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If $f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right)$, then $f^{\prime}\left(-\frac{1}{2}\right)$ equals.
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Verified Answer
The correct answer is:
$\sqrt{3} \log _e \sqrt{3}$
$\sqrt{3} \log _e \sqrt{3}$
Since $f(x)=\sin \left(\frac{2 \times 3^x}{1+9^x}\right)$
Suppose $3^x=\tan t$
$$
\begin{aligned}
&\Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \\
&=\sin ^{-1}(\sin 2 t)=2 t=2 \tan ^{-1}(3 x) \\
&\text { So, } f^{\prime}(x)=\frac{2}{1+\left(3^x\right)^2} \times 3^x \cdot \log _e 3 \\
&\therefore \quad f^{\prime}\left(-\frac{1}{2}\right)=\frac{2}{1+\left(3^{-\frac{1}{2}}\right)^2} \times 3^{-\frac{1}{2}} \cdot \log _e 3 \\
&=\frac{1}{2} \times \sqrt{3} \times \log _e 3=\sqrt{3} \times \log _e \sqrt{3}
\end{aligned}
$$
Suppose $3^x=\tan t$
$$
\begin{aligned}
&\Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \\
&=\sin ^{-1}(\sin 2 t)=2 t=2 \tan ^{-1}(3 x) \\
&\text { So, } f^{\prime}(x)=\frac{2}{1+\left(3^x\right)^2} \times 3^x \cdot \log _e 3 \\
&\therefore \quad f^{\prime}\left(-\frac{1}{2}\right)=\frac{2}{1+\left(3^{-\frac{1}{2}}\right)^2} \times 3^{-\frac{1}{2}} \cdot \log _e 3 \\
&=\frac{1}{2} \times \sqrt{3} \times \log _e 3=\sqrt{3} \times \log _e \sqrt{3}
\end{aligned}
$$
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