Put simply, the problem is to find the rate of change of n when D, L, $\sigma$, or T are varied, singly. The equation for this is as follows:


Thus far, I've used the method introduced previously in the book by substituting $n$ with $n + dn$ and the same for $D$ and $D + dD$; however, beyond this point is where I've gotten stuck. After this I tried to following, to no avail:

$n + dn=\frac1{(D+dD)L}\sqrt\frac{gT}{\pi\sigma}$

$n + dn=((DL)^{-1}+(LdD)^{-1})\sqrt\frac{gT}{\pi\sigma}$

$n + dn=\frac1{DL}\sqrt\frac{gT}{\pi\sigma}+\frac1{LdD}\sqrt\frac{gT}{\pi\sigma}$

$dn = \frac1{LdD}\sqrt\frac{gT}{\pi\sigma}$

After this, I realized this looks nothing like the solution presented, which is $\frac{dn}{dD}=-\frac1{LD^2}\sqrt\frac{gT}{\pi\sigma}$

I am quite positive that I must have made an error somewhere, and hopefully after figuring it out, I can proceed with the rest of the book and exercise problems.

Thanks in advance.

  • 2
    $\begingroup$ You made the Freshman's Dream mistake $\frac{1}{a+b}\neq \frac{1}{a}+\frac{1}{b}$ $\endgroup$ Sep 23, 2019 at 1:06
  • $\begingroup$ Does the book expect you to use the binomial approximation ?$$(1+x)^n\to 1+nx \text{ as } x \to 0$$ $\endgroup$
    – WW1
    Sep 23, 2019 at 1:18
  • $\begingroup$ Your are showing the rate of change in $n$ when only $D$ changes. However, your are saying all the other variables could change to. It is important to know all the variables that n depends on. By the way, this looks related:math.stackexchange.com/questions/432383/… $\endgroup$
    – NoChance
    Sep 23, 2019 at 1:18
  • $\begingroup$ WW1 - The book thus far has not mentioned nor used the binomial approximation, only the binomial theorem, used for multiplying binomials. Are they similar? $\endgroup$
    – MrMcblader
    Sep 23, 2019 at 1:21
  • $\begingroup$ NoChance - I didn't realize this was a repeat question, my apologies, I didn't come across it when searching for an answer to this question. Could you clarify what you mean by me saying that all of the other variables could change too? The question states that the variables are varied, singly, and I am only trying to solve for the derivative for one at a time. $\endgroup$
    – MrMcblader
    Sep 23, 2019 at 1:26

2 Answers 2


Thanks to Ninad Munshi for clarifying the major mistake I made. Now I was able to solve the problem, albeit without the constants in the answer, which are not relevant to differentiating a problem like this, as far as I know; nevertheless, here is the work for how I got my answer:


$n + dn=\frac1{(D + dD)L}\sqrt\frac{gT}{\pi\sigma}$

$n + dn=(L(D+dD))^{-1}\sqrt\frac{gT}{\pi\sigma}$

$n + dn=L^{-1}(D^{-1}(1+\frac{dD}D)^{-1})\sqrt\frac{gT}{\pi\sigma}$

$n + dn=L^{-1}(D^{-1}[1-1\frac{dD}D+\frac{(-1-2)}{2!}(\frac{dD}D)^2...])\sqrt\frac{gT}{\pi\sigma}$

$n + dn=L^{-1}(D^{-1}[1-\frac{dD}D])\sqrt\frac{gT}{\pi\sigma}$

$n + dn=L^{-1}(D^{-1}-\frac{dD}{D^2})\sqrt\frac{gT}{\pi\sigma}$

$n + dn=\frac1{DL}-\frac{dD}{LD^2}\sqrt\frac{gT}{\pi\sigma}$

$dn = -dD*\frac1{LD^2}$


I also want to thank Chris Custer for his answer, while it was not exactly what I was looking for, it still answered the question in an easy way, and correctly at that.

  • $\begingroup$ I think it is not right to drop the multiplicative constant as you end up doing above, consider ax^n doesn't drop a when differentiated for x and becomes anx^n-1. See p26-28 in the same chapter. Such a drop only applies if you were adding the square root term. $\endgroup$ Feb 9, 2020 at 19:39

When differentiating with respect to $D$, say, all the other variables are treated as constants. Now $(1/D)'=-(1/D^2)$. Hence the result. This is the power rule for derivatives. Namely, $(x^n)'=nx^{n-1}$.

  • $\begingroup$ I understand the premise of the power rule, but I want to know to work to get there more so than just the answer based on a rule. How would I get from the original equation to the answer through the work, not just by using a rule? I wish to do it this way to get a better understanding for why the power rule works, and not to just use it; however, I do appreciate the answer regardless, thanks. $\endgroup$
    – MrMcblader
    Sep 23, 2019 at 1:17
  • $\begingroup$ Sure. You can re-derive the power rule. But after that you should just commit it to memory, and apply it. $\endgroup$
    – user403337
    Sep 23, 2019 at 1:24

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