All Powers of x are Constant

We are going to prove that for any integer n ≥ 0, (xⁿ)' = 0, where 'prime' means, as is common, the derivative of a function; in this case, f(x) = xⁿ. (We only consider positive x.)

The proof is by induction. To start with, observe that (x⁰)' = 0 because x⁰ = 1 is really a constant. Now assume that the statement has been already proved for n = 0, 1, 2, ..., k, and let n = k + 1. Recollect the formula for the derivative of the product of two functions:

(fg)' = f ' g + fg'.

Take f(x) = x^k and g(x) = x. Then (fg)(x) = x^{k + 1} is the function whose derivative we need to calculate:

(x^{k + 1})' = (x^k)' x + x^k x' = 0,

because both derivatives above vanish according to the inductive assumption.

What went wrong?

References

E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 91

What Went Wrong?

Proof by mathematical induction consists of two,one,two,three,foursteps: the base of induction wherein the statement to be proved is verified for one or more first admissible values of the parameter.

The induction is often compared to a chain of upstanding dominoes,dominoes,cubes,pyramids,Santa's reindeersthat fall down one after another once the first one was set in a motion. The second (inductive) step need to be able to continue from the point verified at the first step. Since on the first step we only verified the statement for n = 0, the inductive step ought to be valid and be in a position to continue starting with n = 1!. But in this case, the formula for the product (x^{k + 1})' = (x^k)' x + x^k x' becomes (x¹)' = (x⁰)' x + x⁰ x' = x', which does not prove the statement for n = 1.

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