if y= x (x2-4a2)1/2 (x2-a2) then dy/dx is Share with your friends Share 1 Aarushi Mishra answered this y=x x2-4a212x2-a2ln y=ln x x2-4a212x2-a2ln y=ln x+ln x2-4a212+ln x2-a2 Since log am=m log aln y=ln x+12ln x2-4a2+ln x2-a2 d dxln y=d dxln x+12ln x2-4a2+ln x2-a21ydy dx=d dxln x+12d dxln x2-4a2+d dxln x2-a21ydy dx=1 x+12x2-4a2×d x2-4a2dx+1x2-a2×d x2-a2dx1ydy dx=1 x+12x2-4a2×2x+1x2-a2×2x1ydy dx=1 x+xx2-4a2+2xx2-a2dy dx=y1 x+xx2-4a2+2xx2-a2Putting value of ydy dx=x x2-4a212x2-a21 x+xx2-4a2+2xx2-a2dy dx=x2-4a212x2-a2+x2 x2-a2x2-4a212+2x2x2-4a212 -2 View Full Answer
