# Talk:Derivative of a constant

I just deleted this paragraph:

- There is no polynomial function
*f*such that*f'*(*x*) =*x*^{-1}; in fact, one such function*f*is the natural logarithm. A polynomial*f*exists such that*f'*(*x*) =*x*^{k}, for*k*= 0, 1, 2, ... . That function has degree*k*+1. In the*x*^{-1}case,*k*+1=0, and this pattern breaks down: the derivative of any function of degree 0 is the derivative of a constant. Which is 0: the anti-derivative of*x*^{-1}can't be a polynomial.

That no antiderivative of *x*^{-1} is a polynomial does not follow from the fact that the derivative of a constant function is zero.
The power rule does not say that any antiderivative of *x*^{-1} must have polynomial degree zero. It says that an antiderivative of *x*^{k} is if . (nothing more).

If you want to show that no polynomial function *f* such that *f'*(*x*) = *x*^{-1} exists, use the fact that the derivative of a polynomial function is a polynomial (which needs the power rule for all nonnegative powers), and prove that *x*^{-1} is not a polynomial. This can be shown in a number of ways, for instance noting that polynomials are continuous everywhere and *x*^{-1} has a non-removable discontinuity at zero.
Sympleko 20:19, 28 Apr 2005 (UTC)

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