Calculus Made Easy Exercise 2 Question 9 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:
$n=\frac1{DL}\sqrt\frac{gT}{\pi\sigma}$
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.
 A: 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:
$n=\frac1{DL}\sqrt\frac{gT}{\pi\sigma}$
$n + dn=\frac1{(D + dD)L}\sqrt\frac{gT}{\pi\sigma}$
$n + dn=(L(D+dD))^{-1}\sqrt\frac{gT}{\pi\sigma}$
$n + dn=L^{-1}(D^{-1}(1+\frac{dD}D)^{-1})\sqrt\frac{gT}{\pi\sigma}$
$n + dn=L^{-1}(D^{-1}[1-1\frac{dD}D+\frac{(-1-2)}{2!}(\frac{dD}D)^2...])\sqrt\frac{gT}{\pi\sigma}$
$n + dn=L^{-1}(D^{-1}[1-\frac{dD}D])\sqrt\frac{gT}{\pi\sigma}$
$n + dn=L^{-1}(D^{-1}-\frac{dD}{D^2})\sqrt\frac{gT}{\pi\sigma}$
$n + dn=\frac1{DL}-\frac{dD}{LD^2}\sqrt\frac{gT}{\pi\sigma}$
$dn = -dD*\frac1{LD^2}$
$\frac{dn}{dD}=-\frac1{LD^2}$
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.
A: 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, $(x^n)'=nx^{n-1}$.
