Question

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

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

Answer 1

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

Since ,
$y=si{n}^{-1}\left(\frac{log{x}^2}{1+(logx{)}^2}\right)=si{n}^{-1}\left(\frac{2logx}{1+(logx{)}^2}\right)$

Let log x $=tan\varphi$
$\therefore y=si{n}^{-1}\left(\frac{2tan\varphi }{1+ta{n}^2\varphi }\right)=si{n}^{-1}\left(sin2\varphi \right)$

$\Rightarrow y=2\varphi =2ta{n}^{-1}\left(logx\right)$

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