Answer:
Let \[y={{\tan }^{-1}}(\sqrt{1+{{x}^{2}}}+x)\] On putting \[x=\cot \,\theta ,\] we get \[y={{\tan }^{-1}}(cosec\theta +cot\theta )\] \[[\because \,\,\text{using}\,\,\text{identity}\,\,\text{1+co}{{\text{t}}^{2}}\theta =cose{{c}^{2}}\theta ]\] \[={{\tan }^{-1}}\left( \frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta } \right)={{\tan }^{-1}}\left( \frac{1+\cos \theta }{\sin \theta } \right)\] \[={{\tan }^{-1}}\left( \frac{2{{\cos }^{2}}\frac{\theta }{2}}{2sin\frac{\theta }{2}\cos \frac{\theta }{2}} \right)={{\tan }^{-1}}\left( \cot \frac{\theta }{2} \right)\] \[={{\tan }^{-1}}\left\{ \tan \left( \frac{\pi }{2}-\frac{\theta }{2} \right) \right\}=\frac{\pi }{2}-\frac{1}{2}\theta =\frac{\pi }{2}-\frac{1}{2}{{\cot }^{-1}}x\] \[\therefore \] \[\frac{dy}{dx}=0-\frac{1}{2}\left( \frac{-\,1}{1+{{x}^{2}}} \right)=\frac{1}{2\,(1+{{x}^{2}})}\]
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