Differentiability implies continuous derivative?

kelsiu

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Jun 15, 2014
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We know differentiability implies continuity, and in 2 independent variables cases both partial derivatives fx and fy must be continuous functions in order for the primary function f(x,y) to be defined as differentiable.

However in the case of 1 independent variable, is it possible for a function f(x) to be differentiable throughout an interval R but it's derivative f ' (x) is not continuous?
 
No, differentiability does NOT imply that the derivative be continuous. A simple example is y= x^2 sin(1/x) if x is not 0, y(0)= 0. That is itself continuous at x= 0 since sin(1/x) is always between -1 and 1 while x^2 goes to 0. For x not 0, the derivative is y'= 2x sin(1/x)- cos(1/x). At x= 0, the difference quotient is \(\displaystyle \frac{h^2 sin(1/h)}{h}= h sin(1/h)\) which goes to 0 as h goes to 0. So the derivative is 2x sin(1/x)- cos(1/x) for x not equal to 0, 0 at x= 0. But 2x sin(1/x)- cos(1/x) has no limit as x goes to 0 so the derivative is not continuous at x= 0. We sometimes use the word "smooth" to denote a function whose derivative is continuous.
 
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