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. $\frac{5}{1+25x^2}$
• B. $\frac{1}{1+25x^2}$
• C. $0$
• D. $\frac{5}{1-25x^2}$
• E. None of these

Answer 1

Answer: (A)

Given, $y={\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{2+3x}{3-2x}\right)$
Or $y={\tan }^{-1}\left(\frac{5x-x}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3}x}\right)$
$\Rightarrow$ $y={\tan }^{-1}(5x)+{\tan }^{-1}(x)$ $+{\tan }^{-1}\left(\frac{2}{3}\right)+{\tan }^{-1}(x)$
$\Rightarrow$ $y={\tan }^{-1}(5x)+{\tan }^{-1}\left(\frac{2}{3}\right)$ On differentiating w.r.t. $x,$ we get $\frac{dy}{dx}=\frac{5}{1+{(5x)}^2}=\frac{5}{1+25x^2}$