Question

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

• A. $\frac{2x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (2+x^2)\cdot \sqrt{1+x^2}}$
  
• B. $\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (2+x^2)\cdot \sqrt{1+x^2}}$
• C. $\frac{1}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (2+x^2)\cdot \sqrt{1+x^2}}$
• D. $\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (1+x^2)\cdot \sqrt{1+x^2}}$

Answer 1

The correct option is B $\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (2+x^2)\cdot \sqrt{1+x^2}}$

$y=\ln \left({\tan }^{-1}\sqrt{1+x^2}\right)$
On differentiation we get,
$\frac{dy}{dx}=\frac{d}{dx}\left(\ln \right({\tan }^{-1}\sqrt{1+x^2}\left)\right)=\frac{1}{{\tan }^{-1}\left(\sqrt{1+x^2}\right)}\frac{d}{dx}\left({\tan }^{-1}\right(\sqrt{1+x^2}\left)\right)=\frac{1}{{\tan }^{-1}\left(\sqrt{1+x^2}\right)}\cdot \frac{1}{1+(\sqrt{1+x^2}{)}^2}\frac{d}{dx}\left(\sqrt{1+x^2}\right)=\frac{1}{\left({\tan }^{-1}\sqrt{1+x^2}\right)}\cdot \frac{1}{1+(\sqrt{1+x^2}{)}^2}\cdot \frac{1}{2\sqrt{1+x^2}}\cdot \frac{d}{dx}\left(x^2\right)=\frac{1}{\left({\tan }^{-1}\sqrt{1+x^2}\right)}\cdot \frac{1}{1+(\sqrt{1+x^2}{)}^2}\cdot \frac{1}{2\sqrt{1+x^2}}\cdot 2x=\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (1+(\sqrt{1+x^2}{)}^2)\cdot \sqrt{1+x^2}}=\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\cdot (2+x^2)\cdot \sqrt{1+x^2}}$