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 ?
Best Answer
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 $f(2)=1$.