# A calculus fallacy

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 $\frac{d(x^2)}{dx}=2x$ Note by Aman Sharma
6 years, 7 months ago

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A 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.

- 6 years, 6 months ago

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.

- 6 years, 7 months ago

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 )

very nicely explained. Thanks a ton mate.

- 5 years, 11 months ago

- 5 years, 11 months ago

In differentiation, we are concerned with the ratio of change of one quantity at a particular instant with respect to another.

- 6 years, 7 months ago