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Let $f:\mathbb{R}^m\to\mathbb{R}^n$. Show that the $i$-th row of the differential, $D_f$ is the gradient of $i$-th function, $\nabla f_i$

I understand it intuitively, because I know that $(D_f)_{ij} = \frac{df_i}{x_j}$

But somehow I wasn't able to write it down formally. I'd be glad for help with that.

Thanks.

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By definition $(\nabla f_i)_j=\partial f_i/\partial x_j$, and as you know this $(D_f)_{ij}$, where the subscript $j$ denotes the $j$-th component.

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  • $\begingroup$ You used the definition, but I think the question asks to prove it. $\endgroup$ – AlonAlon Jan 18 '15 at 20:59
  • $\begingroup$ In that case I think you should make this clear, the way I read your question is that you know this (i.e. the Jacobian represents the differential), but are confused about notation. $\endgroup$ – Mister Benjamin Dover Jan 18 '15 at 21:01

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