Proving differentiability of a function across a domain

[geo222]

Hi,

I'm looking at part b from this question below:



The method that I am using is the definition of differentiability and the step that I can't see working is the part highlighted in green (I believe that it should simplify to -cos(1/x) to give the general differential but I don't see how)



Could I get some help with the highlighted bit, and some clarification that the method I'm doing is the correct way to prove this. Thanks for any help

 

[tkhunny]

Why are you using the definition of a derivative to demonstrate either continuity or differentiability? You should first be working with limits. Keep in mind that you can simplify this by relying on previously-known results. Consider, in particular, \(\displaystyle \lim\limits_{x\rightarrow 0}\dfrac{\sin(x)}{x} = 1\). Does that get us anywhere?

 

[HallsofIvy]

The problem asks you to show that the function is "differentiable in \(\displaystyle R\)" which means "differentiable at every point in \(\displaystyle R\)". I hope it is obvious that, at every point except 0, since it is a product of differentiable functions, it is differentiable there. The only question is whether or not it is differentiable at x= 0. To determine whether or not it is differentiable there, we need to try to find the derivative, using the definition of "derivative", as you say.

That definition is "\(\displaystyle \lim_{h\to 0}\frac{f(0+ h)- f(0)}{h}\)". Here, that gives \(\displaystyle \lim_{h\to 0}\frac{h^2sin(1/h)}{h}= \lim_{h\to 0} hsin(1/h)\).

Does that limit exist? That's all you need to answer. I don't see why you would mention "cos(1/x)" at all!

 

[geo222]

Oh ofcourse I see now, I was trying to prove a general case rather than focusing on x=0.

For part C, do I replicate the same steps just using the differential of the original function? The differential being 2xsin(1/x) - cos(1/x)

 

[tkhunny]

Seems like a good plan.