Question

If $f\left(x\right)={\cos }^{-1}\left(\frac{1-4^x}{1+4^x}\right),$ then $f^{\prime }\left(0\right)=$

• A. $\log 2$
• B. $\frac{2}{5}\log 2$
• C. $2\log 2$
• D. $\frac{4}{5}\log 2$

Answer 1

The correct option is A $\log 2$

$f\left(x\right)={\cos }^{-1}\left(\frac{1-4^x}{1+4^x}\right)f\left(x\right)=2{\tan }^{-1}{2}^x$
$f^{\prime }\left(x\right)=\frac{2}{1+(2^x{)}^2}{2}^x\cdot {\log }_{e}2$
So, $f^{\prime }\left(0\right)=\frac{2}{1+(2^0{)}^2}{2}^0\cdot {\log }_{e}2=\frac{2}{1+1}\log 2$
$f^{\prime }\left(0\right)=\log 2$