The derivative by definition of (x^3)*sin(1/x)

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Homework Statement


define h(x) = (x^3)*sin(1/x) , x not= 0, and h(0) = 0. Show h is differentiable everywhere, h' is continuous everywhere but fails to have a derivative at exactly one point. You can use rules for differentiating, products, sums, and quotients learned in calculus...


Homework Equations


f : D -> R has a derivative at some x0 in D that is also an accumulation point of D iff lim t->0 (f(x0 + t) - f(x0))/t exists iff lim x -> x0 (f(x0) - f(x) )/(x-x0) exists

sinA - sinB = 2sin(1/2(A-B))cos(1/2(A+B))


The Attempt at a Solution


To show that a function is differentiable at all points in its domain, must show the limit (f(x0+t) - f(x0))/t exists as t gets small. I suppose there is a trick with differentiability I am missing. For example, I know I could use the product and chain rules on the function and get the derivative in a couple seconds. But we haven't really gotten there yet, and even though the problem says they are fair game, I am not sure how to show the function is differentiable everywhere by taking the derivative.

Here is what I was thinking, but am not sure how to prove. We also have to show the derivative is continuous everywhere. I can show that I am sure. But if the derivative is continuous everywhere, does that mean that the function is differentiable everywhere?

So, I have two equivalent definitions. I choose the definition involving t. Need to show that the
lim t->0 [(x0 + t)^3sin(1/x0 + t) - x0^3sin(1/x0)] / (x - x0) exists

There is nothing here to really work with so adding and subtracting (x0+t)^3sin(1/x0) seems like a doable choice. This allows one half of the limit to certainly exist.

Let c = (x0 + t)^3sin(1/x) - x0^3sin(1/x) = sin(1/x)*( (x0 + t)^3 - x0)^3)
= sin(1/x0)*[t^3 + 3tx0^2 + 3x0t^2]
Let d = (x0 + t)^3(sin(1/ (x0 + t) - sin(1/x0))
And h is differentiable iff lim t->0 (d - c)/t exists iff lim t->0 d/t exists and lim t->0 -c/t exists.

The lim t->0 -c/t exists. One could easily factor out a t, and then it follows nicely. But what about the limit of d / t ? I tried using the sinA - sinB identity, and things just get messier. I am tempted to believe that things may have to get really messy in this one, because the chain rule is involved in differentiating the actual problem. I just hit a wall.
 

Answers and Replies

  • #2
gabbagabbahey
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So, I have two equivalent definitions. I choose the definition involving t. Need to show that the
lim t->0 [(x0 + t)^3sin(1/x0 + t) - x0^3sin(1/x0)] / (x - x0) exists

You mean [tex]\lim_{t \to 0} \frac{(x+t)^3sin(\frac{1}{x+t})-x^3sin(\frac{1}{x})}{t}[/tex] right?

Start by expanding [itex](x+t)^3[/itex], then divide up the fraction into two terms: (1) a [tex]\frac{x^3\left( sin(\frac{1}{x+t})-sin(\frac{1}{x}) \right)}{t}[/tex] term and (2) the leftovers

For your first term, you will have to use your trig identity and for the second term, just factor out a t and compute the limit directly.
 
  • #3
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You mean [tex]\lim_{t \to 0} \frac{(x+t)^3sin(\frac{1}{x+t})-x^3sin(\frac{1}{x})}{t}[/tex] right?

For your first term, you will have to use your trig identity and for the second term

Yes. that is what I meant.

Now, going on what you said, I get something similar to the problem I originally was trying to solve. I end up with a limit:
[tex]
\lim_{t \to 0} \frac{x^3(2*sin(\frac{-t}{2x(x+t)})*cos(\frac{2x+t}{2x(x+t)}))}{t}
[\tex]

And I just don't know how to get rid of division by 0 in this case, to make the derivative work.
 
  • #4
gabbagabbahey
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Well the cosine term is no problem as t approaches zero, so use the limit of products riule so that you will only need to find:

[tex]\lim_{t \to 0} \frac{sin(\frac{-t}{2x(x+t)})}{t}[/tex]

For which you can use l'Hôpital's rule. (Or, if you really don't want to use derivatives, use the power series representation of sine)
 
Last edited:
  • #5
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Well the cosine term is no problem as t approaches zero, so use the limit of products riule so that you will only need to find:

[tex]\lim_{t \to 0} \frac{sin(\frac{-t}{2x(x+t)})}{t}[/tex]

For which you can use l'Hôpital's rule. (Or, if you really don't want to use derivatives, use the power series representation of sine)

I would use l'Hopitals rule, except "I don't know that yet." The problem is doing this with just the definition of differentiability and the rules for continuity and limits and such. Once we can establish that the limit exists, we can use the product and chain rules to see what the derivative looks like, and show other things, such as where is the derivative not continuous at. You are right though, using the product of limits theorem, I can justify only needing to show
[tex]\lim_{t \to 0} \frac{sin(\frac{-t}{2x(x+t)})}{t}[/tex]
But then, how to show this? I cannot factor out a t. There is no self-apparent way to get rid of the t aka division by 0 of the overall limit. This is essentially the question and I think the problem they are getting at.
 
  • #6
gabbagabbahey
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If you can't use l'Hôpital's rule, then use the power series representation of sine; namely:

[tex]Sin(u)=\sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{(2n+1)!}[/tex]

with [itex]u=\frac{-t}{2x(x+t)}[/itex]....only the n=0 term will be non-zero in the limit, since all others will have some power of t as a factor (after you divide through by t).
 
  • #7
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Brilliant! Thank you.

If you don't mind me asking, what about the problem made you think, power series representation of sine would be appropriate?
 

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