Whether the interval is open or closed, a function needs to be continuous to be differentiable on that interval? Why it needs to be continuous?
Differentiability always implies continuity because you can prove it. The proof is in most calculus books, or you can google it.
Think about it in the other way, If it is not continuous at a given point can we find the derivative of the function ( instant change) in that point?
Adding to the previous informative answers, it is worth noting that in standard calculus or real analysis textbooks, the theorem that differentiability implies continuity is proved by means of the product rule for limits.
If you want to read an epsilon-delta proof of the theorem, i.e. a proof that uses only the epsilon-delta definitions of differentiability and continuity, you may refer to https://q4quantum.files.wordpress.com/2020/07/differentiabilityimpliescontinuity.pdf
If the interval is open, since the definition of the slope of the tangent line depends on the value of the function then it needs to be continuous almost everywhere in the interval.
However, if the interval is closed, you need to be careful in the extrema. But as some colleagues said, access to some book where this topic is seriously considered. Good luck ;)
Otherwise, you may need to move to week derivative or distributional derivative. In that sense, you can find the derivative of discontinuous functions.
They always say in many theorems that function is continuous on closed interval [a,b] and differentiable on open interval (a,b) and an example of this is Rolle's theorem. ... For instance, a function may be differentiable on [a,b] but not at a; and a function may be differentiable on [a,b] and [b,c] but not on [a,c]. http://mathcentral.uregina.ca/QQ/database/QQ.09.09/h/dave4.html
Check https://math.stackexchange.com/questions/126176/differentiablility-over-closed-intervals
Continuous variable is like temperature of a given area which moves from low to maximum and back and forth and can be easily differentiated with that of another area by comparison.
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