This has been confusing me for awhile.(adsbygoogle = window.adsbygoogle || []).push({});

Let's say f(x) is undefined at x=2 and thus f(x) is discontinious at x=2.

And it's derivative is f'(x)=2X. Is the derivative still defined at x=2 or not, because f(x) is undefined at x=2?

If it is defined, wouldn't this mean f(x) is differentiable at x=2 and thus f(x) has to be continuous at x=2 because of the theorem:

If f(x) is differentiable at a then f(x) must be continuous at a.

One last question. Why is (x+1)(x-1)/(x-1) still undefined when x=1? Even though the x-1 cancel out.

Thanks in advance.

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# Derivative question

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