# Derivation of geometric mean? [closed]

It is given the geometric mean:

$$f(x) =(\prod_{i=1}^{k}x_i)^{1/k}$$ on $R_{++}$

The first derivation is a chain rule: $f'(x)=\frac{1}{k}(\prod_{i=1}^{k}x_i)^{\frac{1}{k}-1} \cdot (\text{inner derivation})$

How do I do the inner derivation: $\frac{d}{dx}\left(\prod_{i=1}^{k}x_i\right)$?

## closed as unclear what you're asking by DonAntonio, NCh, Chris Custer, ancientmathematician, Claude LeiboviciApr 2 '18 at 7:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What is $\;x\;$ and how do you differentiate with resepct to it the first expression? – DonAntonio Apr 1 '18 at 11:31
• the best will be to take the logarithm on both sides – Dr. Sonnhard Graubner Apr 1 '18 at 11:39
• @Peter Then your question makes no sense, as what are then the $\;x_i$ 's ?? – DonAntonio Apr 1 '18 at 11:49
• $x_i$ is a scalar > 0 – Peter Apr 1 '18 at 12:01
• Differentiate in elementwise: $$\frac{d}{dx_i} \left(\prod_{j=1}^{k}x_j\right) = \left(\prod_{\substack{j=1 \\ j \ne i}}^{k}x_j\right)$$ – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 12:10

It must be a multivariable ($k$-variable) function: $$f(x)=f(x_1,x_2,\cdots,x_k)=\left(\prod_{i=1}^k x_i \right)^{1/k}.$$ You can take a partial derivative: $$f_{x_i}=\frac{1}{k}\cdot x_i^{\frac{1}{k}-1} \prod_{j=1, j\ne i}^k x_j.$$
For example: $f(x_1,x_2,x_3)=(x_1x_2x_3)^{1/3}:$ $$f_{x_1}=\frac13 x_1^{1/3-1}\cdot x_2^{1/3}x_3^{1/3}$$