If \(y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}^\infty }}}}\) then \(\frac{{dy}}{{dx}}\) is equal to

\(\frac{{{y^2}\tan x}}{{1 + y\;In\;(\cos x)}}\)

\(- \frac{{{y^2}\tan x}}{{1 - y\;In\;(\cos x)}}\)

\(\frac{{{y^2}\tan x}}{{1 - y\;In\;(\sin x)}}\)

\(\frac{{{y^2}\sin x}}{{1 + y\;In\;(\sin x)}}\)

If \(y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}

1.	If \(y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}^\infty }}}}\) then \(\frac{{dy}}{{dx}}\) is equal to
A.	\(- \frac{{{y^2}\tan x}}{{1 - y\;In\;(\cos x)}}\)
B.	\(\frac{{{y^2}\tan x}}{{1 + y\;In\;(\cos x)}}\)
C.	\(\frac{{{y^2}\tan x}}{{1 - y\;In\;(\sin x)}}\)
D.	\(\frac{{{y^2}\sin x}}{{1 + y\;In\;(\sin x)}}\)
Answer» B. \(\frac{{{y^2}\tan x}}{{1 + y\;In\;(\cos x)}}\)

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