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=xn (n≠0), then what is function g?
We want g such that f'.g+f.g'=f'.g' when f=x^n. Substituting, we get n.x^n-1.g+x^n.g'= n.x^n-1.g', which implies n.g+x.g'= n.g', or g'/g=n/(n-x). This is a simple differential equation, and integrating both sides we get ln(g)= -n.ln(n-x)+C, C being any constant. Exponentiating, we finally arrive to g=K/(n-x)^n (K is a constant) which, by the way, works even for n=0.
Differential Solution
