If y=[(tanx)tanx]tanx, then dydx at x=π/4 is
Consider the following differential eq.
y=((tanx)tanx)tanx&x=π4
Taking log both side, we get.
logy=tan2xlog(tanx)
Differentiate both side w.r.t x, we get.
1ydydx=2tanxsec2xlog(tanx)+tan2x1tanxsec2x
1ydydx=((tanx)tanx)tanx(2tanπ4sec2π4log(tanπ4)+tanπ4sec2π4)
dydx=0+2
=2
Hence, this is the required answer.