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Home Partial Differentiation Total Differentials and Tangent Planes Examples Example 1 (Continued)
 
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Example 1 (Continued)

The product function z = xy is smooth for all (x, y). Express Δz in the form

Δz = dz + ε1 Δx + ε2 Δy.

We have

Δz = y Δx + x Δy + Δx Δy,

dz = y Δx + x Δy.

Thus

Δz = dz + Δx · Δy.

The problem has more than one correct answer. One answer is et = 0 and £2 = Δx, so that

Δz = dz + 0 · Δx + Δx · Δy = dz + ε1 Δx + ε1 Δy.

Another answer is ε1 = Δy and ε2 = 0, so that

Δz = dz + Δy · Δx + 0 · Δy - dz + ε1 Δx + ε2 Δy.
Home Partial Differentiation Total Differentials and Tangent Planes Examples Example 1 (Continued)

Last Update: 2010-11-25