If y=log left[tan sqrt{frac{2^x-1}{2^x+1}}rig | vedclass

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(C) Given $y=\log \left[	an \sqrt{\frac{2^x-1}{2^x+1}}ight]$.Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$, $v = \frac{2-1}{2+1} = \frac{1}{3}$.Using the

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$\frac{4 \sqrt{3} \log 2}{9 \sin \left(\frac{2}{\sqrt{3}}\right)}$

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(C) Given $y=\log \left[	an \sqrt{\frac{2^x-1}{2^x+1}}ight]$.Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$, $v = \frac{2-1}{2+1} = \frac{1}{3}$.Using the quotient rule, $\frac{dv}{dx} = \frac{(2^x+1)(2^x \log 2) - (2^x-1)(2^x \log 2)}{(2^x+1)^2} = \frac{2^x \log 2 (2^x+1-2^x+1)}{(2^x+1)^2} = \frac{2 \cdot 2^x \log 2}{(2^x+1)^2} = \frac{2^{x+1} \log 2}{(2^x+1)^2}$.At $x=1$, $\left(\frac{dv}{dx}ight)_{x=1} = \frac{2^2 \log 2}{(2+1)^2} = \frac{4 \log 2}{9}$.Now, $y = \log(	an \sqrt{v})$. Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{	an \sqrt{v}} \cdot \sec^2 \sqrt{v} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}$.Using $\frac{1}{	an \sqrt{v}} \cdot \sec^2 \sqrt{v} = \frac{\cos \sqrt{v}}{\sin \sqrt{v}} \cdot \frac{1}{\cos^2 \sqrt{v}} = \frac{1}{\sin \sqrt{v} \cos \sqrt{v}} = \frac{2}{\sin(2\sqrt{v})}$.So, $\frac{dy}{dx} = \frac{2}{\sin(2\sqrt{v})} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx} = \frac{1}{\sin(2\sqrt{v}) \cdot \sqrt{v}} \cdot \frac{dv}{dx}$.At $x=1$, $v = \frac{1}{3}$, so $\sqrt{v} = \frac{1}{\sqrt{3}}$.$\left(\frac{dy}{dx}ight)_{x=1} = \frac{1}{\sin(2/\sqrt{3}) \cdot (1/\sqrt{3})} \cdot \frac{4 \log 2}{9} = \frac{\sqrt{3} \cdot 4 \log 2}{9 \sin(2/\sqrt{3})} = \frac{4 \sqrt{3} \log 2}{9 \sin(2/\sqrt{3})}$.

If $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right], x>0$, then $\left(\frac{d y}{d x}\right)_{x=1}=$

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