Is it possible to manipulate partial derivatives in this manner?

[SiennaTheGr8]

If I have an expression like ##\dfrac{\partial x}{\partial y}##, and I know that ##x = a + bc## and ##y = f + gh##, then I have:

##\dfrac{\partial (a + bc)}{\partial (f + gh)}##.

I'm wondering what kind of maneuvers are "legal" here. Say that ##b## and ##g## are constants, and the other symbols are variables. Could I, for example, do something like this?

##\dfrac{\partial a + b \, \partial c}{\partial f + g \, \partial h}##

And then even, say, split up the "fraction" and do something like this to each part?:

##\dfrac{\partial a}{\partial f + g \, \partial h} = \dfrac{\partial a \, \partial f - g \, \partial a \, \partial h}{(\partial f)^2 - (g \, \partial h)^2} = \dfrac{\partial a \, \partial f}{\partial f \left( \partial f - \frac{(g \, \partial h)^2}{\partial f} \right) } - \dfrac{g \, \partial a \, \partial h}{\partial h \left( \frac{(\partial f)^2}{\partial h} - g^2 \, \partial h \right) } = \dfrac{\partial a}{\partial f} [+] \dfrac{\partial a}{g \, \partial h}##,

where in the last step I set the only remaining 2nd-order partials to zero.

I'm asking because I had occasion to try using such "methods," and much to my surprise I ended up with the answer I was looking for. I just don't know whether that was coincidence or if this is all actually kosher.

[edit: changed - to +, indicated by square brackets]

 

[gleem]

Your notation is a bit confusing are a, b, f, g, constant and c and h variable? In changing variables you should use the well know "chain rule".

 

[SiennaTheGr8]

No, ##b## and ##g## are the only constants.

 

[gleem]

If a,c,f, and h are independent variables and y does not depend and a and c and x does not depend on f and h then x is not a function of y.
BTW the last expression should have a plus sign.

Is this question a part of a larger problem?

 

[SiennaTheGr8]

BTW the last expression should have a plus sign.

Yes, thank you—fixed.

Is this question a part of a larger problem?

I'm trying to use the transformation of the four-potential ##(\phi, A_x, A_y, A_z)## under a Lorentz boost (along the ##x##-axis) to derive the transformation formulas for the electric and magnetic field components. As an example, I end up with:

##E_y^\prime = - \left( \dfrac{\partial \phi^\prime}{\partial y^\prime} + \dfrac{\partial A_y^\prime}{\partial (ct)^\prime} \right) = - \left( \dfrac{\partial \left( \gamma \phi - \gamma \beta A_x \right)}{\partial y} + \dfrac{\partial A_y}{\partial \left( \gamma ct - \gamma \beta x \right)} \right)##,

which I know is supposed to equal:

##- \gamma \left( \dfrac{\partial \phi}{\partial y} + \dfrac{\partial A_y}{\partial (ct)} \right) - \gamma \beta \left( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \right)##

(##\beta## and ##\gamma=(1 - \beta^2)^{-1/2}## are constants).

I didn't know what to do with those ##\partial (\textrm{foo + bar})## objects, but I forged ahead with the kind of "algebra" shown above in my OP (as if they were ordinary differentials), and I seem to have succeeded. I just don't have experience treating "partial differentials" this way, so I don't know if what I did was valid.

 

[gleem]

If a,c,f, and h are independent variables and y does not depend and a and c and x does not depend on f and h then x is not a function of y.

So, in fact there are only two new independent variables x and t not four. Have you tried the chain rule yet?

 

[SiennaTheGr8]

Yes, thank you, I think I understand now:

##\dfrac{\partial A_y}{\partial ct^\prime (x, ct)} = \dfrac{\partial A_y}{\partial ct} \dfrac{\partial}{\partial ct^\prime} (\gamma ct^\prime + \gamma \beta x^\prime) + \dfrac{\partial A_y}{\partial x} \dfrac{\partial}{\partial ct^\prime} (\gamma x^\prime + \gamma \beta ct^\prime) = \gamma \dfrac{\partial A_y}{\partial ct} + \gamma \beta \dfrac{\partial A_y}{\partial x}##

Been a while since I've worked with partials...