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I have a question:

Consider a function $g:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is differentiable. Find the derivative of the function: $G(x)=[ g\left ( x,x^{2},...,x^{n} \right )]^{2}$ where $x\in \mathbb{R}$.

Here, G is a function of one variable, so I tried to apply the chain rule to find its derivative as follows: $G^{'}\left ( x \right )=2H^{'}\left ( x \right )H(x)$ where $H\left ( u_{1},u_{2},...,u_{n} \right )=g\left ( x,x^{2},...,x^{n} \right )$. Now to find the derivative of $H$, I did the following: $$\frac{d}{dx}H=\frac{\partial H}{\partial u_{1}}\frac{\partial u_{1}}{\partial x}+...\frac{\partial H}{\partial u_{n}}\frac{\partial u_{n}}{\partial x}.$$ Does what I did make sense?

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    Do you really mean to have the $\lfloor \;\;\rfloor$ ([the floor function](http://en.wikipedia.org/wiki/Floor_and_ceiling_functions)) in the definition of $G$? Perhaps you intended to have brackets $[\;\;]$?2012-01-29
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    Provided one replaces $H(u_1,u_2,\ldots,u_n)$ by $H(x)$ and each $\partial H/\partial u_k$ by $\partial g/\partial u_k$, the answer is: yes.2012-01-29
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    @ Zev Chonoles: Yes, I just mean simple brackets only.2012-01-29
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    @pin296: Ah, I see that you actually did have brackets, but my browser was for some reason not displaying them correctly - apologies.2012-01-29
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    Here is my answer based on the above comments: $G^{'}\left ( x \right )=2.g\left ( x,x^{2},...,x^{n} \right ).\left [ \frac{\partial g}{\partial u_{1}}.1+\frac{\partial g}{\partial u_{2}}.2x +...+nx^{n-1}.\frac{\partial g}{\partial u_{n}} \right ]$. where : $u_{i}=x^{i}$. I am not convinced to the presence of the variables $u_{i}=x^{i}$ in my final answer. Does anyone have a better idea on how to solve the problem?2012-01-29

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