<s>[INST] 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 ?
 [/INST] 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
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