# 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 ?

• It seems to me that you solved this as efficiently as possible. Mar 25, 2021 at 23:04
• You can continue to simplify and divide by x both numerator and denominator Mar 25, 2021 at 23:12
• This looks like a ideal candidate for logarithmic differentiation. Mar 25, 2021 at 23:40

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 $$f(2)=1$$.

• Interestingly $f(2)=1$. maybe author of the question expected us to use this method ;) Mar 25, 2021 at 23:53

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 $$f(2)$$).

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

• Oh, I just did it with product rule, and Yes it was easier to work with! Mar 25, 2021 at 23:39