Question

If $f(x)={\sin }^{-1}\left(\frac{2\times 3^x}{1+9^x}\right)$, then $f^{\prime }\left(-\frac{1}{2}\right)$ equals :

• A. $-\sqrt{3}{\log }_e\sqrt{3}$
• B. $\sqrt{3}{\log }_e\sqrt{3}$
• C. $-\sqrt{3}{\log }_{e}3$
• D. $\sqrt{3}{\log }_{e}3$

Answer 1

Answer: (B)

Given:$f(x)={\sin }^{-1}\left(\frac{2.3^x}{1+{\left(3^x\right)}^2}\right)$
Let $3^x=\tan \theta$, then
$f(\theta )={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$
$\Rightarrow f(\theta )={\sin }^{-1}(\sin 2\theta )=2\theta$
$\Rightarrow f(x)=2{\tan }^{-1}\left(3^2\right)$
So $f^{\prime }(x)=2\frac{1}{\left(1+9^x\right)}\cdot 3^x\ln 3$
$f^{\prime }\left(\frac{-1}{2}\right)=\frac{2}{1+9^{-1/2}}{3}^{-1/2}\ln 3=\sqrt{3}\ln \sqrt{3}=\sqrt{3}{\log }_e\sqrt{3}$