Calculating df/dg with Chain Rule: Romeo's Guide

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Mark44 said:
This really doesn't make much sense. You don't calculate the derivative of a function with respect to some other function, but you do calculate the derivative of a function with respect to one of its variables. Here g is a function, not a variable, so df/dg is nonsensical.

For another thing, both functions here have multiple variables, so instead of df/dx, df/dy, and df/dz, you would be working with partial derivatives,
\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \text{and} \frac{\partial f}{\partial z}

Other notation for these partials is fx, fy, and fz.
I disagree strongly with this- you always take the derivative of a function with respect to another function! In basic Calculus , of course, that second function is the identity function, x. But asking for the derivative of f with respect to g is just asking how fast f changes relative to g. If f and g are functions of the single variable, x, then, by the chain rule
\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}

If f and g are functions of the two variables x and y,
\frac{df}{dg}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial g}{\partial x}}+ \frac{\frac{\partial f}{\partial y}}{\frac{\partial g}{\partial y}}
 

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