22.4. C1 FUNCTIONS 469
=n
∑i=1
∫ 1
0
(∂f
∂xi
(x+
n
∑k=i+1
vkek + tviei
)vi −
∂f
∂xi(x)vi
)dt
+n
∑i=1
∫ 1
0
∂f
∂xi(x)vidt =
n
∑i=1
∂f
∂xi(x)vi
+n
∑i=1
∫ 1
0
(∂f
∂xi
(x+
n
∑k=i+1
vkek + tviei
)− ∂f
∂xi(x)
)vidt =
n
∑i=1
∂f
∂xi(x)vi +o(v)
and this shows f is differentiable at x.Some explanation of the step to the last line is in order. The messy thing at the end is
o(v) because of the continuity of the partial derivatives. To see this, consider one term. ByProposition 16.2.2, ∣∣∣∣∣
∫ 1
0
(∂f
∂xi
(x+
n
∑k=i+1
vkek + tviei
)− ∂f
∂xi(x)
)vidt
∣∣∣∣∣≤ √
p∫ 1
0
∣∣∣∣∣∂f∂xi
(x+
n
∑k=i+1
vkek + tviei
)− ∂f
∂xi(x)
∣∣∣∣∣dt |v|
Thus, dividing by |v| and taking a limit as |v| → 0, this converges to 0 due to continuityof the partial derivatives of f . The messy term is thus a finite sum of o(v) terms and istherefore o(v).
Here is an example to illustrate.
Example 22.4.3 Let f (x,y) =(
x2y+ y2
y3x
). Find Df (x,y).
From Theorem 22.4.2 this function is differentiable because all possible partial deriva-tives are continuous. Thus
Df (x,y) =(
2xy x2 +2yy3 3y2x
).
In particular,
Df (1,2) =(
4 58 12
).
Here is another example.
Example 22.4.4 Let f (x1,x2,x3) =
x21x2 + x2
2x2x1 + x3
sin(x1x2x3)
. Find Df (x1,x2,x3).
All possible partial derivatives are continuous, so the function is differentiable. Thematrix for this derivative is therefore the following 3×3 matrix 2x1x2 x2
1 +2x2 0x2 x1 1
x2x3 cos(x1x2x3) x1x3 cos(x1x2x3) x1x2 cos(x1x2x3)
Example 22.4.5 Suppose f (x,y,z) = xy+ z2. Find D f (1,2,3).