1
$\begingroup$

So I've been studying the chain rule for partial derivatives recently and I'm having an extremely difficult time wrapping my head around it as I'm having an incredibly hard time understanding the formulation of the chain rule for partial derivatives in my textbook. In an attempt to help me understannd how it works I've been going over some of the exercises in the back of the book to see if I could at least apply it. I came to an exercise that looks rather simple, but I'm not sure how to solve it. The exercise is as follows:

Consider a differentiable function function $f:\mathbb{R^2}\rightarrow\mathbb{R}$, now find the partial derivative of the function: $F:(x,y)\rightarrow f(2x,3y)$ (State the result in terms of the partial derivatives of $f$)

Now in terms of what I want to find I'm having some doubts around one of the deriavtives. What I find is:

$\frac{\partial F(x,y)}{\partial x}=\frac{\partial f}{\partial y_1}(2x,3y)\frac{\partial(2x)}{\partial x}+\frac{\partial f}{\partial y_2}(2x,3y)\frac{\partial(3y)}{\partial y}$

My issue here is that I can't really figure out what $y_1$ and $y_2$ are supposed to be, and the formulation in my textbook is really confusing, I'd really appreciate if anyone could clarify this for me.

$\endgroup$
2
  • $\begingroup$ These are just the first and second arguments of the function $f(y_1,y_2)$ respectively. $\endgroup$ Commented Jun 16, 2020 at 10:39
  • $\begingroup$ So theywould be 2x and 3y respectively? Or would I require more specific information about f to know these? $\endgroup$ Commented Jun 16, 2020 at 10:43

2 Answers 2

1
$\begingroup$

Maybe some other choice of letters can help. Imagine that $f: \mathbb{R}^2\to \mathbb{R}$ is defined by $(u,v) \mapsto f(u,v)$ and $F(x,y)=f(2x, 3y)$. The chain rule says that $$ \frac{\partial F}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} $$

$$ \frac{\partial F}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}, $$

where $u = 3x, v=3y$.

$\endgroup$
1
$\begingroup$

I think this is the gut-wrenching point most people discover in their math journey. Good, old dependable math; so precise, so clean ... then BAM! partial derivatives. There is no good notation for partial derivatives. Each of the widely different notations has one flaw or another. Once we recover from that blow, we can soldier on.

Here, the $y_1$ (and $y_2$) mean that you're taking the derivative with respect to the first (and second) variable. We can't use $\partial x$ on the bottom, because $x$ is being used for the $2x$ as the first argument. You don't need to write it, but really $y_1 = 2x$ and $y_2 = 3y.$

Your displayed line has an extra "+" in it, by the way.

$\endgroup$

You must log in to answer this question.

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