- #1

SopwithCamel

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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?

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- Thread starter SopwithCamel
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- #1

SopwithCamel

- 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

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

- #3

Simon Bridge

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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?

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