BMAD Posted May 5, 2013 Report Share Posted May 5, 2013 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? Quote Link to comment Share on other sites More sharing options...
0 vigmeister Posted May 5, 2013 Report Share 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. Quote Link to comment Share on other sites More sharing options...
0 dark_magician_92 Posted May 5, 2013 Report Share 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 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 5, 2013 Report Share Posted May 5, 2013 vigmester - nice solve. This is one of the better puzzles of the "find the fallacy" genre. Welcome to the Den! Quote Link to comment Share on other sites More sharing options...
Question
BMAD
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)
= 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 x2, with respect to x, is actually x. Where is the fallacy?
Link to comment
Share on other sites
3 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.