the derivative.

Discussion Status

The discussion is ongoing, with participants sharing differing views on the existence of limits and the applicability of the product rule. Some have suggested revisiting foundational definitions, while others express skepticism about the original poster's findings.

Contextual Notes

There is a focus on the behavior of the function at the point where it is not differentiable, and the constraints of the problem include the specific definitions of the functions involved and their derivatives at that point.

StephenPrivitera
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Can the product rule be applied if one of the functions is not differentiable? For example,
f(x)={g(x)sin(1/x), x not =0
=0, x=0
where g(0)=g'(0)=0.
f'(0)=g'(0)sin(1/0) + g(0)dsin(1/x)/dx
=0sin1/0+0dsin(1/x)/dx=0?
applying the limit definition, I get
f'(0)=g'(0)lim sin(1/h) where h-->0
is this zero?
 
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It turns out that in order to find f'(0) I had to go back to the e-d definition of limit. Anyone see an easier way?
 
The limit you're describing does not exist.a

Product rule:
f(x)g(x)=f'(x)g(x)+f(x)g'(x)

You need derivatives of both.

Now we have
g(x)sin(1/x)

the derivative of sin(1/x) is
x-2cos(1/x)
and not defined at zero (no limit at zero either)
you'd need g(x) to grow at better than x2 to have a potential derivative there.
 
Originally posted by NateTG
The limit you're describing does not exist.a
Hi NateTG, I have found the limit. See attached.
I had to go back to epsilons and deltas. I was wondering if anyone knows an easier way to find the derivative.
 

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