Chain Rule (partial derivatives): basic interpretation question

[nomadreid]

TL;DR Summary

If z is a a continuous function of x,y, then dz=(∂z/∂x)dx+(∂z/∂y)dy is a basic formula whose intuition escapes me, unless one treats dx and dy as vectors, which doesn't seem right.

The proof for the above ubiquitous formula (as in the summary) in "Chain rule for one independent variable" at the beginning of
https://math.libretexts.org/Bookshe...5:_The_Chain_Rule_for_Multivariable_Functions
is something that I need to work through, but I don't see the forest for all the trees: that is, along with the formal proof I would like to have a rough intuition here, which fails me when I consider that
(∂z/∂x)dx seems to be the (infinitesimal) change in z in the xz plane, and similarly
(∂z/∂y)dy seems to be the change in z in the yz plane,
which would seem to indicate that the corresponding change in z would be (if dx, dy and dz are scalars) the Pythagorean combination rather than the simple addition as if they were vectors.

Obviously I am looking at it wrongly; any pointers to correct this would be greatly appreciated.

 

[PeroK]

You can see it from a first-order Taylor series expansion:
$$z(x + \Delta x, y + \Delta y) \approx z(x, y + \Delta y) + \frac{\partial z}{\partial x}\Delta x \approx z(x, y) + \frac{\partial z}{\partial y}\Delta y + \frac{\partial z}{\partial x}\Delta x$$So:
$$\Delta z = z(x + \Delta x, y + \Delta y) - z(x, y) \approx \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y$$

 

[fresh_42]

TL;DR Summary: If z is a a continuous function of x,y, then dz=(∂z/∂x)dx+(∂z/∂y)dy is a basic formula whose intuition escapes me, unless one treats dx and dy as vectors, which doesn't seem right.

Obviously I am looking at it wrongly; any pointers to correct this would be greatly appreciated.

What makes you think you are wrong? The total differential is vector in the cotangent space $T^*_pU=\{T_pU\longrightarrow \mathbb{R}\}$ and $\dfrac{\partial z}{\partial x_k}$ its coordinates in the standard basis $dx_k.$ You can look at it as
$$
(dz)_p(X) =\left. \dfrac{d}{dt}\right|_{t=0} z(\gamma(t))=\bigl\langle \operatorname{grad}z,X \bigr\rangle
$$
with a path $\gamma(t)$ through $p$ along the vector field $X,$ i.e. $\gamma(0)=p\, , \,\dot\gamma (0)=X.$

 

[nomadreid]

You can see it from a first-order Taylor series expansion:
$$z(x + \Delta x, y + \Delta y) \approx z(x, y + \Delta y) + \frac{\partial z}{\partial x}\Delta x \approx z(x, y) + \frac{\partial z}{\partial y}\Delta y + \frac{\partial z}{\partial x}\Delta x$$So:
$$\Delta z = z(x + \Delta x, y + \Delta y) - z(x, y) \approx \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y$$

PeroK: That is a much nicer proof than the one I cited from the website, thank you.

What makes you think you are wrong? The total differential is vector in the cotangent space $T^*_pU=\{T_pU\longrightarrow \mathbb{R}\}$ and $\dfrac{\partial z}{\partial x_k}$ its coordinates in the standard basis $dx_k.$ You can look at it as
$$
(dz)_p(X) =\left. \dfrac{d}{dt}\right|_{t=0} z(\gamma(t))=\bigl\langle \operatorname{grad}z,X \bigr\rangle
$$
with a path $\gamma(t)$ through $p$ along the vector field $X,$ i.e. $\gamma(0)=p\, , \,\dot\gamma (0)=X.$

Super! Thanks, fresh_42.