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I have a simple problem im looking at and I dont understand some concepts.

Let me define a multivariable polynomial $f(x_1,\ldots x_n,y_1,\ldots y_n)$. If I need to compute the value of the derivative $h'(\alpha)$ in terms of f (possibly in terms of the partial derivatives of f) would it just be gradient(F)*(individial partial derivatives vector)?

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  • $\begingroup$ Use the chain rule. $h$ is $f\circ V$ whith $V:\mathbb R \rightarrow \mathbb R^{2n},\ t \mapsto (t,\ldots,t,1,\ldots,1)$. $\endgroup$ – jflipp Nov 11 '14 at 21:43
  • $\begingroup$ if you dont mind, could you expand it? would it just be d(f)/d(x_1)+d(f)/d(x_2)+....? $\endgroup$ – user191754 Nov 11 '14 at 22:09
  • $\begingroup$ Yes. $dV/dt$ is the constant vector $(1,\ldots,1,0,\ldots,0)$. So $d(f\circ V)/dt = grad\ f \cdot (dV/dt) = \partial f/\partial x_1 + \cdots + \partial f / \partial x_n$. $\endgroup$ – jflipp Nov 11 '14 at 22:16
  • $\begingroup$ Is is generally true that for any $g:R \rightarrow {R}^n$, $g'(a)=\partial(g)/\partial(x_1)(a)+\ldots \partial(g)/\partial(x_n)(a)$ $\endgroup$ – user191754 Nov 12 '14 at 13:02

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