BMAD 62 Report post Posted May 5, 2013 The derivative of x^{2}, with respect to x, is 2x. However, suppose we write x^{2} 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 x^{2}, with respect to x, is actually x. Where is the fallacy? Share this post Link to post Share on other sites

0 vigmeister 0 Report post Posted May 5, 2013 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. Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted May 5, 2013 (edited) 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 May 5, 2013 by dark_magician_92 Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted May 5, 2013 vigmester - nice solve. This is one of the better puzzles of the "find the fallacy" genre. Welcome to the Den! Share this post Link to post Share on other sites

The derivative of x

^{2}, with respect to x, is 2x. However, suppose we write x^{2}as the sum of x x's, and then take the derivative:Let f(x) = x + x + ... + x (x times)

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

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

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

This argument appears to show that the derivative of x

^{2}, with respect to x, is actually x. Where is the fallacy?## Share this post

## Link to post

## Share on other sites