If y=tan ^{-1}left(frac{5 x+1}{3-x-6 x^2}righ | vedclass

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(A) Given $y=	an ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}ight)$.We can rewrite the expression inside the inverse tangent function as:$y=	an ^{-1}\left(\

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$\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}$

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(A) Given $y=	an ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}ight)$.We can rewrite the expression inside the inverse tangent function as:$y=	an ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}ight)$.Using the identity $	an ^{-1}(A)+	an ^{-1}(B)=	an ^{-1}\left(\frac{A+B}{1-AB}ight)$,we get:$y=	an ^{-1}(3 x+2)+	an ^{-1}(2 x-1)$.Now,differentiating with respect to $x$ using the chain rule:$\frac{d y}{d x}=\frac{d}{d x}(	an ^{-1}(3 x+2))+\frac{d}{d x}(	an ^{-1}(2 x-1))$.$\frac{d y}{d x}=\frac{1}{1+(3 x+2)^2} 	imes 3+\frac{1}{1+(2 x-1)^2} 	imes 2$.$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+4+1}+\frac{2}{4 x^2-4 x+1+1}$.$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{2}{4 x^2-4 x+2}$.Simplifying the second term:$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}$.

If $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$,then $\frac{d y}{d x}=$

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