Oz sin 0 Use the Chain Rule to find du when 1 . = ue 6v cos # + Vuz + 12 2 = e cos 0 O 2 . az u sin 0 au = e (6v cos 0 and Vu2+ 12 r = 6uv , 0 = Vu2+ 02. O az = e" (60 cose + u sin 0 3 . 2Vu2+ 02. O az sin 0 4 . au = e (6v cos 0 + Vuz+ 82 sin 0 5 . du = e (6v cos 0 2Vu2+ 02.Use the Chain Rule to find the partial 1 . = 4 as aw derivative for O 2 . = 6 w =x2+y2+22, x =st, y = scost, z = ssint 0 3 . aw = 5 as when s = 2, t = 0. O 4 . aw = 1 as 0 5. aw = 8 asFind the directional derivative, fv, of 0 1. fv = f(x, y) = V6x - 2y at the point (2, -3) in the direction 0 2. fv = v = i+j. O 3. fv = 0 4. fv = 5. fv =Find the directional derivative, fv, of the 0 1. fv = 5 function f(x, y) = 6+3xvy 0 2. fv = 73 15 at the point P(1, 9) in the direction of the 0 3. fv = 24 vector 5 V = (3, -4) . O 4. fv = 76 15 0 5. fv = 74 15Find the directional derivative, Dvf, of 1. Dvflp 12 f(x, y, z) = xtan O 2. Duflp = 091 # at the point P = (1, 1, 1) in the direction of the vector 3. Dvflp = TT 12 v = i+ 2j + 2k . O 4. Dyfp = 0 5. Dvflp = 091 H O 6. Dvfl p =