If y=Sinx^Sinx^Sinx.......infinity then prove that dy/dx=2ycotx/1-ylog(sinx).?

Answer 1

interesting function

If we drop the first level, the remaining part is still equal to (sinx)^(sinx)^sinx ...
and indistinguishable from the original function
So we say
y = (sinx)^y
take ln of both sides
ln y = ln ((sinx)^y)
ln y = y ln(sinx)

(dy/dx) / y = y)(cosx/sinx) + (dy/dx)ln(sinx)
multiply both sides by y
dy/dx = y^2 cotx + y(dy/dx)ln(sinx)
dy/dx - y(dy/dx)ln(sinx) = y^2 cotx
dy/dx( 1 - y ln(sinx) ) = y^2 cots

dy/dx = y^2 cotx / (1 - y ln(sinx) )

I think you have a typo in your answer, it should have been y^2cotx... instead of 2ycotx....