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A function f is not differentiable if it is not continuous. While your reasoning is correct for x∈N, it does not apply to x∈R∖N; hence, f is not continuous and therefore not differentiable.
When you are defining x^2=x+x+..+x(x times), you are tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) cannot be differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point of the set N. Here lies the fallacy.
(last modified 1 hour ago )
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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A function f is not differentiable if it is not continuous. While your reasoning is correct for x∈N, it does not apply to x∈R∖N; hence, f is not continuous and therefore not differentiable.
Try writing π2 as π a π number of times. Or more simply 3.52 as a sum of 3.5 3.5′s . To make it more obvious just write 25 as 5 added 5 times.
When you are defining x^2=x+x+..+x(x times), you are tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) cannot be differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point of the set N. Here lies the fallacy. (last modified 1 hour ago )
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very nicely explained. Thanks a ton mate.
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Happy to help you..:-)
In differentiation, we are concerned with the ratio of change of one quantity at a particular instant with respect to another.
See this lovely note by the great @Agnishom Chattopadhyay
@Aman Sharma