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For f to be differentiable at x=0, prove thatis finite
I am having trouble with this problem. It asks to: Prove that the function f(x)= x^2*sin(1/x) if x doesn't equal zero and 0 if x=0 is differentiable at x=0 and to compute the derivative at zero.
This is what i've done so far:
Noted f(0)=0 so its defined at 0.
Proved that the limit of f(x) as x->0 is 0 through the squeeze theorem.
-Through this f(x) is continuous*
I think i need to prove that the limit of the derivative from the left and right as x->0 = some number correct? or am i wrong? this way it would prove no sharp turns and the derivative at 0? Can i get some hints or help? Thank you!
To prove that a function is differentiable at some point, prove the existence and continuity of partial derivatives (or derivative) at this point.
You must thus compute the derivative for x not equal to 0 (say a) and the derivative at x = 0 (say b) and prove that limit when x goes to 0 of a = b. Just the way you did it to prove continuity on the function.
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i was looking at the wrong graph. I did most of that and the graph didn't look the same so i assumed i was wrong, but i just made a typo in my graph. Thanks hahah 
