We all are fimiler with the following well known results in calculus:-

**(1)** \(\frac{d(x^n)}{dx}=nx^{n-1}\)

**(2)** \(\frac{d}{dx}(f_1(x)+f_2(x))=f'_1(x)+f'_2(x)\)

Lets try to find the derivative of \(x^2\) by using property (2):-

\[x^2=\underbrace{x+x+x+...........+x}_\text{x times}\]

Differentiating both sides of above equation:-

\[\frac{d(x^2)}{dx}=\underbrace{1+1+1+......+1}_\text{x times}\]

\[\frac{d(x^2)}{dx}=x\]

Which is horribly wrong,because \(\frac{d(x^2)}{dx}=2x\)

I can't identify the fallacy in it ,please help me

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## Comments

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TopNewestA function \(f\) is not differentiable if it is not continuous. While your reasoning is correct for \(x \in \mathbb{N}\), it does not apply to \(x \in \mathbb{R} \setminus \mathbb{N}\); hence, \(f\) is not continuous and therefore not differentiable.

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Try writing \(\pi ^2 \) as \(\pi \) a \(\pi \) number of times. Or more simply \( 3.5^2 \) as a sum of \( 3.5 \text{ } 3.5 ' \)s . To make it more obvious just write \(25\) as \( 5 \) added \( 5 \) times.

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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..:-)

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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

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