I have two set of variables, which are related: $\left\{ \alpha, \beta, \gamma \right\}$ and $\left\{ v_0, v_1, v_2 \right\} = \left\{ \alpha, \beta, \beta \gamma \right\}$

Now, I want to compute the partial derivative $\frac{\partial \beta}{\partial \left(\alpha \beta \gamma \right)}$.

My first approach: as $\alpha$ and $\gamma$ do not depend on $\beta$,

$\frac{\partial \beta}{\partial \left(\alpha \beta \gamma \right)} = \frac{1}{\alpha \gamma} \frac{\partial \beta}{\partial \beta} = \frac{1}{\alpha \gamma} = \frac{v_1}{v_0 v_2}$.

But also, one could thinks as follows:

$\frac{\partial \beta}{\partial \left(\alpha \beta \gamma \right)} = \frac{\partial v_1}{\partial \left( v_0 v_2 \right)} = 0$

Something should be meshed up related with the dependencies, but I cannot see it. Can anyone give me a hand?


In your second approach, you write

$$\frac{\partial \beta}{\partial \left(\alpha \beta \gamma \right)} = \frac{\partial v_1}{\partial \left( v_0 v_2 \right)} = 0 \tag{1}\label{eq1}$$

This only works if $v_1$ doesn't change as $v_0 v_2$ does, which is what I believe you are assuming. However, since $v_0 v_2 = \alpha \beta \gamma$ and $v_1 = \beta$, the $2$ expressions are not generally independent of each other due to $\beta$ being in both expressions (note the exception is if $\alpha \gamma$ has a dependency on $\beta$ only as a fraction with a factor of $\beta$ in the denominator, but you say "as $\alpha$ and $\gamma$ do not depend on $\beta$", then this is not true). Thus, if for example, you change $v_0 v_2$ by changing $\beta$, then $v_1$ will be changing as well, so you cannot state that $v_1$ doesn't change at all while $v_0 v_2$ is changing, so this means that

$$\frac{\partial v_1}{\partial \left( v_0 v_2 \right)} \neq 0 \tag{2}\label{eq2}$$

More generally, note that expressing a value in a different way doesn't change any result, or other behavior, which uses that value. For example, $6 = 3 \times 2$, but $\frac{30}{6}$ and $\frac{30}{3 \times 2}$ are both equal to $5$. The important thing is what particular numeric values are being used, regardless of how they're represented. In this situation, expressing $\beta$ as $v_1$ and $\alpha \beta \gamma$ as $v_2$ doesn't change their numeric values, or relationship, at any point, so their extent of (in)dependence is still the same, and any partial derivatives involving them are also the same.

I hope that I've interpreted what you are doing correctly and that this addresses the issue.

  • $\begingroup$ Great, it seems clear. So would the following approach be consistent? $\frac{\partial v_1}{\partial \left( v_0 v_2 \right)} = \frac{1}{\frac{\partial \left( v_0 v_2 \right)}{\partial v_1}} = \frac{1}{v_0 \frac{\partial v_2 }{\partial v_1} + v_2 \frac{\partial v_0 }{\partial v_1}} = \frac{1}{v_0 \frac{\partial v_2 }{\partial v_1} } = \frac{1}{\alpha \gamma}$, as $\frac{\partial v_2 }{\partial v_1} = \gamma$ $\endgroup$ – Kikolo Feb 23 '19 at 11:18
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    $\begingroup$ @Kikolo The approach in your comment appears to be consistent & correct as you used the rules of derivatives, such as the product rule, and what the variables represent, all correctly. Thus, as you see, you got the same answer as before which, as far as I can tell, is correct, based on the assumption that $\alpha$, $\beta$ and $\gamma$ are independent of each other. $\endgroup$ – John Omielan Feb 23 '19 at 17:16

I do not see the role of the second set of variables. If I intepret correctly, you are trying to compute $\frac{\partial \beta}{\partial(\alpha,\beta,\gamma)}$, that is the vector $(0,1,0)$.

  • $\begingroup$ No, there is no comma there; it is the product of the three variables: $\alpha \cdot \beta \cdot \gamma$. No vectors involved $\endgroup$ – Kikolo Feb 22 '19 at 16:11
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    $\begingroup$ What is your definition of derivative of a function with respect to a function? $\endgroup$ – Alessio Del Vigna Feb 22 '19 at 16:22
  • $\begingroup$ math.stackexchange.com/questions/291376/… I think about it as a change of variable: $\frac{\partial \beta}{\partial q}$, with $q = \alpha \beta \gamma$ $\endgroup$ – Kikolo Feb 22 '19 at 17:31
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    $\begingroup$ Following that approach I would say $1=\frac{\partial \beta}{\partial\beta} = \frac{\partial \beta}{\partial{\alpha\beta\gamma}}\frac{\partial \alpha\beta\gamma}{\partial\beta}=\frac{\partial \beta}{\partial{\alpha\beta\gamma}}\cdot \alpha\gamma$, so that $\frac{\partial \beta}{\partial{\alpha\beta\gamma}}=\frac{1}{\alpha\gamma}$. $\endgroup$ – Alessio Del Vigna Feb 22 '19 at 17:33
  • $\begingroup$ That is certainly right, but where is the mistake in the second approach? $\endgroup$ – Kikolo Feb 22 '19 at 18:06

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