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...
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.