TITLE: Fastest way to find derivative of the function at the given point
QUESTION [8 upvotes]: What is derivative of $f(x)=\left(\dfrac{\sqrt[3]{x^2+2x}}{x^2-x}\right)^3$ at the point $x=2$? $$1)-\frac34\qquad\qquad2)-\frac54\qquad\qquad3)-\frac52\qquad\qquad4)-\frac{15}4$$


This is a problem from an timed Exam, so I am looking for the fastest way to solve this. Here is my solution:
We have $f(x)=\dfrac{x^2+2x}{(x^2-x)^3}$ . by using Quotient rule, derivative at  $x=2$ is:
$$\dfrac{6\times8-(3\times3\times4)\times8}{8^2}=\frac34-\frac92=-\frac{15}4$$
Although the way I putted the values of functions instead of writing the whole derivative of function seems to be fast , it is hard to avoid algebraic mistakes and find the correct answer in the exam condition (having time pressure, stress  and so on).
After all, is there a better approach (faster) to solve this problem ?

REPLY [3 votes]: Here is a way, not sure how much faster you'll find it, using Taylor series.
You have
$$
f(x) = \frac{x^2+2x}{(x^2-x)^3} = \frac{x+2}{x^2(x-1)^3}
$$
You want to look at things around $x=2$, so set $u:= x-2$:
$$
\frac{x+2}{x^2(x-1)^3} = \frac{u+4}{(u+2)^2(u+1)^3}
= \frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3}
$$
Now you want to look at things around $u=0$: use $(1+u)^a = 1+au + O(u)$ for any constant $a\in\mathbb{R}$ to get
$$
\frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3} = \left(1+\frac{u}{4}\right)\left(1-u + o(u)\right)\left(1-3u+ o(u)\right)
$$
Expand, only keeping the first two terms:
$$
\frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3} = 1+\left(\frac{1}{4}-1-3\right)u+o(u)
= 1 + \boxed{\frac{-15}{4}}u+o(u)
$$
You get the derivative $f'(2)=-\frac{15}{4}$ you wanted here. ((Also, the first term, $1$, is $f(2)$).