If \(y=\tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is equal to

\(\begin{aligned} y & =\tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right) \\ & =\tan ^{-1}\left(\frac{\frac{3}{2}+x}{1-\frac{3}{2} x}\right)+\tan ^{-1}\left(\frac{4 x-x}{1+(4 x)(x)}\right)\end{aligned}\)
\(\begin{aligned} & =\tan ^{-1}\left(\frac{3}{2}\right)+\tan ^{-1} x+\tan ^{-1} 4 x-\tan ^{-1} x \\ \therefore \quad y & =\tan ^{-1}\left(\frac{3}{2}\right)+\tan ^{-1}(4 x)\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=0+\frac{4}{1+(4 x)^2} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4}{1+16 x^2}
\end{aligned}\)