- #1

- 2

- 0

For example,

The derivative of f(x)=(x^2)(sinx) at x=0 is 0 (using limit definition). Is that all the proof needed to show that the function is differentiable at x=0?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter SopwithCamel
- Start date

- #1

- 2

- 0

For example,

The derivative of f(x)=(x^2)(sinx) at x=0 is 0 (using limit definition). Is that all the proof needed to show that the function is differentiable at x=0?

- #2

lavinia

Science Advisor

Gold Member

- 3,269

- 654

try to convince yourself that the function is automatically continuous at the point

- #3

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,656

If one uses the limit definition of a derivative (lim of (f(x)-f(a)) / (x-a)) as x approaches a) on a function and you get a value (ie. it is not undefined) does that mean the derivative of the function at that point exists?

http://www.mathcs.org/analysis/reals/cont/derivat.html

You'd normally just say $$f^\prime(a)=\lim_{(x-a)\rightarrow 0}\frac{f(a+(x-a))-f(a)}{x-a}$$... follows from the definition of a derivative. If the limit exists then the function is differentiable at point a

Well, in each of those cases, the limit won't converge will it? Well... the above is basically a one-sided limit: see below.In other words, even if the limit definition of the derivative works, do you still need to determine whether the function is continuous, smooth and non-vertical at x=a in order to know that the function is differentiable at x=a?

In this case, yep.The derivative of f(x)=(x^2)(sinx) at x=0 is 0 (using limit definition). Is that all the proof needed to show that the function is differentiable at x=0?

However, it gets conceptually hairy when we include things like the Cantor function.

Is the Cantor function "continuous"? Is it differentiable?

Last edited:

Share: