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