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…

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
  1. $\frac{1}{1+16 x^2}$
  2. $\frac{4}{1+16 x^2}$
  3. $\frac{1}{1+4 x^2}$
  4. $\frac{4}{1+4 x^2}$

Solution

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

Asked in: MHT CET 2024 (03 May Shift 2)

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