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

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• 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: $f(x_1,x_2,x_3)=(x_1x_2x_3)^{1/3}:$ $$f_{x_1}=\frac13 x_1^{1/3-1}\cdot x_2^{1/3}x_3^{1/3}$$