464 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

Let f : U → Rq where U is an open subset of Rp and f is differentiable. It was justshown that

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

∂f(x)∂x1

· · · ∂f(x)∂xp

)v+o(v) .

Taking the ith coordinate of the above equation yields

fi (x+v) = fi (x)+p

∑j=1

∂ fi (x)

∂x jv j +o(v) ,

and it follows that the term with a sum is nothing more than the ith component of J (x)vwhere J (x) is the q× p matrix

∂ f1∂x1

∂ f1∂x2

· · · ∂ f1∂xp

∂ f2∂x1

∂ f2∂x2

· · · ∂ f2∂xp

......

. . ....

∂ fq∂x1

∂ fq∂x2

· · · ∂ fq∂xp

 .

Thusf (x+v) = f (x)+ J (x)v+o(v) , (22.6)

and to reiterate, the linear transformation which results by multiplication by this q × pmatrix is known as the derivative.

Sometimes x,y,z is written instead of x1,x2, and x3. This is to save on notation and iseasier to write and to look at although it lacks generality. When this is done it is understoodthat x = x1,y = x2, and z = x3. Thus the derivative is the linear transformation determinedby  f1x f1y f1z

f2x f2y f2zf3x f3y f3z

 .

Example 22.2.7 Let A be a constant m×n matrix and consider f (x) = Ax. Find Df (x)if it exists.

f (x+h)−f (x) = A(x+h)−A(x) = Ah= Ah+o(h) .

In fact in this case, o(h) = 0. Therefore, Df (x) = A. Note that this looks the same as thecase in one variable, f (x) = ax.

Example 22.2.8 Let f (x,y,z) = xy+ z2x. Find D f (x,y,z).

Consider f (x+h,y+ k,z+ l)− f (x,y,z). This is something which is easily computedfrom the definition of the function. It equals

(x+h)(y+ k)+(z+ l)2 (x+h)−(xy+ z2x

)Multiply everything together and collect the terms. This yields(

z2 + y)

h+ xk+2zxl +(hk++2zlh+ l2x+ l2h

)