Chain Rule and Total Diﬀerentials 1. Find the total diﬀerential of w = x3 yz + xy + z + 3 at (1, 2, 3). Answer: The total diﬀerential at the point (x0 , y0 , z0 ) is dw = wx (x0 , y0 , z0 ) dx + wy (x0 , y0 , z0 ) dy + wz (x0 , y0 , z0 ) dz. In our case, wx = 3x2 yz + y, wy = x3 z + x, wz = x3 y + 1. Substituting in the point (1, 2, 3) we get: wx (1, 2, 3) = 20, wy (1, 2, 3) = 4, wz (1, 2, 3) = 3. Thus, dw = 20 dx + 4 dy + 3 dz. 2. Suppose w = x3 yz + xy + z + 3 and x = 3 cos t, y = 3 sin t, z = 2t. dw and evaluate it at t = π/2. dt Answer: We do not substitute for x, y, z before diﬀerentiating, so we can practice the chain rule. dw ∂w dx ∂w dy ∂w dz = + + dt ∂x dt ∂y dt ∂z dt Compute = (3x2 yz + y)(−3 sin t) + (x3 z + x)(3 cos t) + (x3 y + 1)(2). At t = π/2 we have x = 0, y = 3, z = π, sin π/2 = 1, cos π/2 = 0. Thus, dw = 3(−3) + 3(0) + (1)2 = −7. dt π/2 3. Show how the tangent approximation formula leads to the chain rule that was used in the previous problem. Answer: The approximation formula is Δw ≈ ∂f ∂x Δx + o ∂f ∂y Δy + o ∂f ∂z Δz. o If x, y, z are functions of time then dividing the approximation formula by Δt gives Δw ∂f ≈ Δt ∂x o Δx ∂f + Δt ∂y o Δy ∂f + Δt ∂z o Δz . Δt In the limit as Δt → 0 we get the chain rule. Note: we use the regular ’d’ for the derivative dw dt because in the chain of computations t → x, y, z → w the dependent variable w is ultimately a function of exactly one independent variable t. Thus, the derivative with respect to t is not a partial derivative.