# 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) = xk and g(x) = x. Then (fg)(x) = xk + 1 is the function whose derivative we need to calculate:

(xk + 1)' = (xk)' x + xk x' = 0,

because both derivatives above vanish according to the inductive assumption.

What went wrong?

### References

1. E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 91 Copyright © 1996-2018 Alexander Bogomolny

### 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 (xk + 1)' = (xk)' x + xk x' becomes (x1)' = (x0)' x + x0 x' = x', which does not prove the statement for n = 1. Copyright © 1996-2018 Alexander Bogomolny

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