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I read an answer of Partial derivative of a composite function on math.stackexchange.com. The writer claimed that to solve a partial derivative of a multiple-variable function

$$ f(x,y) \ = \ \varphi (\ \underbrace{\frac yx}_u \ , \ \underbrace{ x^2-y^2}_v \ , \ \underbrace{y-x}_w \ ) \ , $$

we can use the multivariate extension of the Chain Rule

$$ \frac{\partial f}{\partial x} \ = \ \frac{\partial \varphi}{\partial u}\frac{\partial u}{\partial x} \ + \ \frac{\partial \varphi}{\partial v}\frac{\partial v}{\partial x} \ + \ \frac{\partial \varphi}{\partial w}\frac{\partial w}{\partial x} $$

I know the derivatives and partial derivatives are defined by

$$ \frac{df}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} $$

and

$$ \frac{\partial f(x, y)}{\partial x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} $$

I also what to write the formula of the (partial) chain rule to the definition form. I got four versions below. Which is the correct expression without derivative operators for the equation got by the chain rule?

$$ \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} = \frac{\varphi (u(x + \Delta x, y), v, w) - \varphi (u(x, y), v, w)}{u(x + \Delta x, y) - u(x, y)}\frac{u(x + \Delta x, y) - u(x, y)}{\Delta x} + \cdots \tag{1} $$ $$ \lim_{\Delta x, \Delta y \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} = \frac{\varphi (u(x + \Delta x, y), v, w) - \varphi (u(x, y), v, w)}{u(x + \Delta x, y + \Delta y) - u(x, y)}\frac{u(x + \Delta x, y) - u(x, y)}{\Delta x} + \cdots \tag{2} $$ $$ \lim_{\Delta x, \Delta y \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} = \frac{\varphi (u(x + \Delta x, y + \Delta y), v, w) - \varphi (u(x, y), v, w)}{u(x + \Delta x, y) - u(x, y)}\frac{u(x + \Delta x, y) - u(x, y)}{\Delta x} + \cdots \tag{3} $$ $$ \lim_{\Delta x, \Delta y \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} = \frac{\varphi (u(x + \Delta x, y + \Delta y), v, w) - \varphi (u(x, y), v, w)}{u(x + \Delta x, y + \Delta y) - u(x, y)}\frac{u(x + \Delta x, y) - u(x, y)}{\Delta x} + \cdots \tag{4} $$

Updated

Here is the full expansion of (1)

$$ \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} = \frac{\varphi (u(x + \Delta x, y), v, w) - \varphi (u(x, y), v, w)}{u(x + \Delta x, y) - u(x, y)} \frac{u(x + \Delta x, y) - u(x, y)}{\Delta x} \\ + \frac{\varphi (u, v(x + \Delta x, y), w) - \varphi (u, v(x, y), w)}{v(x + \Delta x, y) - v(x, y)} \frac{v(x + \Delta x, y) - v(x, y)}{\Delta x} \\ + \frac{\varphi (u, v, w(x + \Delta x, y)) - \varphi (u, v, w(x, y))}{w(x + \Delta x, y) - w(x, y)} \frac{w(x + \Delta x, y) - w(x, y)}{\Delta x} $$

gyro
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  • Suppose $f$ only depends on $x$, and likewise $u,v,w$. Are you OK with the chain rule in the good old form $$ \frac{d f}{d x} \ = \ \frac{\partial \varphi}{\partial u}\frac{d u}{d x} \ + \ \frac{\partial \varphi}{\partial v}\frac{d v}{dx} \ + \ \frac{\partial \varphi}{\partial w}\frac{d w}{d x},? $$ – Kurt G. Dec 10 '22 at 12:12
  • I’m voting to close this question because OP does not respond. – Kurt G. Dec 11 '22 at 07:22
  • Wait! We are living in different time zones! And I don't have much time to stay online every hour. If my question is not clear enough, please point out that instead of closing the question. Once the question is closed I will not have chance to let others get the knowledge. For the total derivative form, yes. The operators are different. However it will be great if anyone can also show the explicit form without derivative operators for the total derivative form – gyro Dec 12 '22 at 05:07
  • The correct expressions for the chain rule are the one in my previous comment, and if $f$ also depends on the constant $y$, your expression $$\frac{\partial f}{\partial x} \ = \ \frac{\partial \varphi}{\partial u}\frac{\partial u}{\partial x} \ + \ \frac{\partial \varphi}{\partial v}\frac{\partial v}{\partial x} \ + \ \frac{\partial \varphi}{\partial w}\frac{\partial w}{\partial x}$$ in OP. It is hard to tell what you are asking. What is a correct expression without derivative operators that has $+\dots$ in it ? – Kurt G. Dec 12 '22 at 08:35
  • I want to know how to convert the partial chain rule back to the definition of derivatives, like four equations I listed. If I want to convert $\partial$ to limit, which is correct? – gyro Dec 13 '22 at 10:17
  • I updated my question to be clearer. – gyro Dec 13 '22 at 10:31
  • Still not much clearer. What I can say is that (2),(3),(4) look wrong since -as you should know- $y$ is a constant and you don't take the limit $\Delta y\to 0$. Concentrate on (1) write it out ( no $\dots$ !) and we shall see. – Kurt G. Dec 13 '22 at 10:48
  • The terms skipped by ... are the same structure on v and w. – gyro Dec 15 '22 at 06:11

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