Product rule

StephenPrivitera

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?

Related Introductory Physics Homework Help News on Phys.org

StephenPrivitera

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?

NateTG

Homework Helper
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.

StephenPrivitera

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

• 41 KB Views: 99

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving