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Question:

\begin{equation} \dfrac{dB}{dC} = ? \end{equation}

Context

The present problem might be considered a practical exercise with respect to the ongoing discussion in [1]. In addition, [2] offers a related question in that both the question here and there are derivatives with respect to a function.

Given:

\begin{align} A & = f{\left(A_1, \ldots, A_M\right)} = \sum_{k = 1}^{M}{A_k} = \textrm{constant }\\ B & = g{\left(A_1, \ldots, A_M\right)} = \sum_{k = 1}^{M}{B_k(A_k)} \\ C & = h{\left(A_1, \ldots, A_M\right)} = \sum_{k = 1}^{M}{C_k(A_k)} \end{align}

Approach I: Ratio of Two Differentials:

The differentials of $A$, $B$, and $C$, are respectively written as \begin{align} dA & = \sum_{k = 1}^{M}{dA_k} = 0 \quad\quad \textrm{Eq. 1} \\ dB & = \sum_{k = 1}^{M}{\dfrac{\partial{B_k}}{\partial{A_k}}} \, dA_k \quad\quad \textrm{Eq. 2} \\ dC & = \sum_{k = 1}^{M}{\dfrac{\partial{C_k}}{\partial{A_k}}} \, dA_k \quad\quad \textrm{Eq. 3} \end{align}

Next, arbitrarily singling out the $M$th differential, $dA_M$, we rewrite Eq. 1 as \begin{align} dA_M & = -\sum_{k = 1}^{M-1}{dA_k} \end{align} and continue to rewrite Eqs. 2 and 3 as \begin{align} dB & = \sum_{k = 1}^{M-1}\left[\dfrac{\partial{B_k}}{\partial{A_k}} - \dfrac{\partial{B_M}}{\partial{A_M}} \right] \, dA_k \\ dC & = \sum_{k = 1}^{M-1}\left[\dfrac{\partial{C_k}}{\partial{A_k}} - \dfrac{\partial{C_M}}{\partial{A_M}} \right] \, dA_k \end{align}

Naively or not, I write the derivative as the ratio of the differentials. My solution is thus \begin{equation} \dfrac{dB}{dC} = \dfrac{\sum_{k = 1}^{M-1}\left[\dfrac{\partial{B_k}}{\partial{A_k}} - \dfrac{\partial{B_M}}{\partial{A_M}} \right] \, dA_k}{\sum_{k = 1}^{M-1}\left[\dfrac{\partial{C_k}}{\partial{A_k}} - \dfrac{\partial{C_M}}{\partial{A_M}} \right] \, dA_k}. \end{equation}

Approach II: Chain Rule:

From [3], let $A=A(A_1, A_2, \ldots, A_M) = \textrm{constant}$, $B = B(A_1, A_2, \ldots, A_M)$, and $C = C(A_1, A_2, \ldots, A_M)$, then the partial derivative of $C$ with respect to $B$, holding $A$ constant, written $\left(\dfrac{\partial{C}}{\partial{B}}\right)_A$, can be expressed as

\begin{align} \left(\dfrac{\partial{B}}{\partial{C}}\right)_A = \sum\limits_{k=1}^{M} { \left(\dfrac{\partial{B}}{\partial{A_k}}\right)_{\substack{A_{1,\ldots, j, \ldots, M}\\ k \neq j }} \, \left(\dfrac{\partial{A_k}}{\partial{C}}\right)_A } \end{align}

The subscripts on the parenthesis above indicate which variables are considered constants with respect to the given partial derivative. In particular, $\substack{A_{1,\ldots, M}\\ k \neq j }$ indicates that all of the $A_j; j\in 1,\ldots, j, \ldots, M$ are considered constants, save the partial derivative in question (i.e, the $k^\textrm{th}$ one).

Discussion

Question: Why did I start by asking for a a derivative and end up with a partial derivative?

Answer: The initial question appears to be ill-posed. As opposed to finding the derivative, what I really saught all along was $\left(\dfrac{\partial\left\{B(A_1, \ldots, A_M)\right\}}{\partial\left\{C(A_1, \ldots, A_M)\right\}}\right)_A$. I am purposefully leaving the title as a misnomer because like myself, there might be others who would not think to search for a constrained partial derivative.

Conclusion

Beyond the solution to my question, this question offers a demonstration that the ratio of two differentials is not a derivative.

Referencess

[1] Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

[2] Differentiating with respect to a function

[3] CRC Math Encyclopedia, Edition XX, 336.

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Solution

I write the solution as a constrained partial derivative as \begin{align} \left(\dfrac{\partial{B}}{\partial{C}}\right)_A = \sum\limits_{k=1}^{M} { \left(\dfrac{\partial{B}}{\partial{A_k}}\right)_{\substack{A_{1,\ldots, M}\\ k \neq j }} \, \left(\dfrac{\partial{A_k}}{\partial{C}}\right)_A } \end{align}

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