Question

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

• A. 1.5
• B. 2.5
• C. 1
• D. 10

Answer 1

The correct option is B 2.5

$y=ta{n}^{-1}\left(\frac{5x-x}{1+5x.x}\right)+ta{n}^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3}x}\right)=ta{n}^{-1}5x+ta{n}^{-1}\frac{2}{3}$
$\frac{dy}{dx}=\frac{5}{1+25.x^2}\Rightarrow atx=\frac{1}{5},\frac{dy}{dx}=\frac{5}{2}$