Find dy/dx of these functions: 1. y=sin-1(1/3x) dy/dx=1/(√1-(1/3x)2 (1/3) dy/dx=1/(3√1-(1/9x2) However the book answer is 1/√9-x2). How did they get this? 2. y=sin-1(1/x) dy/dx=1/√1-(1/x)2 (-1/x2) If i tried to simplified it by multiply it by x^2/x^2then dy/dx= -x^2/x4√x2-1 which equal -1/x2√x2-1 However the book answer is -1/absx√x2-1. Can you explain to me how they get this? 3. find dy/dx by implicit differentiation sin-1(xy)=cos-1(x-y) so d/dx[sin^-1(xy)=cos^-1(x-y)] 1/√1-(xy)2 (y+x) (dy/dx)=-1/√1-(x-y)2 (0) (y+x)/√1-(xy)^2 (dy/dx)=0 dy/dx= (- y+x)/(√1-(xy)2) The formula i used for all these problems are d/dx[sin-1u]=1/√1-u2 (du/dx) d/dx[cos-1u]=-1/√1-u2 (du/dx) ~Thank you so much in advance. I really need help before my big test!