Question

If $y={\tan }^{-1}\left(\sqrt{1+x^2}-x\right)$ , then $\frac{dy}{dx}$ equlas

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

Answer 1

Answer: (C)

We have, $y={\tan }^{-1}(\sqrt{1+x^2}-x)$
$\Rightarrow \tan y=\sqrt{1+x^2}-x...(i)$
On differentiating both sides w.r.t. ${}^{\prime }{x}^{\prime }$ we get
${\sec }^{2}y\frac{dy}{dx}=\frac{1}{2}(1+x^2{)}^{\frac{1}{2}-1}\times 2x-1$
$\Rightarrow (1+{\tan }^{2}y)\frac{dy}{dx}=\frac{1}{2}(1+x^2{)}^{-1/2}\cdot 2x-1$
$[\because {\sec }^2\theta =1+{\tan }^2\theta ]$
$\Rightarrow [1+(\sqrt{1+x^2}-x{)}^2]\frac{dy}{dx}=\frac{x}{\sqrt{1+x^2}}-1$
$\Rightarrow [1+1+x^2+x^2-2x\sqrt{1+x^2}\frac{dy}{dx}$
$=\left(\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right)$
$\Rightarrow \left[2+2x^2-2x\sqrt{1+x^2}\right]\frac{dy}{dx}=\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}$
$\Rightarrow 2\left[1+x^2-x\sqrt{1+x^2}\right]\frac{dy}{dx}=\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}$
$\Rightarrow 2\left[{\left(\sqrt{1+x^2}\right)}^2-x\sqrt{1+x^2}\right]\frac{dy}{dx}=\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}$
$\Rightarrow 2\sqrt{1+x^2}\left[\sqrt{1+x^2}-x\right]\frac{dy}{dx}=-\frac{\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}}$
$\Rightarrow \frac{dy}{dx}=\frac{-\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}\cdot 2\sqrt{1+x^2}\left(\sqrt{1+x^2-x}\right)}$
$=\frac{-1}{2\left(1+x^2\right)}$