James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

appendix F Proofs of Theorems A49
Referring to Figure 4, we write
1 Dz − f f sa 1 Dx, b 1 Dyd 2 f sa, b 1 Dydg 1 f f sa, b 1 Dyd 2 f sa, bdg
y
(a, b+Îy)
(a+Îx, b+Îy)
(u, b+Îy)
(a, b)
(a, √)
R
FIGURE 4
0
x
Observe that the function of a single variable
tsxd − f sx, b 1 Dyd
is defined on the interval fa, a 1 Dxg and t9sxd − f x sx, b 1 Dyd. If we apply the Mean
Value Theorem to t, we get
tsa 1 Dxd 2 tsad − t9sud Dx
where u is some number between a and a 1 Dx. In terms of f, this equation becomes
f sa 1 Dx, b 1 Dyd 2 f sa, b 1 Dyd − f x su, b 1 Dyd Dx
This gives us an expression for the first part of the right side of Equation 1. For the
second part we let hsyd − f sa, yd. Then h is a function of a single variable defined on
the interval fb, b 1 Dyg and h9syd − f y sa, yd. A second application of the Mean Value
Theorem then gives
hsb 1 Dyd 2 hsbd − h9svd Dy
where v is some number between b and b 1 Dy. In terms of f, this becomes
f sa, b 1 Dyd 2 f sa, bd − f y sa, vd Dy
We now substitute these expressions into Equation 1 and obtain
Dz − f x su, b 1 Dyd Dx 1 f y sa, vd Dy
− f x sa, bd Dx 1 f f x su, b 1 Dyd 2 f x sa, bdg Dx 1 f y sa, bd Dy
1 f f y sa, vd 2 f y sa, bdg Dy
− f x sa, bd Dx 1 f y sa, bd Dy 1 « 1 Dx 1 « 2 Dy
where
« 1 − f x su, b 1 Dyd 2 f x sa, bd
« 2 − f y sa, vd 2 f y sa, bd
Since su, b 1 Dyd l sa, bd and sa, vd l sa, bd as sDx, Dyd l s0, 0d and since f x and f y
are continuous at sa, bd, we see that « 1 l 0 and « 2 l 0 as sDx, Dyd l s0, 0d.
Therefore f is differentiable at sa, bd.
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