The correct option is C 3(d2ydx2)2
∵y=(ax+bcx+d)
or c xy+dy=ax+b
Differentiating both sides w.r.t. x, then
c{xdydx+y.1}+ddydx=a
or xdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t. x, then
or xd2ydx2+dydx+dydx+(dc)d2ydx2=0
or x+2dydx(d2ydx2)+dc=0
Again differentiating both sides w.r.t. x, then
1+(d2y.2d2ydx2−2dydx.d3ydx3)(d2ydx2)2∴2dydx.d3ydx3=3(d2ydx2)2