If $y = ({\log _{\cos x}}\sin x)({\log _{\sin x}}\cos x) + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}},$ then $\frac{{{\rm{d}}y}}{{{\rm{d}}x}}$ at $x = \frac{\pi }{2}$ is equal to

$\frac{8}{{\left( {4 + {\pi ^2}} \right)}}$

$ - \frac{8}{{\left( {4 + {\pi ^2}} \right)}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

y = ({\log _{\cos x}}\sin x)({\log _{\sin x}}\cos x) + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}},

then

\frac{{\mathrm{d}y}}{{\mathrm{d}x}}

x = \frac{\pi }{2}

is equal to

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the function is not defined at

x = \frac{\pi}{2}

, so the derivative cannot be determined.

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