Suppose we have a function f (x , y) where x and y are functions of u and v i.e. x (u,v) and y (u,v).
Then, from the chain rule, the following relationship holds $$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}$$
However, on the RHS of the equation, can the two $\partial x$ and $\partial y$ not cancel like a fraction, which would then give $\frac{\partial f}{\partial u} = 2\frac{\partial f}{\partial u}$, which is a contradiction?
In fact, when is it that the derivatives can be cancelled out like in a fraction, and when can they not?
Thank you for your help.
Note that they don’t cancel because they represent product of derivatives, for instance
$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u}$$
$$\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial v}$$
To better understand the concept you should consider that chain rule is just obtained by matrix product of gradient and/or jacobians.
In the example given:
$$\nabla f(u,v)= \begin{bmatrix}f_u\\f_v\end{bmatrix}= \begin{bmatrix}x_u&y_u\\x_v&y_v\end{bmatrix}\cdot \begin{bmatrix}f_x\\f_y\end{bmatrix}$$