Derivative of a Polynomial Function

Question

Show that there is no polynomial function that has a derivative of x^-1.

Answer 1

Jim's proof is correct but it uses the fact that integer exponents of x are linearly  independent over the field of reals. A shorter argument would be the following: A derivative of a polynomial is a polynomial and thus defined at x=0. On the other hand x^-1 is not defined at zero.

 

 

Answer 2

^ Just to elaborate on the above, n may be 0 for a term (in which case the term is simply a constant); the above analysis is used for terms with n>0, but for n=0, the derivative of a constant is simply 0.

 

so one proof, as suggested by above, is to use contradiction and explain that constant terms are zeroed out while we equate one of the non-constant terms (i.e. with n>0) (if we assume the incorrect conclusion to be true) with x^-1, which would lead us to conclude that n=0 thereby contradicting our assumption that n>0 for that term.

Answer 3

Hi Shreya,

 

A polynomial is always the sum of terms like axⁿ where a a constant and n is an integer >0. The derivative is ax^n-1 so if this were to be x^-1 then n-1=-1 or n=0 but n>0 QED

 

Jim