Q.Find $\dfrac{dy}{dx}$ if $y = \tan^{-1}\left(\dfrac{5x+1}{3-x-6x^2}\right)$.

Differentiate $g(x)=\dfrac{5x+1}{3-x-6x^2}$ by the quotient rule, then use $\dfrac{d}{dx}\tan^{-1}g = \dfrac{g'}{1+g^2}$.

Let $g(x) = \dfrac{5x+1}{3-x-6x^2} = \dfrac{N}{D}$ with $N=5x+1$, $D=3-x-6x^2$, so $y=\tan^{-1}g(x)$.

$N'=5$, $D'=-1-12x$.

By the quotient rule:

$g'(x) = \frac{N'D-ND'}{D^2} = \frac{5(3-x-6x^2)-(5x+1)(-1-12x)}{D^2}$

Compute the numerator:

$5(3-x-6x^2) = 15-5x-30x^2$

$(5x+1)(-1-12x) = -5x-60x^2-1-12x = -60x^2-17x-1$

$N'D-ND' = (15-5x-30x^2)-(-60x^2-17x-1) = 15-5x-30x^2+60x^2+17x+1 = 30x^2+12x+16$

So $g'(x) = \dfrac{30x^2+12x+16}{D^2}$.

Now, $\dfrac{dy}{dx} = \dfrac{g'(x)}{1+g(x)^2} = \dfrac{g'(x)}{\frac{D^2+N^2}{D^2}} = \dfrac{30x^2+12x+16}{D^2+N^2}$.

Compute $D^2+N^2$:

$D^2 = (3-x-6x^2)^2 = 36x^4+12x^3-35x^2-6x+9$

$N^2 = (5x+1)^2 = 25x^2+10x+1$

$D^2+N^2 = 36x^4+12x^3-10x^2+4x+10$

Therefore: …