22.2. THE DERIVATIVE 463

Proof: Suppose T1 is another linear transformation which works. Thus, letting t be asmall positive real number,

f (x+th) = f (x)+Tth+o(th) , f (x+th) = f (x)+T1th+o(th)

Now o(th) = o(t) and so, subtracting these yields Tth−T1th= o(t) . Divide both sidesby t to obtain Th− T1 h= o(t)

t . It follows on letting t → 0 that Th = T1h. Since h isarbitrary, this shows that T = T1. Thus the derivative is well defined. So what is the matrixof this linear transformation? From Theorem 18.1.6, this is the matrix whose ith column isTei. However, from the definition of T, letting t ̸= 0,

f (x+ tei)−f (x)

t=

1t(T (tei)+o(tei)) = T (ei)+

o(tei)

t= T (ei)+

o(t)t

Then letting t → 0, it follows that Tei =∂f∂xi

(x) . Recall from theorem 18.1.6 this showsthat the matrix of the linear transformation is as claimed.

Other notations which are often used for this matrix or the linear transformation aref ′ (x) ,J (x), and even ∂f

∂xor df

dx . Also, the above definition can now be written in the form

f (x+v) = f (x)+p

∑j=1

∂f (x)

∂x jv j +o(v)

orf (x+v)−f (x) =

(∂f(x)

∂x1· · · ∂f(x)

∂xn

)v+o(v)

Here is an example of a scalar valued nonlinear function.

Example 22.2.5 Suppose f (x,y) =√

xy. Find the approximate change in f if x goes from1 to 1.01 and y goes from 4 to 3.99.

We can do this by noting that

f (1.01,3.99)− f (1,4) ≈ fx (1,2)(.01)+ fy (1,2)(−.01)

= 1(.01)+14(−.01) = 7.5×10−3.

Of course the exact value is√(1.01)(3.99)−

√4 = 7.4610831×10−3.

Notation 22.2.6 When f is a scalar valued function of n variables, the following is oftenwritten to express the idea that a small change in f due to small changes in the variablescan be expressed in the form

d f (x) = fx1 (x)dx1 + · · ·+ fxn (x)dxn

where the small change in xi is denoted as dxi. As explained above, d f is the approximatechange in the function f . Sometimes d f is referred to as the differential of f .