Why can't this expression be factorised without using trigonometrical identities?

Question

The question specified

Find $\frac{dy}{dx}$ of $\tan^{-1}(\frac{3x - x^3}{1 - 3x^2})$

However, we were asked to try the sum without using the trig identities of $\tan(3x)$. I used chain rule and the general formula of $\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1 + x^2}$ and after few steps I get the expression $\frac{6x^2 + 3x^4 + 3}{3x^4 + 3x^2 + x^6 + 1}$ and this is where I cannot further simplify. Can someone please point out how to proceed or is the only way to do this sum using the trig identities ?

Answer 1

From the binomial theorem we know \begin{align} (a+b)^2 &= a^2 + 2ab +b^2\\ (a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \end{align} Now notice that \begin{align} \frac{6x^2 + 3x^4 + 3}{3x^4 + 3x^2 + x^6 + 1} & = 3\frac{ x^4 +2x^2 + 1}{ x^6 +3x^4 + 3x^2 + 1}\\ & = 3\frac{ (x^2)^2 +2(x^2)(1) + 1^2}{ (x^2)^3 +3(x^2)^2(1) + 3(x^2)(1)^2 + 1^3} \end{align} Can you conclude the problem from here?

Answer 2

One can also avoid requiring trigonometric identities, other than the basic definitions, in this way. Writing the equation to be differentiated implicitly as $ \ \tan y \ = \ \frac{x^3 - 3x}{3x^2 - 1} \ \ , \ $ use this to describe a right triangle with "legs" $ \ u \ = \ x^3 - 3x \ $ and $ \ 3x^2 - 1 \ = \ u' + 2 \ \ . $ The hypotenuse of this conceptual triangle is given by $$ H^2 \ \ = \ \ (x^3 \ - \ 3x)^2 \ + \ (3x^2 \ - \ 1)^2 \ \ = \ \ (x^6 \ - \ 6x^4 \ + \ 9x^2) \ + \ (9x^4 \ - \ 6x^2 \ + \ 1) $$ $$ = \ \ x^6 \ + \ 3x^4 \ + \ 3x^2 \ + \ 1 \ \ = \ \ (x^2 \ + \ 1)^3 \ \ , $$ as Robert Lee shows. We then re-arrange the equation as $$ \tan y \ = \ \frac{u}{u' \ + \ 2} \ \ \Rightarrow \ \ (u' \ + \ 2) · \tan y \ \ = \ \ u $$ $$ \Rightarrow \ \ (u' \ + \ 2)' · \tan y \ \ + \ \ (u' \ + \ 2) · (\tan y)' \ \ = \ \ u' $$ $$ \Rightarrow \ \ u'' · \tan y \ \ + \ \ (u' \ + \ 2) · (\sec^2 y) \ y' \ \ = \ \ u' $$ $$ y' \ \ = \ \ \frac{u' \ - \ u'' · \tan y }{(u' \ + \ 2) · \sec^2 y} \ \ = \ \ \frac{u' \ - \ u'' · \left(\frac{u}{u' \ + \ 2} \right)}{u' \ + \ 2 } \ · \ (\cos^2 y) $$ $$ = \ \ \frac{u' · (u' \ + \ 2) \ - \ u'' · u}{(u' \ + \ 2)^2 } \ · \ \left[\frac{ (u' \ + \ 2) }{H } \right]^2 \ \ = \ \ \frac{(3x^2 \ - \ 3) · (3x^2 \ - \ 1) \ - \ 6x · (x^3 - 3x)}{H^2 } $$ $$ = \ \ \frac{3·(3x^4 \ - \ 3x^2 \ - \ x^2 \ + \ 1 \ - \ 2x^4 \ + \ 6x^2)}{H^2 } \ \ = \ \ \frac{3·(x^4 \ + \ 2x^2 \ + \ 1 )}{H^2 } $$ $$ = \ \ \frac{3·(x^2 \ + \ 1)^2}{(x^2 \ + \ 1)^3 } \ \ = \ \ \frac{3}{x^2 \ + \ 1 } \ \ , $$ again employing the binomial theorem. (This approach was taken to postpone manipulation of polynomials as late as manageable. You probably cannot evade dealing with the rational function you obtained: the problem seems to be constructed to produce it.)