Question

Let $y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),x\ge 0,y\ge 0$ and $xy<1$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1+y^2}{x^2(1+x^2)}$
• B. $\frac{1+y^2}{y^2(1+x^2)}$
• C. $\frac{1+x^2}{x^2(1+y^2)}$
• D. $\frac{1+x^2}{y^2(1+y^2)}$

Answer 1

The correct option is B $\frac{1+y^2}{y^2(1+x^2)}$

Given : $y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),x\ge 0,y\ge 0$ and $xy<1$
$\Rightarrow y={\tan }^{-1}x+{\tan }^{-1}y$
Differentiating both sides w.r.t. $x,$
$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}+\frac{1}{1+y^2}\cdot \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1+y^2}{y^2(1+x^2)}$