I'm working with the equation for the Clayton Skill Score, which is as follows: $\frac{a}{a+b}-\frac{c}{c+d}$, and with a certain entity $U_a$ which is approximated by a partial derivative of it.
In the example I give below I'm taking the partial derivative holding variables $b$, $c$, and $d$ constant. Adapting an approach from WolframAlpha I have written below an approach using the quotient rule which works, in the sense that it delivers the right answer. However it's been suggested to me that I could have solved the problem more efficiently (presumably, using fewer lines of explanation) using another method. Is that so, and if so how can I do that?
$U_a\approx\frac{\partial}{\partial a}\left(\frac{a}{a+b}-\frac{c}{c+d}\right)$
Differentiating the sum term by term:
$U_a\approx\frac{\partial}{\partial a}\left(\frac{a}{a+b}\right)+\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)$
Using the quotient rule $\frac{\partial}{\partial a}\frac{u}{v}=\frac{v \frac{\partial u}{\partial a}-u \frac{\partial v}{\partial a}}{v^2}$ where $u=a$ and $v=a+b$:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{(a+b)\frac{\partial}{\partial a}(a)-a\frac{\partial}{\partial a}(a+b)}{(a+b)^2}$
The derivative of $a$ is $1$:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{-\left(a\left(\frac{\partial}{\partial a}(a+b)\right)\right)+1(a+b)}{(a+b)^2}$
Simplifying the expression to eliminate the $1$ and move the positive elements in the numerator to the left:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{a+b-\left(a\left(\frac{\partial}{\partial a}(a+b)\right)\right)}{(a+b)^2}$
Differentiate the sum term by term in the numerator:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{a+b-(a\left(\frac{\partial}{\partial a}(a)+\frac{\partial}{\partial a}(b)\right)}{(a+b)^2}$
The derivative of $a$ is $1$:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{a+b-(a\left(1+\frac{\partial}{\partial a}(b)\right)}{(a+b)^2}$
The derivative of $b$ is $0$:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{a+b-(a\left(1+0\right)}{(a+b)^2}$
Simplifying the expression:
$U_a\approx\frac{\partial}{\partial a}\left(-\frac{c}{c+d}\right)+\frac{b}{(a+b)^2}$
The derivative of $-\frac{c}{c+d}$ is $0$:
$U_a\approx 0+\frac{b}{(a+b)^2}$
Simplifying the expression:
$U_a\approx \frac{b}{(a+b)^2}$
I don't know if this was the method in mind, but you can avoid using a cumbersome quotient rule by rewriting
$$\frac{a}{a+b}=1-\frac{b}{a+b}$$
The here point is to eliminate a variable in the numerator. Still I think the way you have done it is just fine.
We have $$\begin{align}U_a&\approx\frac{\partial}{\partial a}\left(\frac{a}{a+b}-\frac{c}{c+d}\right) \\&=\frac{\partial}{\partial a}\left(1-\frac{b}{a+b}-\frac{c}{c+d}\right) \\&=\frac{\partial}{\partial a}\left(1\right)-\frac{\partial}{\partial a}\left(\frac{b}{a+b}\right)-\frac{\partial}{\partial a}\left(\frac{c}{c+d}\right) \\&=0-\frac{\partial}{\partial a}\left(\frac{b}{a+b}\right)-0 \\&=-\left(-\frac{b}{(a+b)^2}\right) \\&=\frac{b}{(a+b)^2} \end{align}$$