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I have a doubt regarding the multivariate chain rule PDE.

Consider an arbitrary function $\phi(x+y+z,x^2+y^2-z^2)=0$. We have to eliminate the function & form a PDE.

The solution as follows:

Let $u=x+y+z$ & $v=x^2+y^2-z^2$. Then the original equation becomes $\phi(u,v)=0$

Differentiating wrt 'x' partially gives $\frac{\partial \phi}{\partial u}\left(\frac{\partial u}{\partial x}+p\frac{\partial u}{\partial z} \right) + \frac{\partial \phi}{\partial v}\left(\frac{\partial v}{\partial x}+p\frac{\partial v}{\partial z} \right)=0$

Here in $\frac{\partial u}{\partial x}$ & $\frac{\partial v}{\partial x}$, should we consider the z term in u & v as constant or should we derive it partially again ?

Will $\frac{\partial u}{\partial x}$ be 1 or 1+p and $\frac{\partial v}{\partial x}$ be 2x or 2x-2pz ?

If z is taken to be constant why ?

Note : $p=\frac{\partial z}{\partial x}$

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    $\begingroup$ What is $p$ here? $\endgroup$ Commented Nov 11 at 8:10

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Partial derivatives are always derivatives for places. For a function $u(x,...)$ depending on $x$ and other variables, $\frac{\partial u}{\partial x}$ is always the derivative for (changes in) the first place. Thus here $\frac{\partial u}{\partial x}=1$, $\frac{\partial v}{\partial x}=2x$.

Reconstructing the task so that most of your remarks are sensible, it would have to be that $y=y_0$ is considered as constant and the equation $\psi(x,y_0,z)=\phi(u(x,y_0,z),v(x,y_0,z))=0$ is to be solved for $z=f(x)$ as function of $x$. Using implicit differentiation one can obtain an ordinary differential equation for this implicit function, which is a classical method for approaching the implicit function theorem.

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  • $\begingroup$ So, if i were to follow the differential tree diagram, then I only need to find the derivative independent to other variables. In this example u is a function of x,y,z and z is a function of x & y. By following the differential tree, for partial derivative of u wrt x all other variables including z are considered as constant.. ? @lutz-lehmann $\endgroup$ Commented Nov 11 at 11:34
  • $\begingroup$ Yes, that is generally true for partial derivatives, all other positions in the argument/variable list are considered constant. It gets a little less clear in composite functions, as often the argument variable list of the composite expression is not explicit, but "can easily be seen". But here the "inner" functions $u,v$ have the same argument list. To be thorough, one would have to name the composite function to write down the complete equation of the chain rule. But these parts are correctly used in OP, so that might be redundant. $\endgroup$ Commented Nov 11 at 11:50

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