If y=loglogx, then eydydx=
1x
1xlogx
1logx
ey
Find the dydx:
Given that, y=loglogx
Differentiate y with respect to x using chain rule we get:
dydx=1logx·1x
And y=loglogx⇒ey=logx
By substituting value we have:
dydx=1ey·1xeydydx=1x
Hence, option A is correct.