Theorems on derivatives.

10. More about derivativesThe main result is:Theorem 10.1. Let A ⊂ Rn be an open subset and let f : A −→ Rm be a function. If the partial derivatives ∂fi ,∂xj exist and are continuous, then f is differentiable. We will need: Theorem 10.2 (Mean value theorem). Let f :[a,b] −→ R is continnuou and differentiable at every point of (a,b), then we may find c ∈ (a,b) such that f(b) − f(a)= f�(c)(b − a). Geometrically, (10.2) is clear. However it is surprisingly hard to give a complete proof. Proof of (10.1). We may assume that m = 1. We only prove this in the case when n = 2 (the general case is similar, only notationally more involved). So we have f : R2 −→ R. Suppose that P =(a,b) and let −→ı + h2ˆLetPQ = h1ˆj. P0 =(a,b) P1 =(a + h1,b) and P2 =(a + h1,b + h2)= Q. Now f(Q) − f(P )=[f(P2) − f(P1)] + [f(P1) − f(P0)]. We apply the Mean value theorem twice. We may find Q1 and Q2 such that ∂f ∂f f(P1) − f(P0)= (Q1)h1 and f(P2) − f(P1)= (Q2)h2. ∂x ∂y Here Q1 lies somewhere on the line segment P0P1 and Q2 lies on the line segment P1P2. Putting this together, we get ∂f ∂f f(Q) − f(P )= ∂x (Q1)h1 + ∂y (Q2)h2. 1 Thus (∂f (Q1) − ∂f (P ))h1 +(∂f (Q2) − ∂f (P ))h2|f(Q) − f(P ) − A · −→| = |∂x ∂y ∂y |PQ∂x �−→PQ� �−→PQ� (∂f ∂f (∂f ∂f ∂x (Q1) − ∂x (P ))h1|∂y (Q2) − (P ))h2|∂y ≤|�−→| + �−→PQ� PQ� (∂f (Q1) − ∂f (P ))h1(∂f (Q2) − ∂f (P ))h2≤|∂x h1∂x ||∂y h2∂y |+ || ||∂f ∂f ∂f ∂f = |(∂x(Q1) − ∂x(P ))| + |( ∂y (Q2) − ∂y (P ))|. Note that as Q approaches P , Q1 and Q2 both approach P as well. As the partials of f are continuous, we have PQ∂f lim |f(Q) − f(P ) − A · −→| ≤ lim(( ∂f (Q1)− ∂f (P ))+( ∂f (Q2)− (P ))) = 0. QP PQ� QP |∂x ∂x||∂y ∂y |→�−→→Therefore f is differentiable at P , with derivative A. � Example 10.3. Let f : A −→ R be given by x f(x,y)= � , x2 + y2 where A = R2 Then−{(0, 0)}.∂f =(x2 + y2)1/2 − x(2x)(1/2)(x2 + y2)−1/2 = y2 .∂x x2 + y2 (x2 + y2)3/2 Similarly ∂f xy . ∂y = −(x2 + y2)3/2 Now both partial derivatives exist and are continuous, and so f is differenttiable with derivative the gradient, �f = ( (x2 + y2 y2)3/2 , −(x2 + xyy2)3/2 )= (x2 +1 y2)3/2 (y 2 , −xy). Lemma 10.4. Let A =(aij ) be an m × n matrix. If �v ∈ Rn then �A�v�≤ K��v�, where � K =( a 2 )1/2 .ij i,j 2 Proof. Let �a1,�a2,...,�am be the rows of A. Then the entry in the ith row of A�v is �ai · �v. So, �A�v�2 =(�a1 · �v)2 +(�a2 · �v)2 + +(�an · �v)2 ··· +≤��a1�2��v�2 + ��a2�2��v�2 ··· + ��an�2��v�2 =(��a1�2 + ��a2�2 + ··· + ��an�2)��v�2 = K22 .��v�Now take square roots of both sides. � Theorem 10.5. Let f : A −→ Rm be a function, where A ⊂ Rn is open. If f is differentiable at P , then f is continuous at P . Proof. Suppose that Df(P )= A. Then lim f(Q) − f(P ) − A · −→=0. PQ Q→P �−→PQ This is the same as to require �f(Q) − f(P ) − A −→lim · PQ� =0. Q→P �−→PQ But if this happens, then surely lim P �f(Q) − f(P ) − A PQ� =0.−→Q→· So �f(Q) − f(P )� = �f(Q) − f(P ) − A −→PQ + A −→PQ�·· ≤�f(Q) − f(P ) − A · −→PQ� + �A · −→PQ� ≤�f(Q) − f(P ) − A · −→PQ�.PQ� + K�−→Taking the limit as Q approaches P , both terms on the RHS go to zero, so that lim P �f(Q) − f(P )� =0, Q→and f is continuous at P . � 3 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.