Fastest way to find derivative of the function at the given point 
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 ?
 A: 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 $f(2)$).
A: Perhaps slightly...but very slightly...easier: using that $\;(x^2-x)^3=x^3(x-1)^3\;$ and also  $\;x=2\implies (x-1)^n=1\;$
$$f(x)=\frac{x^2+2x}{(x^2-x)^3}=\frac{x+2}{x^2(x-1)^3}=\frac1{x(x-1)^3}+\frac2{x^2(x-1)^3}\implies$$
$$f'(x)=\left(\frac1{x(x-1)^3}\right)'+2\left(\frac1{x^2(x-1)^3}\right)'=$$
$$=-\frac{(x-1)+3x}{x^2(x-1)^4}-2\frac{2(x-1)+3x(x-1)}{x^3(x-1)^4}\implies$$
$$f'(2)=-\frac{1+6}{4\cdot1}-2\frac{2+6}{8\cdot1}=-\frac74-2=-\frac{15}4$$
Another Option: Write your function multiplicatively and use the product rule, which surely will be easier than before:
$$f(x)=x^{-2}(x+2)(x-1)^{-3}\implies $$
$$f'(x)=-2x^{-3}(x+2)(x-1)^{-3}+x^{-2}(x-1)^{-3}-3x^{-2}(x+2)(x-1)^{-4}\implies$$
$$f'(2)=-2\cdot\frac18\cdot4+\frac14-\frac34\cdot4=-1+\frac14-3=-4+\frac14=-\frac{15}4$$
I'd go with this last option. As thumb rule, product rule nice than quotient rule
A: Cancelling out the $x$ yields
$$\dfrac{x^2+2x}{(x^2-x)^3}=\dfrac{x^2+2x}{x^3(x-1)^3}=\dfrac{x+2}{x^2(x-1)^3}.$$
If we take the logarithm on both sides we get
$$\log f(x)=\log(x+2)-\left[2\log x+3\log(x-1)\right].$$
Now taking the derivative of this expression is trivial,
$$\frac{f'(x)}{f(x)}=\frac{1}{x+2}-\frac 2 x-\frac{3}{x-1}\Rightarrow \frac{f'(2)}{f(2)}=\frac{1}{4}-1-3=-\frac{15}{4}.$$
Last step was obtained by $f(2)=1$.
