#5**+5 **

As one of my calculus teachers would have said.....differentiability implies continuity, but continuity does not imply differentiability.....

To see why, look at the absolute value function at x = 0. Note that, it is continuous at that point but the derivative is undefined. (What is the "slope" of the function at that point .....1 or -1 ??)

We can see this in one more way......the absolute value of a number "x" is defined as √(x^{2}).....and taking the derivative of this gives us ....... (1/2) (x^{2})^{-1/2} *(2x) = x / √(x^{2}) = x / lxl ........and note that, this derivative is undefined at x = 0 !!!!

CPhill
Sep 3, 2014

#2**+5 **

False. This answers your question:

http://oregonstate.edu/instruct/mth251/cq/Stage5/Lesson/diffVsCont.html

AzizHusain
Sep 3, 2014

#4**+5 **

I'm just thinking.

If f(x) is continuous in an open region does that mean it could be a relation that is not a function.

Like a circle or an s shape?

Relations that are not functions are not differentiable either.

Melody
Sep 3, 2014

#5**+5 **

Best Answer

As one of my calculus teachers would have said.....differentiability implies continuity, but continuity does not imply differentiability.....

To see why, look at the absolute value function at x = 0. Note that, it is continuous at that point but the derivative is undefined. (What is the "slope" of the function at that point .....1 or -1 ??)

We can see this in one more way......the absolute value of a number "x" is defined as √(x^{2}).....and taking the derivative of this gives us ....... (1/2) (x^{2})^{-1/2} *(2x) = x / √(x^{2}) = x / lxl ........and note that, this derivative is undefined at x = 0 !!!!

CPhill
Sep 3, 2014