# 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
3 years, 8 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.

- 3 years, 8 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.

- 3 years, 8 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 )

- 3 years, 2 months ago

very nicely explained. Thanks a ton mate.

- 3 years, 1 month ago

- 3 years, 1 month ago

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

- 3 years, 8 months ago