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If $y=\log _{\sin x} \tan x$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}$ has the value
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Verified Answer
The correct answer is:
$\frac{-4}{\log 2}$
$$
\begin{aligned}
& y=\frac{\log \tan x}{\log \sin x} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right) \cdot \sec ^2 x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x)^2}
\end{aligned}
$$
At $x=\frac{\pi}{4}$
$$
\begin{aligned}
& \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2})^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{\left[\log \left(\frac{1}{\sqrt{2}}\right)\right]^2} \\
& =\frac{-2 \times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2)^2} \quad \ldots[\because \log 1=0] \\
& =\frac{-4}{\log 2}
\end{aligned}
$$
\begin{aligned}
& y=\frac{\log \tan x}{\log \sin x} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right) \cdot \sec ^2 x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x)^2}
\end{aligned}
$$
At $x=\frac{\pi}{4}$
$$
\begin{aligned}
& \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2})^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{\left[\log \left(\frac{1}{\sqrt{2}}\right)\right]^2} \\
& =\frac{-2 \times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2)^2} \quad \ldots[\because \log 1=0] \\
& =\frac{-4}{\log 2}
\end{aligned}
$$
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