Students in calculus learn the product rule of differentiation as (f*g)'= f'*g+f*g', but a common mistake is taking the product rule as (f*g)'= f'*g'.
Usually that answer is wrong, but there are pairs of functions f and g where the wrong product rule generates the right answer.
If f,g is one of those special pairs and f=x^{n} (n≠0), then what is function g?
Let g(x) = C/(nx)^n.
We want to show that
[x^n]'*g'(x) = x^n*g'(x) + [x^n]'*g(x)
or
([x^n]'  x^n)*g'(x) = [x^n]'*g(x)
([x^n]'  x^n)*g'(x) = (n*x^(n1)  x^n)*[n*C/(nx)^(n+1)]
= n*x^(n1)*C/(nx)^n
= [x^n]'*g(x)

Posted by Bractals
on 20051130 10:16:56 