MHT CET · Maths · Differentiation
If \(y=\log _{\cos x} \sin x\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \cos x)^{2}}\)
- B \(\frac{(\tan x \log \cos x+\cot x \log \sin x)}{(\log \cos x)^{2}}\)
- C \(\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \sin x)^{2}}\)
- D None of the above
Answer & Solution
Correct Answer
(A) \(\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \cos x)^{2}}\)
Step-by-step Solution
Detailed explanation
Given, \(y=\log _{\cos x} \sin x=\frac{\log \sin x}{\log \cos x}\)
On differentiating w.r.t. \(x\), we get
\(
\frac{d y}{d x}=\frac{\cot x \cdot \log \cos x+\tan x \cdot \log \sin x}{(\log \cos x)^{2}}
\)
On differentiating w.r.t. \(x\), we get
\(
\frac{d y}{d x}=\frac{\cot x \cdot \log \cos x+\tan x \cdot \log \sin x}{(\log \cos x)^{2}}
\)
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