Question

If $y={\cos }^{-1}\frac{1-(\log x{)}^2}{1+(\log x{)}^2},$ then $f^{\prime }\left(e\right)=$

• A. $e$
• B. $\frac{1}{e}$
• C. $1$
• D. $\text{does not exist}$

Answer 1

The correct option is B $\frac{1}{e}$

$y={\cos }^{-1}\frac{1-(\log x{)}^2}{1+(\log x{)}^2}=2{\tan }^{-1}\left(\log x\right)$
$\frac{dy}{dx}=2\cdot \frac{1}{1+(\log x{)}^2}\cdot \frac{1}{x}$
Put $x=e$
so, $\frac{dy}{dx}=f^{\prime }\left(e\right)=\frac{2}{e(1+(\log e{)}^2)}$
$\Rightarrow f^{\prime }\left(e\right)=\frac{2}{e(1+(1{)}^2)}=\frac{1}{e}$