Question

If $y={\sin }^{-1}\left(\frac{\log x^2}{1+{\left(\log x\right)}^2}\right)$, then the value of $\frac{dy}{dx}$ is

• A. $\frac{2}{x(1+\log x)}$
• B. $\frac{2}{x{[1+{\left(\log x\right)}^2]}^{}}$
• C. $\frac{1}{x{[1+{\left(\log x\right)}^2]}^{}}$
• D. $\frac{3}{x{[1+{\left(\log x\right)}^2]}^{}}$

Answer 1

Answer: (B)

$\frac{2}{x{[1+{\left(\log x\right)}^2]}^{}}$

Given, $y={\sin }^{-1}\left(\frac{\log x^2}{1+{\left(\log x\right)}^2}\right)$
$={\sin }^{-1}\left(\frac{2\log x}{1+{\left(\log x\right)}^2}\right)$
Let $\log x=\tan \varphi$
Therefore, $y={\sin }^{-1}(\frac{2\tan \varphi }{1+{\tan }^2\varphi }={\sin }^{-1}\left(\sin 2\varphi \right))=2\varphi =2{\tan }^{-1}\left(\log x\right)$
$\Rightarrow \frac{dy}{dx}=\frac{2}{1+{\left(\log x\right)}^2}\frac{1}{x}=\frac{2}{x[1+{\left(\log x\right)}^2]}$