If $y=\log _{\sin x} an x$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x} ight)_{x=\frac{\pi}{4}}$ has the value

Question

If $y=\log _{\sin x} 	an x$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=\frac{\pi}{4}}$ has the value

Aicharya · Accepted Answer

$\begin{aligned} & y=\frac{\log 	an x}{\log \sin x} \ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(\log \sin x)\left(\frac{1}{	an x}ight) \cdot \sec ^2 x-(\log 	an x)\left(\frac{1}{\sin x}ight)(\cos x)}{(\log \sin x)^2} \end{aligned}$ At $x=\frac{\pi}{4}$ $\begin{aligned} & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=\frac{\pi}{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}ight)\left(\frac{1}{1}ight)(\sqrt{2})^2-(\log 1)\left(\frac{\sqrt{2}}{1}ight)\left(\frac{1}{\sqrt{2}}ight)}{\left[\log \left(\frac{1}{\sqrt{2}}ight)ight]^2} \ & =\frac{-2 	imes \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2)^2} \quad \ldots[\because \log 1=0] \ & =\frac{-4}{\log 2} \end{aligned}$

If $y=\log _{\sin x} \tan x$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}$ has…

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