Partial Differentiation of an equation using implicit differentiation confusion

I wanted to ask a question about implicit differentiation in partial differentiation.

When I was at school, I remember partial differentiation as something like this:

When you have a function composed of $$x$$ and $$y$$'s and you run into a $$y$$ term, differentiate with respect to $$y$$ and multiply by $$\frac{dy}{dx}$$ i.e. $$\frac{d}{dx} = \frac{d}{dy} \times \frac{dy}{dx}$$

Now, I read a problem on the Physics section yesterday afternoon that is relevant to my Chemistry course and I couldn't work it out.

If I have the equation

\begin{aligned} \frac{F\left(N_{A}, N_{B}\right)}{k T}=& N_{A} \ln \left(\frac{N_{A}}{N}\right)+N_{B} \ln \left(\frac{N_{B}}{N}\right) \\ &+\left(\frac{z w_{A A}}{2 k T}\right) N_{A}+\left(\frac{z w_{B B}}{2 k T}\right) N_{B}+\chi_{A B} \frac{N_{A} N_{B}}{N} \end{aligned}

I can get a quantity called the Chemical potential $$\mu_{A}$$ by differentiating the above equation with respect to $$N_A$$ while keeping $$N_B$$ and $$T$$ constant.

$$\mu_{A}=\left(\frac{\partial F} {\partial N_{A}}\right)_{T, N_{B}}$$

(The above equation is called the Free Energy Equation)

$$N$$ is the total number of molecules in the system, $$N_A$$ and $$N_B$$ are the number of molecules of A and B respectively. $$N$$ is not constant. They are related quite simply as via the sum:

$$N = N_A + N_B$$

which makes sense.

This equation also shows us that a small change in $$N_A$$ will also yield a small change in $$N$$ hence why $$N$$ is not constant, as mentioned above.

So I wanted to get from this equation, $$\mu_{A}$$

$$\frac{\mu_{A}}{k T}=\left[\frac{\partial}{\partial N_{A}}\left(\frac{F}{k T}\right)\right]_{T, N_{B}}$$

which makes sense.

The result given from the book (page 7), however, confused me. They gave the result as

$$=\ln \left(\frac{N_{A}}{N}\right)+1-\frac{N_{A}}{N}-\frac{N_{B}}{N}+\frac{z w_{A A}}{2 k T}+\chi_{A B} \frac{\left(N_{A}+N_{B}\right) N_{B}-N_{A} N_{B}}{\left(N_{A}+N_{B}\right)^{2}}$$

and only the last two terms made sense.

The OP asked where the terms

$$-\frac{N_{A}}{N}-\frac{N_{B}}{N}$$

came from and the answer given was as follows:

You missed the $$N$$ in the logarithm. Since $$N_{B}$$ is kept constant while changing $$N_{A}$$, the total number of particles $$N=N_{A}+N_{B}$$ changes as well. The missing term is $$\dfrac{\partial N}{\partial N_{A}}\dfrac{\partial}{\partial N}\left[N_{A}\ln\left(\dfrac{N_{A}}{N}\right)+N_{B}\ln\left(\dfrac{N_{B}}{N}\right)\right]=-\dfrac{N_{A}}{N}-\dfrac{N_{B}}{N}$$

and in the comments the value of $$\dfrac{\partial N}{\partial N_{A}}$$ was clarified to be $$1$$, again using the equation $$N = N_A + N_B$$. I understood this step.

but I then thought, if that was the case, how did the terms $$\ln \left(\frac{N_{A}}{N}\right)+1$$ arise then?

My original thought, similar to the OP was to differentiate:

$$\dfrac{\partial}{\partial N_{A}}\left[N_{A}\ln\left(\dfrac{N_{A}}{N}\right)+N_{B}\ln\left(\dfrac{N_{B}}{N}\right)\right]$$

but as explained, this assumed $$N$$ was constant, which it is not.

How are the terms $$\ln \left(\frac{N_{A}}{N}\right)+1$$

yielded by this application of implicit differentiation, if $$\dfrac{\partial N}{\partial N_{A}} = 1$$ using $$\dfrac{\partial N}{\partial N_{A}}\dfrac{\partial}{\partial N}$$?

• You can easily achieve those terms by differentiating $N_A \ln (N_A/N)$ with respect to $N_A$ – user673903 May 25 at 0:15

We want to compute \begin{aligned} \frac{\partial }{\partial N_A}\left( N_{A} \ln \left(\frac{N_{A}}{N}\right)+N_{B} \ln \left(\frac{N_{B}}{N}\right) +\left(\frac{z w_{A A}}{2 k T}\right) N_{A}+\left(\frac{z w_{B B}}{2 k T}\right) N_{B}+\chi_{A B} \frac{N_{A} N_{B}}{N}\right) \end{aligned}

First note that $$\frac{\partial N }{\partial N_A}= \frac{\partial}{\partial N_A}(N_A+N_B)=1+0=1$$

Now, let's compute partial derivative term by term.

\begin{align}\frac{\partial}{\partial N_A}\left(N_A\ln \left( \frac{N_A}{N}\right) \right) &= \left( \frac{\partial N_A}{\partial N_A}\right) \ln \left( \frac{N_A}{N}\right) + N_A \frac{\partial }{\partial N_A}\left( \ln \left( \frac{N_A}{N}\right)\right)\\ &= \ln \left( \frac{N_A}{N}\right) + N_A \frac{\partial }{\partial N_A}\left( \ln N_A - \ln N \right) \\ &=\ln \left( \frac{N_A}{N}\right)+ N_A (\frac1{N_A}-\frac1N \frac{\partial N}{\partial N_A})\\ &= \ln \left( \frac{N_A}{N}\right)+1-\frac{N_A}{N}\tag{1}\end{align}

where in the first line I have used product rule; in the second line, I have used a logarithm identity; in the third line, I used chain rule.

\begin{align}\frac{\partial}{\partial N_A}\left(N_B\ln \left( \frac{N_B}{N}\right) \right) &= \frac{\partial}{\partial N_A}\left(N_B(\ln \left( N_B)- \ln(N)\right) \right)\\ &= N_B(-\frac1N \frac{\partial N}{\partial N_A})\\ &=- \frac{N_B}{N}\tag{2}\end{align}

\begin{align}\frac{\partial}{\partial N_A}\left(\left(\frac{zw_{AA}}{2kT}\right)N_A\right) =\left(\frac{zw_{AA}}{2kT}\right)\tag{3}\end{align}

\begin{align}\frac{\partial}{\partial N_A}\left(\left(\frac{zw_{BB}}{2kT}\right)N_B\right) =0\tag{4}\end{align}

\begin{align}\frac{\partial}{\partial N_A}\left(\chi_{AB}N_B\left(\frac{N_A}{N}\right)\right) &= \chi_{AB}N_B\left(\frac{N-N_A \frac{\partial}{\partial N_A}N}{N^2} \right) \\&=\chi_{AB}\left(\frac{N_B}{N}\right)^2\tag{5}\end{align}

We just have to sum these terms up to get the result. Equation $$(1)$$ is of particular interest to you.

You're right that $$N$$ is not a constant; in fact, it's shorthand for a function $$N(N_A, N_B) = N_A + N_B$$.

$$N$$ is substituted for convenience. If you want to differentiate the expression $$F/kB$$ with respect to $$N_A$$, one option is to replace $$N$$ with $$N_A+N_B$$ wherever it occurs. $$N$$ is defined to be $$N_A+N_B$$ so this replacement is always okay, and when you differentiate the overall substituted expression with respect to $$N_A$$, you will get the right answer.

The full expression, with substitution, is:

\begin{aligned}F(a,b)/kB = & a \ln \left(\frac{a}{a+b}\right)+b \ln \left(\frac{b}{a+b}\right) \\ &+\left(\frac{z w_{A A}}{2 k T}\right) a+\left(\frac{z w_{B B}}{2 k T}\right) b+\chi_{A B} \frac{ab}{a+b} \end{aligned}

which consists of five terms. You can differentiate each of them separately with respect to $$a$$ then add up the results. For example, you can work out that the derivative of the first term is $$\frac{b}{a+b} + \ln \left(\frac{a}{a+b}\right)$$ and the derivative of the second term is $$\frac{-b}{a+b}$$.

Added together, these two terms give you $$\ln(\frac{a}{a+b}).$$ Actually this is equal to the first four mysterious terms $$\ln(N_A/N)+1 - N_A/N - N_B/N$$ in your answer—note that the four terms are actually simpler than they appear because the last three cancel:

$$1 - \frac{N_A}{N} - \frac{N_B}{N} = 1 - \frac{N_A+N_B}{N} = 1 - \frac{N}{N} = 1 - 1 = 0$$

so those four mysterious terms are equivalent to simply

$$\ln\left(\frac{N_A}{N}\right)$$

Instead of using the substitution route, you can also just apply the chain rule. Done properly, this must give you the same result.

Let's try it just on a simple term like $$\ln(N_A/ N)$$. We have:

\begin{align*}\partial_a \ln\left(\frac{a}{n}\right) &= \frac{1}{a/n} \cdot \partial_a \frac{a}{n} & \text{\{chain rule for ln\}}\\ &= \frac{1}{a/n} \cdot \frac{\partial_a(n)a - \partial_a(a)n}{n^2}&\text{\{quotient rule\}}\\ &= \frac{1}{a/n} \cdot \frac{1\cdot a - 1\cdot n}{n^2}\\& = \frac{n}{a}\frac{a-n}{n^2} \\&= \frac{1}{a}\frac{a-n}{n} \\&= \frac{1}{a}\frac{a}{n} - \frac{1}{a}\frac{n}{n} \\&= \frac{1}{n} - \frac{1}{a}\end{align*}

So when we compute the derivative of a more complex term like $$N_A\ln(N_A/N)$$, the product rule says that this is:

$$N_A \cdot \partial_{N_A} \ln\left(\frac{N_A}{N}\right) + \partial_{N_A}[N_A] \cdot \ln\left(\frac{N_A}{N}\right)$$ $$=N_A\left(\frac{1}{N} - \frac{1}{N_A}\right) + 1\cdot \ln\left(\frac{N_A}{N}\right) = \left(\frac{N_A}{N}-1\right) + \ln\left(\frac{N_A}{N}\right)$$