MHT CET · Maths · Differentiation
If \(y=\log _{\sin x} \tan x\), then \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}\) has the value
- A \(\frac{4}{\log 2}\)
- B \(-3 \log 2\)
- C \(\frac{-4}{\log 2}\)
- D \(3 \log 2\)
Answer & Solution
Correct Answer
(C) \(\frac{-4}{\log 2}\)
Step-by-step Solution
Detailed explanation
\(y=\frac{\log \tan x}{\log \sin x} \)
\( \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right) \cdot \sec ^2 x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x)^2}\)
At \(x=\frac{\pi}{4}\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2})^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{\left[\log \left(\frac{1}{\sqrt{2}}\right)\right]^2} \)
\( =\frac{-2 \times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2)^2} \ldots[\because \log 1=0] \)
\( =\frac{-4}{\log 2}\)
\( \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right) \cdot \sec ^2 x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x)^2}\)
At \(x=\frac{\pi}{4}\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2})^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{\left[\log \left(\frac{1}{\sqrt{2}}\right)\right]^2} \)
\( =\frac{-2 \times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2)^2} \ldots[\because \log 1=0] \)
\( =\frac{-4}{\log 2}\)
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