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I am attempting to model the following equation with 3 variables (H, M, and S) and 1 constant (C):

D = .05*H*M + .05*S*(M+C)

or, when factored

D = .05*M (H+S) + .05*C(S)

It'd be a simple matter to calculate the value of D at any point in the formula (just plug in the variables), but I'm more interested in the relationship between S, M, and H; I'd like to be able to determine how much D will change (dD) when I increase S, M, or H (or some combination thereof).

It's been a while since I've used partial derivatives, but I believe that's what I'd need to use to answer this?

I think I've got a handle for single values (i.e., partial derivative with respect to M only), but how do I calculate the expected change of both (for instance) increasing H by 1 and decreasing S by 1?

Any help?

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  • $\begingroup$ First order derivatives measure sensitivity to one parameter while holding others constant. Perhaps you would need to consider the second order mixed partial derivative $\dfrac{\partial^2 D}{\partial H \partial S}$. $\endgroup$
    – gt6989b
    Jul 3 '13 at 15:41
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If $D(H,M,S) = .05 M (H+S) + .05 C(S)$, then $$\frac{\partial D(H,M,S)}{\partial H} = 0.05 M, \frac{\partial D(H,M,S)}{\partial M} = 0.05(S+H), \frac{\partial D(H,M,S)}{\partial S} = 0.05(M+C)$$

If you are trying to estimate the change due to 'small' changes in the parameters, you can use:

$$D(H+h,M+m,S+s) \approx D(H,M,S) + \frac{\partial D(H,M,S)}{\partial H}h + \frac{\partial D(H,M,S)}{\partial M}m+\frac{\partial D(H,M,S)}{\partial S}s$$

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  • $\begingroup$ This doesn't answer the question - how do you account for both $H$ and $S$ changing? $\endgroup$
    – gt6989b
    Jul 3 '13 at 15:42
  • $\begingroup$ @gt6989b: The question changed after I started answering it. I have modified my answer. $\endgroup$
    – copper.hat
    Jul 3 '13 at 15:45
  • $\begingroup$ i would certainly include at least the second order effects... $\endgroup$
    – gt6989b
    Jul 3 '13 at 18:38
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    $\begingroup$ @gt6989b: That would work, but since the formula is already a quadratic, you will just get the original formula. This is why I presumed the OP wanted a simpler model... $\endgroup$
    – copper.hat
    Jul 3 '13 at 18:41

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