1
$\begingroup$

I'm trying to understand the Proof of Chain Rule for functions of 1 independent variable and 2 intermediate variables.

Here's my reasoning, step-by-step:

  1. The book reasons that the proof "consists of showing that if $x$ and $y$ are differentiable at $t=t_0$, then $w$ is differentiable at $t_0$, then $\frac{d w}{d t}\left(t_{0}\right)=\frac{\partial w}{\partial x}\left(P_{0}\right) \frac{d x}{d t}\left(t_{0}\right)+\frac{\partial w}{\partial y}\left(P_{0}\right) \frac{d y}{d t}\left(t_{0}\right)$". That makes exactly zero sense to me. Why do we have to start like this? Is there no geometric proof, akin to 3Blue1Brown's beautiful, intuitive picture of sliders for the independent, intermediate, and dependent variables?
  2. OK, say I buy into that -- that $w$ is differentiable if its inner functions are differentiable. Now the book moves onto $\Delta w=\frac{\partial w}{\partial x}\left(P_{0}\right) \Delta x+\frac{\partial w}{\partial y}\left(P_{0}\right) \Delta y+\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y$, where $\Delta x, \Delta y, \& \Delta w$. What?! Where does the $\varepsilon_{1} \Delta x+\varepsilon_{2}\Delta y$ even come from? There must be a geometrical way to understand this.

All else afterwards is trivial, save for the final statement that

$\frac{d w}{d t}=\frac{\partial w}{\partial x} \frac{d x}{d t}+\frac{\partial w}{\partial y} \frac{d y}{d t}$

Why do we use $\frac{d w}{d t}$ even though $w=f(x(t),y(t))$ and is a function of two variables, not one? Shouldn't it be $\frac{\partial w}{\partial t}$

And as always, constructive criticism > downvotes. I think MSE understands this better than SO. 
$\endgroup$
7
  • $\begingroup$ For 2.: $\varepsilon$s comes from definition of differentiability and geometrically means approximation by plane function in given point. $\endgroup$
    – zkutch
    Jun 22, 2021 at 23:03
  • $\begingroup$ SO has not been overrun with people asking for free answers to their HW problems. I would say they have done something right. $\endgroup$ Jun 22, 2021 at 23:09
  • $\begingroup$ Hi @zkutch, is there a diagram that will help me understand this better? $\endgroup$
    – rb3652
    Jun 22, 2021 at 23:47
  • $\begingroup$ One here people.ucalgary.ca/~aswish/MATH267L33_DiffLinAppr.pdf on first page. $\endgroup$
    – zkutch
    Jun 23, 2021 at 0:08
  • $\begingroup$ Thanks for this @zkutch. Trying to wrap my mind around the diagram there. I have an arbitrary surface $z=f(x,y)$, shown as the parabolic curve. I intend to find $df$ at $(a,b)$. Now why exactly is the linearization formula $f(a,b)+A(x-a)+B(y-b)$. Why is there an $A$ in front of the $(x-a)$ term? Likewise, why the $B$? Also, what does the tangent plane represent? $\endgroup$
    – rb3652
    Jun 23, 2021 at 1:14

1 Answer 1

1
$\begingroup$

By definition, the derivative of a function $f$ at a point $x$ is the unique linear continuous function $u$ such that for all $h$ close to zero, $f(x + h)$ can be expanded linearly in terms of $h$ by means of $u$ with a negligible error; that is, $$ f(x + h) = f(x) + u(h) + o(h), $$ where $o$ is a function defined on a neighbourhood of zero (eventually punctured) such that $\dfrac{o(h)}{\|h\|}$ tends to zero as $h \neq 0$ tends to zero. The continuity of $u$ is taken by definition and the uniqueness ought to be proven (if $u$ and $v$ both satisfy this definition, with respective $o_u$ and $o_v,$ then you can fix a unit vector $e$ and consider $h = r e$ for small radii $r > 0$ and then $(u - v)(e) = \dfrac{o_v(r e) - o_u(r e)}{r},$ this tends to zero as $r \downarrow 0$ but the LHS is unchanged, meaning that $u(e) = v(e)$ for all unit $e,$ and by linearity, everywhere). (This definition applies with greater generality that is usually taught, you only need everything defined on normed vector spaces.) Next, suppose $f$ and $g$ are two functions such that the composition $g \circ f$ makes sense. If $f$ is differentiable at $x$ and $g$ is differentiable at $y = f(x),$ that is, if $f$ has a unique linear expansion around $x,$ the image of $f$ at $x$ falls inside a small neighbourhood of $y = f(x)$ and $g$ also possess a unique linear expansion around $y,$ then the composition $g \circ f$ possesses a unique linear expansion around $x$ and the linear function in this expansion is none other that the composition of the unique linear functions corresponding to the expansions of $f$ and $g.$ This is the chain rule. It just so happens that when the vector spaces where $f$ and $g$ are defined happen to be $\mathbf{R}^p$ and $\mathbf{R}^q$ ($f:\mathbf{R}^p \to \mathbf{R}^q$ and $g:\mathbf{R}^q \to \mathbf{R}^r$), then every linear function is canonically representable by a matrix of the canonical bases of the spaces under consideration. And here comes the conundrum, the matrix expansion of a linear function is not the linear function itself, it is just a very common way to represent the linear function. With the chain rule the same sort of confusion is common, the chain rule states that the composition of derivatives is the derivative of composition, but for resasons that are beyond my intellect, people seems to always present the chain rule using the partial derivative expansion thus killing this intuition. I hope this clarifies some doubts.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .