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## Question

The derivative of x2, with respect to x, is 2x. However, suppose we write x2 as the sum of x x's, and then take the derivative:

Let f(x) = x + x + ... + x (x times)

Then f'(x)

= d/dx[x + x + ... + x] (x times)

= d/dx[x] + d/dx[x] + ... + d/dx[x] (x times)

= 1 + 1 + ... + 1 (x times)

= x

This argument appears to show that the derivative of x2, with respect to x, is actually x. Where is the fallacy?

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Derivative measures the rate at which a given quantity changes with a change in x. When you write x*x as a sum of x x's, one of them becomes a constant. for instance, if x = 5, we start with 25. But if you increase it to 6, x^2 becomes 36, but (x+x+x+x+x) becomes just 30 (with the plus notation, what you differentiate does not mathematically capture the 'x times' part because you do not magically add a d/dx(x) term when you increase x.

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above poster seems right. It is not right to write f(x)=x+x+x+.....(x times) as x is a variable, so the number of times x comes is also not defined, till we give x a value.

Edited by dark_magician_92
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vigmester - nice solve.

This is one of the better puzzles of the "find the fallacy" genre.

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