Question

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

• A. $\frac{1}{1+x^2}$
• B. $\frac{1}{1-x^2}$
• C. $\frac{1}{2(1+x^2)}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2(1+x^2)}$

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

Let $x=\tan \theta$

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

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

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

$y={\cos }^{-1}\left[\cos \theta /2\right]$

$y=\frac{\theta }{2}$

put $\theta ={\tan }^{-1}x$

$y=\frac{1}{2}{\tan }^{-1}x$

differentiating gives:

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