Given ƒ(1) = 1, ƒ'(1) = 3 Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x)) 2 On differentiating both sides with respect to x we get, ${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x) Now at x = 1, ${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1) = ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1) = ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1) = 3$\times$3$\times$3 + 2$\times$3 = 33