The correct option is
D 2ydydxGiven
y=tanx∴dydx=sec2x.......(1)
d2ydx2=2secx.(secx.tanx)=2sec2xtanx .....(2)
substituting the value of sec2x from (1) in (2) we get,
⇒d2ydx2=2(dydx)tanx
Also we have y=tanx, substituting we get
⇒d2ydx2=2ydydx
Hence, the answer is 2ydydx.