(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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...

2. Relevant 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))

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: The derivative by definition of (x^3)*sin(1/x)

**Physics Forums | Science Articles, Homework Help, Discussion**