- #1

SiennaTheGr8

- 492

- 193

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]

##\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]

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