Question

If $y =\tan^{-1} \left(\frac{4x}{1+5x^{2}}\right) + \tan^{-1} \left(\frac{2+3x}{3-2x}\right) $ then $\frac{dy}{dx} $ is equal to

• A. $\frac{1}{ 1 +x^2}$
• B. $\frac{5}{ 1 + 25 x^2}$
• C. 1
• D. none of these.

Answer 1

Answer: (B)

$\tan^{-1} \frac{4x}{1+5x^2} = \tan^{-1} \left(\frac{5x-x}{1+5x.x}\right)$
$ =\tan^{-1} 5x -\tan^{-1}x $
and $\tan^{-1} \left(\frac{2+3x}{3-2x}\right) =\tan^{-1} \left(\frac{\frac{2}{3} +x}{ 1- \frac{3}{2}x}\right)$
$ =\tan^{-1} \frac{2}{3} + \tan^{-1} x$
$ \therefore y =\tan^{-1} 5x +\tan^{-1} \frac{2}{3} $
$\Rightarrow \frac{dy}{dx} = \frac{1}{1+25x^{2}}.5 = \frac{5}{1+25 x^{2}}$