22.2. THE DERIVATIVE 465
It follows easily the last term at the end is o(h,k, l) and so the derivative of this function isthe linear transformation coming from multiplication by the matrix
((z2 + y
),x,2zx
)and
so this is the derivative. It follows from this and the description of the derivative in termsof partial derivatives that
∂ f∂x
(x,y,z) = z2 + y,∂ f∂y
(x,y,z) = x,∂ f∂ z
(x,y,z) = 2xz.
Of course you could compute these partial derivatives directly.Given a function of many variables, how can you tell if it is differentiable? In other
words, when you make the linear approximation, how can you tell easily that what is leftover is o(v). Sometimes you have to go directly to the definition and verify it is differen-tiable from the definition. For example, here is an interesting example of a function of onevariable.
Example 22.2.9 Let f (x) ={
x2 sin( 1
x
)if x ̸= 0
0 if x = 0. Find D f (0).
f (h)− f (0) = 0h+h2 sin(
1h
)= o(h) ,
and so D f (0) = 0. If you find the derivative for x ̸= 0, it is totally useless information ifwhat you want is D f (0). This is because the derivative turns out to be discontinuous. Tryit. Find the derivative for x ̸= 0 and try to obtain D f (0) from it. You see, in this exampleyou had to revert to the definition to find the derivative.
It isn’t really too hard to use the definition even for more ordinary examples.
Example 22.2.10 Let f (x,y) =(
x2y+ y2
y3x
). Find Df (1,2).
First of all, note that the thing you are after is a 2×2 matrix.
f (1,2) =(
68
).
Then
f (1+h1,2+h2)−f (1,2) =(
(1+h1)2 (2+h2)+(2+h2)
2
(2+h2)3 (1+h1)
)−(
68
)after some simplification,
=
(4 58 12
)(h1h2
)+
(2h1h2 +2h2
1 +h21h2 +h2
212h1h2 +6h2
2 +6h22h1 +h3
2 +h32h1
)
=
(4 58 12
)(h1h2
)+o(h) .
Therefore, the matrix of the derivative is(
4 58 12
). You let the o(h) terms be the higher
order terms, polynomials in the components of h which have higher degree than 1.