We wish to develop a multivariable First Derivative Test, but, first, let’s recall whatthe single-variable First Derivative Test says.Suppose that we havef=f(x) which is continuous on the open interval (a, b), andthatpis a point in (a, b) such thatf0(x)>0 for allxin the interval (a, p), andf0(x)<0for allxin the interval (p, b). Then, even iffis not di↵erentiable atp,fincreases onthe interval (a, p] and decreases on the interval [p, b); thus,fattains a local (actually,global, on (a, b)) maximum value atp. Iff0is first negative, then positive, there is theanalogous result for a local minimum value.How do we generalize this to a statement for multivariable functions?At first, itmay seem difficult to produce a multivariable statement that corresponds to the factthat the sign of the derivative “switches asxpasses throughp”.Recall that, for a functionf=f(x) of a single variable,dxf(v) =f0(x)·v. Thus, wecan write the condition that “x < pimpliesf0(x)>0” as “x < pimpliesdxf(p-x)>0”,and the condition that “p < ximpliesf0(x)<0” as “p < ximpliesdxf(p-x)>0”.Therefore, we can state the single-variable First Derivative Test for a local maximum as:Suppose thatfis continuous on the open interval(a, b), thatpis in(a, b), andthat, for allx6=pin(a, b),dxf(p-x)>0. Then,fattains a local maximumvalue atp.In words, what this says is thatfattains a local maximum value atpprovided that,at all pointsxnear (but unequal to)p, the instantaneous rate of change off, as youhead fromx, straight towardsp, is positive. This seems like it should also be true formultivariable functions and, in fact, it is.We give the proof of this theorem, since it is not too difficult, and it is instructiveto see how single-variable results lead to multivariable results.Note that, in particular, sincefattains an extreme value atp,pmust be a critical pointoff. However, it may be thatfis not di↵erentiable atp.Theorem 2.9.19.(First Derivative Test)Letf=f(x)be a continuous functionon an open ballUinRn, whereUis centered at a pointp.Suppose that, for allx6=pinU,dxfexists anddxf(p-x)>0(respectively,dxf(p-x)<0). Then,fattains a strict global maximum (respectively, minimum)value atp.Proof.We shall prove the maximum case; the minimum case is completely analogous.