From (i), we have
y=tanxy
Taking log on both sides
logy=ylogtanx
Differentiating on both sides, we get
1ydydx=ysec2xtanx+logtanxdydx
Put x=π4 and y=1
⇒11dydx=(1)sec2π4tanπ4+logtanπ4dydx
⇒dydx=21+log1dydx
⇒dydx=2+0dydx
⇒dydx=2
So, correct answer is 2.