Approximation of a partial derivative

Question

I am looking for a useful approximation for a partial derivative of the form

partial derivative of lnP with respect to lnA at constant S.

Suppose that P=1000 and A=200.

Suppose that I choose deltaP = 5 and deltaA = 0.5 and solve for P=1005 and A=200.5 at constant S.

Would the following then be a good approximation:

ln(P/(P+deltaP)) / ln(A/(A+deltaA)) and in which, in this example, would be

ln(1000/1005) / ln(200/200.5)

?

Answer 1

Hi Will,

Thanks for your question!

Here's how I'd think about the problem. Presumably, P = P(A,S), but since S is constant, we have P = P(A). Now I'm going to write the functional relationship P = P(A) using an intermediate variable that I'll call u:

P = P(u(A)), (Equation 1)

where u = ln(A). As a result, we can think of expressions like dP/du as being equivalent to dP/d(ln(A)). This keeps our notation nice and clean.

Let's suppose now that I vary A by a small amount δA (so that |δA| << |A|). Then, P necessarily changes by a small amount δP (with |δP| << |P|) so that Equation 1 becomes

P + δP = P(u(A + δA)). (Equation 2)

I'm now going to Taylor series expand the function u about A, giving us

u(A + δA) = u(A) + u'(A)δA + ..., (Equation 3)

where the prime on u denotes a derivative with respect to A. Substituting the expansion in Equation 3 into Equation 2, we get

P + δP = P(u(A) + u'(A)δA + ...). (Equation 4)

Now I'm going to Taylor series expand the function P about u(A), giving us

P(u(A) + u'(A)δA + ...) = P(u(A)) + P'(u(A))(u'(A)δA + ...) + ..., (Equation 5)

where the prime on P denotes a derivative with respect to u. Substituting the expansion in Equation 5 into Equation 4, we get

P + δP = P(u(A)) + P'(u(A))(u'(A)δA + ...) + ... . (Equation 6)

The P and P(u(A)) cancel from both sides of Equation 6 by Equation 1. If I drop terms of order (δA)² and smaller on the RHS of Equation 6, I find the approximation

δP ≈ P'(u(A))(u'(A)δA), (Equation 7)

which can be written as

P'(u(A)) ≈ δP/(u'(A)δA). (Equation 8)

We're on the home stretch now! Suppose I divide Equation 8 on both sides by P(u(A)). Then, using the fact that d(ln(P))/du = 1/P(u)dP/du, Equation 8 becomes

(ln(P(u(A))))' ≈ δP/(P(u(A))u'(A)δA). (Equation 9)

Now we use the fact that u = ln(A) to see that the LHS of Equation 9 is nothing but d(lnP)/d(lnA) when the functional dependencies are dropped and the RHS cleans up to (AδP)/(PδA) since u'(A) = (ln(A))' = 1/A. We then obtain an approximation for the derivative d(lnP)/d(ln A), which is

d(lnP)/d(lnA) ≈ (AδP)/(PδA). (Equation 10)

Note that we could have arrived at a similar answer more formally by noting that d(lnP) = dP/P and d(ln(A)) = dA/A, but it's nice to see how this formal argument is justified more precisely.