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

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The discussion centers on the application of the chain rule to calculate the derivative of one function with respect to another, specifically when both functions depend on multiple variables. Participants express confusion over whether it is valid to compute df/dg, with some arguing that it is nonsensical since g is a function, not a variable. Others clarify that while derivatives are typically taken with respect to variables, it is possible to derive one function concerning another under certain conditions, using partial derivatives and the chain rule. The conversation highlights the complexity of multivariable calculus and the need for proper assumptions about the functions involved. Ultimately, the consensus suggests that while challenging, it is feasible to consider derivatives of functions relative to one another when appropriately framed.
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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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