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Let $a,b : I \to \mathbb{R}$ be differentiable functions, and $m: I \to \mathbb{R}^2$ be a function such that $m(t)=(a(t),b(t))$.

Further, let there be a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, and let $z:I\to\mathbb{R}$, $z(t)=f(a(t),b(t))$.

We wish to find $\frac{dz}{dt}$.

We know that $\frac{dz}{dt}=D(z)(t)$. (Total derivative)

So $$D(z)(t)=D(f\circ m)(t)=D(f)(m(t)).D(m)(t)$$ (Chain rule) $$D(f)(m(t))=[\frac{\partial f}{\partial x}(m(t)) \space\space\space\frac{\partial f}{\partial y}(m(t))]$$ By this I mean, partial derivative of $f$ w.r.t $x$ at $m(t)$. $$ D(m)(t)= \left[ {\begin{array}{c} \frac{da}{dt} \\ \frac{db}{dt} \\ \end{array} } \right] $$. Multiplying both gives, $$\frac{dz}{dt}=\frac{\partial f}{\partial x}.\frac{da}{dt}+\frac{\partial f}{\partial y}.\frac{db}{dt}$$ Now how to express this in terms of $z$ only not $f$?

According to the book I have, the answer is $$\frac{dz}{dt}=\frac{\partial z}{\partial a}.\frac{da}{dt}+\frac{\partial z}{\partial b}.\frac{db}{dt}$$ Also, what is the meaning of $\frac{\partial z}{\partial a}$ according to the definition?

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    $\begingroup$ I don't know but your notation here and the book's notation seem to be different. If I try to reconcile the 2, it seems like the book, from your perspective, is using notation such as $z = z(a,b)$ where $a = a(t)$ and $b = b(t)$. That is, they give the 3 functions a name $z, a , b$ then call the output of the function, the same as the function name. This is commonly done in science classes, but causes confusion sometimes. It's better to do as in this question, to call the output of $a(t) = x$, the output $b(t)$ as $y$. $\endgroup$ Commented Oct 4, 2018 at 13:57
  • $\begingroup$ @DWade64, can you please elaborate a little bit more? Maybe an answer? I took $x$ and $y$ as the variables which go inside $f$. That is, $f(x,y)$. $\endgroup$ Commented Oct 4, 2018 at 14:10
  • $\begingroup$ Thank you for the reply but I'm not sure what more to say since I'm getting confused by what seems to be 2 different notations between you and the book. Your work looks correct. You seem to call the composition, you give it the name, $f \circ m = : z$. Which I think is good. The outside function is $f$, the inside function is $m$, giving altogether a new function $z$. However the book seems to call the outside function $z$, with direct input variables labeled as $a$ and $b$. I wish I could help more, but I'm getting confused as well $\endgroup$ Commented Oct 4, 2018 at 14:36
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    $\begingroup$ @DWade64 Okay. Well I could wait for an answer which tells what is correct then. :) $\endgroup$ Commented Oct 4, 2018 at 14:37

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The formula

$$\frac{dz}{dt}=\frac{\partial z}{\partial a} \cdot \frac{da}{dt}+\frac{\partial z}{\partial b} \cdot \frac{db}{dt}$$

given in your book does not make sense. In fact, $z$ is a function of one variable $t$ and there are no partial derivatives $\frac{\partial z}{\partial a} ,\frac{\partial z}{\partial b}$ with respect to two variables $a, b$. The correct formula is nothing else than

$$\frac{dz}{dt}=\frac{\partial f}{\partial x} \cdot \frac{da}{dt}+\frac{\partial f}{\partial y} \cdot \frac{db}{dt}$$

or, written pointwise,

$$\frac{dz}{dt}(t_0) =\frac{\partial f}{\partial x}(a(t_0),b(t_0)) \cdot \frac{da}{dt}(t_0)+\frac{\partial f}{\partial y}(a(t_0),b(t_0)) \cdot \frac{db}{dt}(t_0) .$$

It is impossible to eleminate $f$ from this formula. The only "explanation" is that your book abuses notation and writes

$$z(t) = z(a(t),b(t))$$

which is not correct.

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  • $\begingroup$ Some programming languages support the concept of "polymorphic functions". See for example en.wikipedia.org/wiki/Ad_hoc_polymorphism. Mathematically it is an abuse of notation to denote different things by the same symbol, but perhaps it is what your book intends. $\endgroup$ Commented Oct 4, 2018 at 15:03
  • $\begingroup$ This answer was useful to me, reminding me that we need unambiguous notation in mathematics as also seen here in this question. Compositions should always be given a different function name as you point out. Also the wikipedia page on the chain rule is horrendous in this regard. If I say $y = f(x)$, technically, I should not write down $dy/dx$ as y is a coordinate. Because on my $xy$ coordinate plane I could have $y = f(x)$ and $y = g(x)$ making $\endgroup$ Commented Oct 4, 2018 at 15:14
  • $\begingroup$ $dy/dx$ ambiguous $\endgroup$ Commented Oct 4, 2018 at 15:14
  • $\begingroup$ Thanks so much for the answer! $\endgroup$ Commented Oct 4, 2018 at 17:34

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