​$$\frac{d}{dx}\left[\log_{2}\sqrt{x^{2}+1}\right]$$ equals to :​

Correct option is C

Given:
​$$\frac{d}{dx}\left[\log_{2}\sqrt{x^{2}+1}\right]$$​​

Concept used:
​$$\log_{a} b = \frac{\ln b}{\ln a}$$​​
and $$\frac{d}{dx}(\ln u) = \frac{1}{u}\frac{du}{dx}$$​​

Solution:
​$$y = \log_{2}\sqrt{x^{2}+1} \\ = \frac{\ln(\sqrt{x^{2}+1})}{\ln 2} \\ \ \\= \frac{1}{\ln 2} \cdot \frac{1}{2}\ln(x^{2}+1) \\ \ \\\frac{dy}{dx} = \frac{1}{2\ln 2} \cdot \frac{1}{x^{2}+1} \cdot 2x \\ \ \\= \frac{x}{(x^{2}+1)\ln 2}$$​​
Correct answer is (c) $$\frac{x}{(x^{2}+1)\log 2}$$​​