Let yyy......∞=loge(x+loge(x+.....∞))=u
Then , u=yu and u=loge(x+u)
Taking, u=yu
logu=ulogy
1u=u(1y)dydu+logy
1u−logy=uydydu
yu2−yulogy=dydu
⇒yu2(1−ulogy)=dydu
⇒u(1/u)−2(1−logu)=dydu.....(1)
Now, taking u=loge(x+u)
1=1(x+u)(dxdu+1)
dxdu=x+u−1
⇒dxdu=eu−1...(2)
Since y=u1/u, when y=√2, u=2 or u=4
Also since x=eu−u, when x=e2−2, u=2.
The value of u which satisfies both is u=2
From (1) and (2) dydx at the given values of 'x' and 'y' is
2−3/2(1−log2)e2−1
=log(e/2)2√2(e2−1)