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Unformatted text preview: Chain rule and total diﬀerentials
1. Find the total diﬀerential of w = z e(x+y) at (0, 0, 1).
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 = z e(x+y) , wy = z e(x+y) , wz = e(x+y) Substituting in the point (0, 0, 1) we get: wx (0, 0, 1) = 1,
dw = dx + dy + dz. wy (0, 0, 1) = 1, wz (0, 0, 1) = 1. dw
and evaluate it when
2. Suppose w = z e(x+y) and x = t, y = t2 , z = t3 . Compute
t = 2. Answer: We use the chain rule: dw
dt = ∂ w dx ∂ w dy ∂ w dz
∂ x dt
∂ y dt
∂ z dt = (z e(x+y) )(1) + (z e(x+y) )(2t) + (e(x+y) )(3t2 ).
At t = 2 we have x = 2, y = 4, z = 8. Thus,
� = 8e6 + 8e6 (4) + e6 (12) = 52e6 . dt �2 MIT OpenCourseWare
http://ocw.mit.edu 18.02SC Multivariable Calculus
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This document was uploaded on 10/26/2011 for the course FKP bmfp at UTEM Chile.
- Fall '11