Question

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

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

Answer 1

Answer: (A)

Given $f(x)={\sin }^{-1}\left(\frac{2\times 3^x}{1+9^x}\right)$
Let $3^x=\tan (t)\Rightarrow t={\tan }^{-1}\left(3^x\right)$
So, $f(x)={\sin }^{-1}\left(\frac{2\tan (t)}{1+{\tan }^2(t)}\right)$
We know that, $\sin (2t)=\frac{2\tan (t)}{1+{\tan }^2(t)}$
$\Rightarrow f(x)={\sin }^{-1}(\sin (2t))$
$\therefore f(x)=2t=2{\tan }^{-1}\left(3^x\right)$
$\Rightarrow \frac{df(x)}{dx}=\frac{2}{1+{\left(3^x\right)}^2}\times 3^x\cdot {\log }_{e}3$
At $x=\frac{1}{2},\frac{df}{dx}=\frac{2}{1+{\left(3^{\frac{1}{2}}\right)}^2}\times 3^{\frac{1}{2}}\cdot {\log }_{e}3$
$=\frac{1}{2}\times \sqrt{3}\times {\log }_{e}3$
$=\sqrt{3}\times {\log }_e\sqrt{3}$