`(dy)/dx=cosx+tanx`

dy/dx-y = cosx

(dy)/(dx)+2tanx=0

The solution of the differential equation (dy)/(dx)=e^x+cosx+x+tanx is

If y=(cosx)^((((cosx)^(((cosx)^((...oo)) prove that (dy)/(dx)=-(y^2tanx)/(1-ylogcosx)

If y=(cosx)^((((cosx)^(((cosx)^((...oo)) prove that (dy)/(dx)=-(y^2tanx)/(1-ylogcosx)

If y=cosx^(cosx^(cosx^dotoo)) , prove that (dy)/(dx)=-(y^2tanx)/((1-ylogcosx))

If y=(cosx)^(cosx)^(cosx)^^((((oo)))),p rov et h a t(dy)/(dx)=-(y^2tanx)/((1-ylogcosx)dot

"If "y=(x)^(cosx)+(sinx)^(tanx)", prove that "(dy)/(dx)=x^(cosx){(cosx)/(x)-(sinx)logx}+(sinx)^(tanx).{1+(log sinx)sec^(2)x}.

Find (dy)/(dx) if y=tanx .