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 ${\frac {x^{k}}{k+1}}$ if $k\neq 0$. (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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