Question

lf $f\left(x\right)={\log }_{x^2}\left(\log x\right)$ , then $f^{\prime }\left(x\right)$ at $x=e$ is

• A. $0$
• B. $1$
• C. $1/e$
• D. $1/2e$

Answer 1

Answer: (D)

$1/2e$

$f\left(x\right)=\frac{Log\left(Logx\right)}{2Logx}$
$f^{{}^{\prime }}\left(x\right)=\frac{1}{2}\left(\frac{\left(\frac{1}{xLogx}\right)Logx-Log\left(Log\right(x\left)\right)\left(\frac{1}{x}\right)}{{\left(Log\right(x\left)\right)}^2}\right)$
$=\frac{1}{2x}\left(\frac{1-Log\left(Log\right(x\left)\right)}{{\left(Log\right(x\left)\right)}^2}\right)$
$f^{{}^{\prime }}(x{)}_{x=e}=\frac{1}{2e}$