# Differentiation of geometric mean

I've just encountered the following problem. Define the following geometric mean $$Y_t = \left( \prod_{k \neq i} X^k_t \right)^{1/n}$$ where $$i = 1,2,...,n$$ and $$t$$ means time. My question is how to take following differentiation $$\frac{dY_t}{Y_t}?$$ I tried several times but got stuck. I appreciate any hint or answer. Many thanks!

• $\frac{dY_t}{Y_t}$? What is this notation supposed to mean? Also, show your attempts – Brevan Ellefsen May 27 at 12:41
• If I understand correctly, I believe that $\frac{dY_t}{Y_t}$ means the differentiation of $Y_t$ devided by $Y_t$. I wrote $\prod_{k \neq i} X^k_t = X^1_t X^2_t ... X^{i-1}_t X^{i+1}_t ... X^n_t$ and then tried to get $\frac{dY_t}{Y_t}$ but getting overwhelmed since there are many variables. – Min Haw May 27 at 12:50

$$\log Y_t=\frac1n\log\left(\prod_{k\ne i} X_t^k\right)=\frac1n\sum_{k\ne i}\log X_t^k$$
$$\frac{dY_t}{Y_t}=\frac1n\sum_{k\ne i}\frac{d X_t^k}{ X_t^k}.$$
• Thanks for your response. Do you mean $d \log Y_t = \frac{dY_t}{Y_t}$? I am not sure whether we can treat $Y_t$ as a normal variable in this case or not. – Min Haw May 27 at 13:14