Product rule

  • #1
StephenPrivitera
363
0
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?
 

Answers and Replies

  • #2
StephenPrivitera
363
0
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?
 
  • #3
NateTG
Science Advisor
Homework Helper
2,452
7
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.
 
  • #4
StephenPrivitera
363
0
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.
 

Attachments

  • mthhw7.doc
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