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Who wants to chew on my Calculus question?


Sarahisme

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It would depend on te function.

 

 

sqrt(x) on the positive reals (ie including zero) is not differentiable. On the strictly positives it is differentiabel at all points.

 

You simply show that for any given point of the domain the function is differentiable. Unless you have specific example in mind there is nothing more that anyone can say to you.

 

alext87's post is absolutely wrong.

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Like matt grime said, it depends on the function. You can check a few things though. If you want it to be differentiable at every real x then it has to be continous at every real x. Also, make sure there are no sharp points, like in the absolute value function at 0.

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or you could use the fact that |x|^3 is just x^3 if x is positive, and -x^3 if x is negative, and thus clearly differentiable at all points but 0. so it suffices to check the derivative at 0, from the right you get the derivative being as for x^3, and the left as -x^3, both of which give 0 at the origin, so it is differentiable.

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Guest eKo.RavanDi

first of all the function should be continous at all of the points in its domain. because there is no derivation in the end-points. then you should check for the sharp points. each point is NOT a sharp point if the left and right derivatives are equal. f'+(a)=f'-(a). the left and right derivatives are the left and right limits of f(x)-f(a)/x-a when x gets near to a.

sorry for bad english, ;)

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