# Thread: Finding out if a function can be derived at certain point

1. ## Finding out if a function can be derived at certain point

Hello,

One question that the professor of my calculus course likes to include in his exams is, if a certain function f(x) can be derived at a certain point (e.g. x=0).

In order to find this out, do I always have to find the limit $\lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}$ or can I also just find the derivative and then look at the limit $\lim\limits_{x\rightarrow x_0}f'(x)$?

2. Originally Posted by thomasdotnet
Hello,

One question that the professor of my calculus course likes to include in his exams is, if a certain function f(x) can be derived at a certain point (e.g. x=0).

In order to find this out, do I always have to find the limit $\lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}$ or can I also just find the derivative and then look at the limit $\lim\limits_{x\rightarrow x_0}f'(x)$?
The latter would seem OK. But you're best advised to go ask your professor - he's the one who will be marking your exam, not me.

3. You don't always have to use the limit

consider f(x) = x^(1/3)

then f ' (x) =1/3 x^(-2/3) = 1/[3x^(2/3)] so f is not differentiable at 0

Often you simply compute f ' and see where it does not exist.

In other cases like f(x) = |x|

then at 0 then f ' = 1 if x>0 and f ' = -1 x< 0

here you use lim f ' at 0 to see it is not differentiable at 0

hope this helps

By the way we talk about whether or not a function can be differentiated
at apoint not derived.

4. Originally Posted by Calculus26
You don't always have to use the limit

consider f(x) = x^(1/3)

then f ' (x) =1/3 x^(-2/3) = 1/[3x^(2/3)] so f is not differentiable at 0

Often you simply compute f ' and see where it does not exist.

In other cases like f(x) = |x|

then at 0 then f ' = 1 if x>0 and f ' = -1 x< 0

here you use lim f ' at 0 to see it is not differentiable at 0

hope this helps

By the way we talk about whether or not a function can be differentiated
at apoint not derived.
I think that last point depends upon whether you speak "British" English or "American" English!

5. What ? The British speak English ?

6. Thank you guys for your help! Just to clarify this once more, is $\lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}=\lim\limits_{x\rightarrow x_0}f'(x)$?

Originally Posted by Calculus26
Often you simply compute f ' and see where it does not exist.
But couldn't there be a case, where the function f' isn't defined at some point, but f is still differentiable there?

Regarding differentiable vs. derivable: Since the Americans vote for one and the British vote for the other, I guess we'll have to let the Australians decide, which one we should use.

7. What the above quote is saying is that f ' (x) not only exists but is continuous which is a more stringent condition.

In fact I was in advanced calculus before i even saw an example of a function which was differentiable at a point but the derivative was not continuous at that point.

Again you can think of functions which are defined at a point but are not

continuous for eg f(x) = |x|/x if x does not equal 0 and = 1 if x = 0

f(0) = 1 but f is not continuous at 0.

But couldn't there be a case, where the function f' isn't defined at some point, but f is still differentiable there?
the definition of f being differentiable at a point is that f ' exists at that point

8. Thank you Calculus26 for that explanation, now it's a lot clearer to me.

Originally Posted by Calculus26
Again you can think of functions which are defined at a point but are not

continuous for eg f(x) = |x|/x if x does not equal 0 and = 1 if x = 0

f(0) = 1 but f is not continuous at 0.
Isn't $f(x)=\frac{|x|}{x}$ undefined at x=0?