If $y = {\log _{\cos x}}\sin x,$ then $\frac{{dy}}{{dx}}$ is equal to

$\frac{{(\tan x\log \cos x + \cot x\log \sin x)}}{{{{(\log \cos x)}^2}}}$

$\frac{{(\cot x\,{\rm{log cos}}\,x + \tan x\log \sin x)}}{{{{(\log \sin x)}^2}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\log _{\cos x}}\sin x,

then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: A

\frac{{(\cot x\log \cos x + \tan x\log \sin x)}}{{{{(\log \cos x)}^2}}}

Option A is correct because it correctly applies the quotient rule and change of base formula for logarithms.

Practice All Chapter MCQs →

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