Question

y=cosx^cosx^cosx....up to infinity

Answer 1

I dont know
It's right or wrong

Answer 2

Answer:

$y ={{{(\cos(x)) }^{( \cos(x)) } }^{( \cos(x))... \infty } } \\ y = { (\cos(x) )}^{y} \\ log(y) = log( { \cos(x) )}^{y} \\ log(y) = y log( \cos(x) ) \: \: \: ...(i) \\ On \: differentiating \: both \: sides \: of \: (i)\:w.r.t. \: x, \: we \: get \\ \frac{1}{y}.\frac{dy}{dx} = log( \cos(x) ) \frac{dy}{dx} + y \frac{( - \sin(x)) }{ \cos(x) } \\ \frac{1}{y}.\frac{dy}{dx} = log( \cos(x) ) \frac{dy}{dx} - y \tan(x) \\ \frac{dy}{dx} = y log( \cos(x) ) \frac{dy}{dx} - {y}^{2} \tan(x) \\ \frac{dy}{dx} - y log( \cos(x) ) \frac{dy}{dx}= - {y}^{2} \tan(x)\\ (1 - y log( \cos(x) )) \frac{dy}{dx} = - {y}^{2} \tan(x) \\ \boxed{\frac{dy}{dx} = \frac{ - {y}^{2} \tan(x) }{(1 -y log( \cos(x) )) } }✓$