Why is $d\phi=\nabla\phi.d\vec r$ in multivariable calculus in both the Cartesian system and the new curvilinear coordinates ($u_1, u_2, u_3$), provided $u_1, u_2, u_3$ are ortogonal to each other?
I actually tried evaluating each term out but I arrived at a contradiction:
$\vec r = x\widehat i+y\widehat j +z\widehat k$, where x, y, z are functions of $u_1, u_2, u_3$, the new coordinate system.
$$d\vec r= \frac{\partial \vec r}{\partial u_1}du_1+\frac{\partial \vec r}{\partial u_2}du_2+\frac{\partial \vec r}{\partial u_3}du_3$$
$$d\vec r= (\widehat u_1 |\frac {\partial \vec r}{\partial u_1}|du_1)+(\widehat u_2 |\frac {\partial \vec r}{\partial u_2}|du_2)+(\widehat u_3 |\frac {\partial \vec r}{\partial u_3}|du_3)$$
Let $h_1=|\frac{\partial \vec r}{du_1}|$ and so on...
$$d\vec r= \widehat u_1 h_1du_1+\widehat u_2 h_2du_2+\widehat u_3 h_3du_3$$
Now, for $\nabla \phi$ we simply apply the definition of $\nabla$,
$$ \nabla \phi=\frac{\partial \phi}{\partial u_1}\widehat u_1+\frac{\partial \phi}{\partial u_2}\widehat u_2+\frac{\partial \phi}{\partial u_3}\widehat u_3$$
$$\nabla\phi.d\vec r=h_1du_1\frac{\partial \phi}{\partial u_1}+h_2du_2\frac{\partial \phi}{\partial u_2}+h_3du_3\frac{\partial \phi}{\partial u_3}$$
Now this is clearly a contradiction as:
$$d\phi=du_1\frac{\partial \phi}{\partial u_1}+du_2\frac{\partial \phi}{\partial u_2}+u_3\frac{\partial \phi}{\partial u_3}$$
I'm sure I'm making a mistake somewhere as $d\phi=\nabla\phi.d\vec r$ is a well-established result for transformation from Cartesian coordinates to any orthogonal curvilinear coordinates. I know this proof is used for the deviation of the formula of the nabla operator ($\nabla$) for the new curvilinear coordinates. Yet, I arrive at a contradiction. Can someone help me resolve this?