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. $'x'$ 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}}\cdot2\sqrt{1+x^{2}}\left(\sqrt{1 + x^{2} -x}\right)} $
$ = \frac{-1}{2\left(1+x^{2}\right)}$