# Derivation of Flory's Model for Linear Condensation of Polymers

I'm having some trouble understanding a part of Flory's model for linear polycondensation reactions. In his 1936 paper, he derives an expression for the probability of the existence of a particular $$x$$-mer as

$$\Pi_x = xp^{x-1}(1 - p)^2\tag{2}$$

Later, he states the following:

In order to locate the $$x$$ value which gives a maximum in $$\Pi_x,$$ let

$$\partial\Pi_x/\partial x = (1 - p)^2(p^{x - 1} + xp^{x - 1}\ln p) = 0$$

Not being the most mathematically savvy person, can someone explain to me how he got to this expression?

### Reference

1. Flory, P. J. Molecular Size Distribution in Linear Condensation Polymers. J. Am. Chem. Soc. 1936, 58 (10), 1877–1885. DOI: 10.1021/ja01301a016.
• "Not being the most mathematically savvy person" - How much do you know about differentiation? Does mytutor.co.uk/answers/5998/A-Level/Maths/… Make any sense? Do you know the Product Rule for differentiation? Mar 4, 2023 at 10:07
• @Justus Have a look at my answer and if it answers your question, please click on the tick mark next to the answer. You also get +2 points on doing this :) Mar 5, 2023 at 11:00

Given $$\Pi_{x}$$= $$𝑥*𝑝^{𝑥−1}$$ *$$(1−𝑝)^2$$

$$\partial$$ refers to partial differentiation.

$$\frac{\partial \Pi_{x}}{\partial x}$$ means differentiating $$\Pi_{x}$$ with respect to x.

Therefore $$(1−𝑝)^2$$ is a constant when differentiating with respect to x. So during differentiation the constant stays as a constant multiplier.

Now see that there are two terms involving x, therefore we need to use the product rule of differentiation.

$$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$$ = $$(1−𝑝)^2$$ * ($$𝑝^{𝑥−1}* \frac{\partial x}{\partial x} + x*\frac{\partial p^{x-1}}{\partial x}$$)

$$\frac{\partial x}{\partial x}$$ = 1 and for calculating $$\frac{\partial p^{x-1}}{\partial x}$$ look at exponentiation rule of differentiation as suggested by @Ian Bush in the comments.

Therefore we finally have :

$$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$$ = $$(1−𝑝)^2$$ * ($$𝑝^{𝑥−1}* 1 + x*ln(p)*p^{x-1}$$)

And since he is trying to locate the 𝑥 value which gives a maximum in $$\Pi_{x}$$, he equates the equation to 0.

$$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$$ = $$(1−𝑝)^2$$ * ($$𝑝^{𝑥−1}* 1 + x*ln(p)*p^{x-1}$$) = 0