Question

If $f(x) = \sin^{-1} \left[ \frac{2^{x+1}}{1+4^x} \right]$, then f'(0) =

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

Answer 1

Answer: (D)

$f\left(x\right)=sin^{-1} \left[\frac{2.2^{x}}{1+\left(2^{x}\right)^{2}}\right]$
$Put 2^{x}=tan \theta$
$f\left(x\right)=sin^{-1}\left[\frac{2 tan\theta}{1+tan^{2}\theta}\right]$
$=sin^{-1}\left[sin 2 \theta\right]$
$f\left(x\right)=2\theta$
$f\left(x\right)=2tan^{-1}\left(2^{x}\right)$
$f '\left(x\right)=\frac{2}{1+\left(2^{x}\right)^{2}}.2^{x} log_{e}^{2}$
$f '\left(0\right)=log_{e}2$