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This is is pretty straightforward. I have a function, say $X = X(q)$. And $q=q_1+q_2(q_1)$. So X is a function of q, and q is a function of $q_1,q_2$. But $q_2$ is also a function of $q_1$. Now calculate

$\frac{\partial X}{\partial q_1}$

I proceed like:

$\frac{\partial X}{\partial q_1} = \frac{\partial X}{\partial q} \cdot \frac{\partial q}{\partial q_1}= \frac{\partial X}{\partial q} \cdot (1+\frac{\partial q_2}{\partial q_1}) $

Is this correct? Any insight or any more detailed expression that i am missing will be useful.

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  • $\begingroup$ Looks correct to me ... $\endgroup$ – Matti P. Mar 13 at 11:01
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$X(q)$, $X(q_1)$ and $X(q_1,q_2)$ are three different functions which happen to have the same name. Let me expand on that (I am going to assume that $q_1$, $q_2$ and the $X$s all take real values) ...

$X(q_1,q_2)$ is a function from $\mathbb{R}^2$ to $\mathbb{R}$. It is defined for any pair of values $(q_1, q_2)$ in $\mathbb{R}^2$ (or, possibly, in some region $U \subset \mathbb{R}^2$). As such, its has partial derivatives $\frac{\partial X}{\partial q_1}$ and $\frac{\partial X}{\partial q_2}$ at each point $P \in \mathbb{R}^2$ (provided $X$ is continuous and is defined in some open region around $P$).

If we now introduce $q$ then $X(q)$ is a function from $\mathbb{R}$ to $\mathbb{R}$. As such it has a derivative $\frac{dX}{dq}$ (subject to the usual assumptions about continuity etc.) at each point $p \in \mathbb{R}$. If we further say that $q=q_1+q_2$ then we have defined a function $q(q_1,q_2)=q_1+q_2$ from $\mathbb{R}^2$ to $\mathbb{R}$, and we can define an implicit function $Y(q_1,q_2) = (X \circ q)(q_1,q_2) = X(q_1+q_2)$. The connection between the derivative of $X$ and the partial derivatives of $Y$ is

$\frac{\partial Y}{\partial q_1} = \frac{dX}{dq}\frac{\partial q}{\partial q_1}\\ \frac{\partial Y}{\partial q_2} = \frac{dX}{dq}\frac{\partial q}{\partial q_2}\\$

If we then say that $q_2$ is a function $Q_2: \mathbb{R} \to \mathbb{R}$ of $q_1$ then we are restricting our attention to the curve $q_2=Q_2(q_1)$ in $\mathbb{R}^2$ which we are paramterising by the value $q_1$ (which now serves double duty both as a co-ordinate in $\mathbb{R}^2$ and as a parameter of the curve). We now have yet another implicit function $Z:\mathbb{R} \to \mathbb{R}$ defined by $Z(q_1) = Y(q_1, Q_2(q_1)) = X(q_1+Q_2(q_1))$ and

$\frac{dZ}{dq_1} = \frac{\partial Y}{\partial q_1} + \frac{\partial Y}{\partial Q_2}\frac{dQ_2}{dq_1} = \frac{dX}{dq}\left(\frac{\partial q}{\partial q_1}+ \frac{\partial q}{\partial Q_2}\frac{dQ_2}{dq_1} \right)$

which tells us how $X(q)$ varies along the curve $q_2=Q_2(q_1)$.

By convention we often rename $Y(q_1,q_2)$ as $X(q_1,q_2)$ and $Z(q_1)$ as $X(q_1)$ and we depend on context to tell us which $X$ is meant on each occassion that we use the name, but this can be confusing (as you have found).

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This is a problem you encounter early and often in calculus of variations and Lagrangian mechanics. Here is how it is resolved there:

  1. You can consider $X$ as a function of the single variable $q$. In that case, you can ask about $\frac{dX}{dq}=X'(q)$
  2. You can consider $X$ as a function of the two (independent) variables $q_1,q_2$, and in that case, you can ask about $\frac{\partial X}{\partial q_1}$ and $\frac{\partial X}{\partial q_2}$, both of which happen to be equal to $X'(q_1+q_2)$.
  3. You can consider $X$ as a function of the single variable $q_1$, and in that case you can ask about $$\frac{dX}{dq_1}=\frac{\partial X}{\partial q_1}\cdot\frac{dq_1}{dq_1}+\frac{\partial X}{\partial q_2}\cdot \frac{dq_2}{dq_1}\\=X'(q_1+q_2(q_1))\cdot \left(1+q_2'(q_1)\right)$$

It is easy to mix cases 2 and 3, but that makes no sense. It's one or the other.

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Partial differentiation depends on the explicit form of the expression. Consider applying chain rule to compute a full derivative of an example function

$\frac{df(x, y)}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} $

If we were forced to explicitly rewrite $x$ and $y$ in terms of $t$, there would be no difference between a full and a partial derivative.

I would argue that, in your example, the calculation might not be correct, in case $q_2$ is an implicit function of $q_1$. Please have a look at the example from fluid dynamics to get the feeling for the difference between these two types of derivatives

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There are only functions of a single variable here, so partial derivatives are irrelevant. By the ordinary chain rule,

$$\frac{dX}{dq_1}=\frac{dX}{dq}\frac{dq}{dq_1}=X'\left(1+\frac{dq_2}{dq_1}\right)=X'(1+q_2').$$

The $'$ denotes differentiation with respect to the argument.


Technically, you can consider that $q=q_1+q_2$ is a function of two variables and write

$$dq=\frac{\partial q}{\partial q_1}dq_1+\frac{\partial q}{\partial q_2}dq_2=dq_1+dq_2,$$ which justifies

$$\frac{dq}{dq_1}=1+\frac{dq_2}{dq_1},$$

but this is the only place where partials can be used, and in a contrived way.

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It seems correct to me but you can remove the partial sign of $\frac{\partial q_2}{\partial q_1}$ becasuse $q_2$ depends on the single variable $q_1$,

$$\frac{\partial X}{\partial q_1}=\frac{\partial X}{\partial q}(1+\frac{d q_2}{d q_1})$$

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  • $\begingroup$ I believe that, given the broader answers given to the question, this answer may be misleading $\endgroup$ – Aleksejs Fomins Mar 13 at 13:01

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