\[U(u_1 ,u_2 ,u_3) , \: V(v_1 ,v_2 ,v_3)\]
, we can express the derivatives of one in terms of the derivatives of the other.Let
\[(x,y,z)=(x(u_1,u_2,u_3),y(u_1,u_2,u_3),z(u_1,u_2,u_3))\]
and
\[(x,y,z)=(x(v_1,v_2,v_3),y(v_1,v_2,v_3),z(v_1,v_2,v_3))\]
The transformations are onto and one to one, so there exists a transformation from
\[U\]
to \[V\]
.We can write
\[(u_1,u_2,u_3)=(u_1(v_1,v_2,v_3),u_2(v_1,v_2,v_3),u_3(v_1,v_2,v_3))\]
The transformation from
\[V\]
to \[U\]
also exists and is one to one.Differentiating,
\[d \mathbf{r} =\frac{\partial \mathbf{r}}{\partial u_1} d u_1 + \frac{\partial \mathbf{r}}{\partial u_2} d u_2 + \frac{\partial \mathbf{r}}{\partial u_3} d u_3 = \frac{\partial \mathbf{r}}{\partial u_i} d u_i\]
where the repeated use of the index
\[i\]
indicates summation.
Similarly \[d \mathbf{r} = \frac{\partial \mathbf{r}}{\partial v_i}dv_i\]
We can equate these two expressions, so
\[\frac{\partial \mathbf{r}}{\partial u_i}du_i = \frac{\partial \mathbf{r}}{\partial v_i}dv_i\]
\[u_i =u_i (v_1,v_2,v_3) \rightarrow du_i =\frac{\partial u_i}{\partial v_j} dv_j\]
Combining the last two equations gives
\[\frac{\mathbf{r}}{\partial v_j} dv_j = \frac{\mathbf{r}}{\partial u_j}\frac{\partial u_i}{\partial v_j} dv_j \]
Cancelling then gives
\[\frac{\mathbf{r}}{\partial v_j} = \frac{\mathbf{r}}{\partial u_j}\frac{\partial u_i}{\partial v_j} \]