# Show the $i$-th row of $D_f$ is $\nabla f_i$

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.

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.