I 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.

 

[Mozibur Rahman Ullah]

The chain rule is best understood geometrically. To avoid thinking about higher dimensions, we will just consider 2d surfaces. However, the reasoning that follows goes through with minor adjustments for higher dimensions.

Assume we have a surface $M$, then we can attach to it all its tangent planes. We call this the tangent bundle of the surface and denote it as $TM$. Now if we have a map $f : M \rightarrow N$ from this surface $M$ to another surface $N$, then it is intuitively obvious that all the tangent planes of the first shape maps to the tangent planes of the second. We call this map $Tf$ and in symbols, we expect $Tf: TM \rightarrow TN$. Now it is also geometrically obvious that if we had a further map $g: N \rightarrow P$ to another surface $P$, then composing the maps $Tg\circ Tf$ should equal $T(g\circ f)$. This is best understood through a picture, which unfortunately I can't do here. Anyway, this means that $T$ is what is called a functor. Now if we work locally, that is using charts, then it turns out this functorial rule is precisely the chain rule.

 

[nomadreid]

The chain rule is best understood geometrically. To avoid thinking about higher dimensions, we will just consider 2d surfaces. However, the reasoning that follows goes through with minor adjustments for higher dimensions.

Assume we have a surface $M$, then we can attach to it all its tangent planes. We call this the tangent bundle of the surface and denote it as $TM$. Now if we have a map $f : M \rightarrow N$ from this surface $M$ to another surface $N$, then it is intuitively obvious that all the tangent planes of the first shape maps to the tangent planes of the second. We call this map $Tf$ and in symbols, we expect $Tf: TM \rightarrow TN$. Now it is also geometrically obvious that if we had a further map $g: N \rightarrow P$ to another surface $P$, then composing the maps $Tg\circ Tf$ should equal $T(g\circ f)$. This is best understood through a picture, which unfortunately I can't do here. Anyway, this means that $T$ is what is called a functor. Now if we work locally, that is using charts, then it turns out this functorial rule is precisely the chain rule.

Interesting approach, thanks.