$How to derive the term acting like a first derivative with respect to A that I found by accident?$

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$How to derive the term acting like a first derivative with respect to A that I found by accident?$

Dear community,

I would like to analyze how the following function changes, when variables A,B, or C are not zero:

$f(A,B,C) = \frac{Q(1+X)(PA-VB)-FC}{GX}-\frac{A(1+X)(QP(X+A+AX)-QV(X+B+BX)-FC)}{GX(X+A+AX)}$

with

$A,B,C,Q,P,V,F,G∈R$

All other variables besides $A,B,C$ are relevant for the calculation but simply not of interest for my analysis.

Note that $A,B,C$ being zero in the initial state actually leads to $f(A=0,B=0,C=0)=0$, so it holds that

$∆f(A,B,C) =f(A,B,C)-f(A=0,B=0,C=0)= f(A,B,C)-0 = f(A,B,C)$

and the change in $f(A,B,C)$ is generally given by $f(A,B,C)$ itself.

My problem:

Now I noticed by accident in Excel that I can also calculate the value of $∆f(A,B,C)$ equaling $f(A,B,C)$ for any parameter set by

$∆f(A,B,C)=f(A,B,C)=\frac{QV(1+X)}{G(X+A+XA)}*A-\frac{QV(1+X)}{G(X+A+XA)}*B -\frac{F}{G(X+A+XA)}*C$

You can try it yourself by putting in any numbers, the initial equation and this one always have the same result.

Note that this structure is very helpful, because the second and third term are the constant first derivatives with respect to $B$ and $C$, so I can also derive them separately and discuss their effects in my analysis:

$∆f(A,B,C)=f(A,B,C)=\frac{QV(1+X)}{G(X+A+XA)}*A+\frac{df}{dB}*B +\frac{df}{dC}*C$

Questions:

1) The first term in the last function is not $\frac{df}{dA}*A$, because $\frac{df}{dA} = \frac{(1+X)(QV(X+B+XB)+FC)}{G(X+A+XA)^2}$ acc. to an online calculator and is non-constant. Why is this term effectively still working as a "sort of derivative with respect to A", while not actually being the first derivative? Is this a special form of derivative, maybe considering that A is initially zero, that I can derive somehow? How?

2) If deriving the first term as a sort of derivative is not possible: How can I at least transform the initial function step-by-step to reach the shown last form?

Appendix:
Formula in linear form: (Q*(1+X)*(P*A-V*B)-F*C)/(G*X)-(A*(1+X)*(Q*P*(X+A+A*X)-Q*V*(X+B+B*X)-F*C))/ (G*X*(X+A+A*X))

Differential Equations Partial Differential Equations Functional Analysis

Mosi Posi

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The reason is the special form of the function $f$ (which becomes apparent after simplification). The difference between the coefficient of $A$ and $df/dA$ is hidden in the remainder term $R$. The simplification and derivation are in the attached file.

Also, it must be noted that in the approximation of $f$ by its derivative, all the terms $df/dA$, $df/dB$, $df/dC$ must be evaluated at $(0,0,0)$, while the formula for $f$ contains the derivatives $df/dB$ and $df/dC$ at an arbitrary point with variable value for $A$. So that also affects the formulas.

Let us absorb $B,C$ into the coefficient $q$ and write $$ f=\frac{p}{rA+t}A+\frac{q}{rA+t}=\frac{pA+q}{rA+t}$$ Then $$ \frac{df}{dA}=\frac{pt-rq}{(rA+t)^2}$$ At $B=C=0$ we have $q=0$. So $$ f(A)=f(0)+\frac{df(0)}{dA}A+R=0+\frac{pt-rq}{t^2}A+R=\frac{p}{t^2}A+R$$ while $$ f(A)=\frac{p}{rA+t}A=\frac{p}{t}\,\frac{1}{1+rA/t}A=\frac{p}{t}(1-\frac{rA}{t}+\frac{r^2A^2}{t^2}-\dots)A \\ \; \\ = \frac{p}{t}A-\frac{prA^2}{t^2}+\frac{pr^2A^3}{t^3}-\dots$$ So the rest of the series is actually the remainder $R$.

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Martin

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Martin

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Let me know if you have any questions.
- Mosi Posi
  
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  Thanks! 1) Pls answer my 1st question: Is this “coefficient of A” a derivative as well and can I show this in correct language from the final function? Is it the following because it adds constraint B=0,C=0 to the normal derivative? How to show? $\frac{df}{dA} |_{(B=0,C=0)}= \frac{QV(1+X)}{G(X+A+XA)}$ 2) Besides that (1+A) in num is wrong and (1+X) in denom missing in 2nd term, where do the Xs in the nums go after $\frac{-QVX(B-A)}{(X+A+XA)}-\frac{FCX(1+A)}{(X+A+XA)}$
- Martin
  
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  The "X" got canceled on both sides (with the one before "f"). You are right about the missing and extra terms, as I added/removed them later and forgot about the ones in the middle.
Martin

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The coefficient of "A" is not a derivative in any sense, even at "B=C=0" the equation you wrote does not hold. I will add an explanation to make this clear.
- Mosi Posi
  
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  Thanks! I don't understand what variables r and t are in your answer. I do understand that the issue arises from A also being included in the coefficients/derivatives of B and C, so when those are not zero, A's coefficient alone doesnt capture its derivative. However, if B and C are zero, and despite your derivation, the first equation under "My Problem" becomes "QV(1+X)/(G*(X+A+XA)) * P" and nothing more, right? Do you agree that the coefficient at least "acts" like a derivative then?
Martin

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r and t are G(X+1) and GX, but their actual value is not important for the calculations to hold. The problem is not merely the dependence on A of the coefficients of B,C. It's also because the coefficient of A depends on A too.
Martin

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The first equation becomes what you wrote when B,C are zero, that is correct. The coefficient seems like a derivative because it is approximately equal to the derivative, so you might say it "acts" like one. In fact at A=0 the derivative and the coefficient have the same value. And for small A they have values close to each other.
Mosi Posi

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When a model assumes A,B,C=0, and all I want is showing what happens if this is not the case, am I not SOLELY analysing the function at A=0? So derivative and coefficient will always have exactly the same value acc. to your words... If I only want to analyse at A=0, shouldn't there be a way to show in precise mathematical terms (eg A=0 as condition?) how the derivative then becomes this coefficient? Plugging in A=0 in the derivative didnt work. Thanks so far, this is my last understanding gap.
- Martin
  
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  Not exactly. When B,C=0 then f=a(A)*A and thus f'=a + a' *A. So at A=0 we have f'(0)=a(0). But for nonzero "A" the two are not exactly equal and are just approximately close when A is small: f'-a = a' *A

$How to derive the term acting like a first derivative with respect to A that I found by accident?$

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