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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}ight)=	an ^{-1} \frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)} \ & \Rightarrow y=	an ^{-1}(3 x+2)+	an ^{-1}(2 x-1) \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+(3 x+2)} 	imes 3+\frac{1}{1+(2 x-1)^2} 	imes 2 \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{9 x^2+12 x+5}+\frac{2}{4 x^2-4 x+2} \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}\end{aligned}$

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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