# Thread: f'(x) - Exists / Continous

1. ## f'(x) - Exists / Continous

In many textbooks I have read a statement like this -
if f'(x) exists at x=x0 and is continuous at x=x0 then => some follow-up logic

My question is

Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.

Also if f'(x) is continuous at x=x0 then it obviously it exists at x=x0.

Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent

Am I correct? Or I am missing something?
Thanks

2. Originally Posted by aman_cc
In many textbooks I have read a statement like this -
if f'(x) exists at x=x0 and is continuous at x=x0 then => some follow-up logic

My question is

Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.

Also if f'(x) is continuous at x=x0 then it obviously it exists at x=x0.

Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent

Am I correct? Or I am missing something?
Thanks
You are PARTIALLY correct.

If a function is differentiable at a point, then that implies it is continuous at that point.

HOWEVER

If a function is continuous at a point, it may OR MAY NOT be differentiable at that point.

E.g. $\displaystyle f(x) = |x|$ at the point $\displaystyle x = 0$ is continuous but NOT differentiable.

3. Originally Posted by aman_cc
My question is
Doesn't the existence of f'(x) (first derivative) at x=x0 imply it is continuous at x=x0? I say that because of the way f'(x) is defined at x=x0.
Hence the two statements: 1. f'(x) exists at x=x0 2. f'(x) is continuous at x=x0 are equivalent
Am I correct? Or I am missing something?
You are missing a simple fact: If $\displaystyle f'(x_0)$ exists it does not mean that derivative, $\displaystyle f'$, itself is continuous at $\displaystyle x_0$.
Of course it implies that $\displaystyle f$ is continuous at $\displaystyle x_0$ but not necessarily $\displaystyle f'$.

4. @Plato - I disagree
From the definition of how f'(x) is defined
Let g(x) = f(x) - f(x0) / x - x0
f'(x0) exists => the limit x->x0 [g(x)] exists and f'(x0) = that limit.
this makes f'(x) continous at x=x0.

Is there an example where f'(x0) exists but f'(x) is not continous at x = x0.

Please note I am not talking of continuity of f(x), existence of which has no relevance on existence of f'(x). @Prove It: You seem to have got confused here

5. Originally Posted by aman_cc
@Plato - I disagree
Is there an example where f'(x0) exists but f'(x) is not continous at x = x0.
This is the standard example.
$\displaystyle f(x) = \left\{ {\begin{array}{cl} {x^2 \sin \left( {\frac{1}{x}} \right),} & {x \ne 0} \\ {0,} & {x = 0} \\ \end{array} } \right.$
$\displaystyle f$ has a derivative everywhere.
But the derivative is not continuous.

6. @Plato - Really appreciate your example. Need time to understand it though. If you have any pointers which I could read-up to understand this better, I would appreciate that. Thanks