What are the steps needed to solve for the derivative of capital Pi?

Question

I have a function that can be generalized as $\Pi_{k=1}^k v(kx^k)$. v is another differentiable function in this case. I am trying to find the derivative of this function with respect to x, and I found the generalized product rule, but I can't figure out how to apply the chain rule to the generalized product rule as given on Wikipedia. Thanks for your time.

Answer 1

Let $f(x,n)=\prod_{k=1}^nv(kx^k)$. Then we can write $$f(x,n)=v(nx^n)\prod_{k=1}^{n-1}v(kx^k)=v(nx^n)f(x,n-1).$$ Then $$f'(x,n)=v'(nx^n)n^2x^{n-1}f(x,n-1)+v(nx^n)f'(x,n-1).$$ Now $f'(x,n-1)$ can be computed in the same way, so $$f'(x,n)=v'(nx^n)n^2x^{n-1}f(x,n-1)+v(nx^n)[v'((n-1)x^{n-1})(n-1)^2x^{n-2}f(x,n-2)+v((n-1)x^{n-1})f'(x,n-2)]$$ $$=n^2v'(nx^n)f(x,n-1)x^{n-1}+(n-1)^2v(nx^n)f(x,n-2)v'((n-1)x^{n-1})x^{n-2}+v(nx^n)v((n-1)x^{n-1})f'(x,n-2)$$ This eventualy ends as a sum of $n$ terms, with the $j$-th term (the one containing $x^{j-1}$) being $$j^2x^{j-1}[v(nx^n)\cdot v((n-1)x^{n-1})\cdots v((j+1)x^{j+1})\cdot v((j-1)x^{j-1})\cdots v(x)]v'(jx^j).$$ The product in the square brackets is almost the entire product $\prod_{k=1}^n v(kx^k)$, however the factor $v(jx^j)$ is omitted. Hence the product in the square brackets we can compactly write as $$\frac{1}{v(jx^j)}\prod_{k=1}^n v(kx^k)=\frac{f(x,n)}{v(jx^j)}.$$ All together we get $$f'(x,n)=\sum_{j=1}^n \frac{j^2x^{j-1}}{v(jx^j)}f(x,n)v'(jx^j)=f(x,n)\sum_{j=1}^n j^2x^{j-1}\frac{v'(jx^j)}{v(jx^j)}.$$