Question

If $y={\cos }^{-1}\sqrt{\frac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}}}$, then $\frac{dy}{dx}=$

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

Answer 1

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

$y={\cos }^{-1}\sqrt{\frac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}}}$

Put $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$

$\therefore y=\sqrt{\frac{\sqrt{1+{\tan }^2\theta }+1}{2\sqrt{1+{\tan }^2\theta }}}={\cos }^{-1}\sqrt{\frac{\sec \theta +1}{2\sec \theta }}$

$={\cos }^{-1}\sqrt{\frac{1+\cos \theta }{2}}={\cos }^{-1}\sqrt{\frac{2{\cos }^2\theta /2}{2}}$

$={\cos }^{-1}\left(\cos \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{{\tan }^{-1}x}{2}$

$\therefore \frac{dy}{dx}=\frac{1}{2(1+x^2)}$