Question

3. (7 points) Consider the function sin f (x, y) = { if (x, y) + (0,0) if (x, y) = (0,0) (a) Prove that f is differentiable at (0,0). (b) Prove that f is not C1 at (0,0). (Hint for part (a): Begin by showing that fx(0,0) and fy(0,0) exist and find their val- ues, and thereby determine Jf(0,0).)

Answer 1

@ fca, y) = 3 (2248) sin (alaya) if (y) +(0,0) ( o if (,4).=(0,0) To prove: f is differentiable at (0:0) we first compute fx (0,0) and fy (0,0). ..lim 5(6th, o) -f(0;o) - Lim feno) -f(0,0) :: nto - h a to ... hi ... :=lim . ? sin (12) - ö Chto.ch = lim insin (12) = 0 (since lim hzo and sin (+2) is bounded in a certain neighbourhood of (0,0) hyo

. fx(0,0) = 0 Now lim f(0, otk)-f(0,0) . Kto the most fro,k)=f(0,0) Kyok ksin (*) - -:om K >0 . to be of (0,0)) Thin k Sin (ka) . . : 0 (since lim kao and Sin (22) . is bounded in a neighbourhood i fy Co,0)=0. Now we use the result that: fir differentiable at a point (wy) if Af-df to as Chok) (0,0) where of = f(a+h, yuk) -f(x,y) and df = fxdxt fydy, the total differential off. Now. Af = f(6+1, 0+k) - f.co.o). . fchok) -frono op (4+x2) Sin Cuttr2) -0. : :*) sin ( wakid) dfah fat kfy = 6.0+k.O=0 . 4f-df (573) sin (närzt.) Jh? Aka w hitkz

nos y smo Esta (since Jsin (what?) si) : LE whenever o< n2+k?<< take sexo Therefore in the sin (not) olce where for each & 7 sto such that O< +k? <s.. le oc(-o)?+ (ko)?<82 na cesti na po sinut to Therefore af-df. Th242 so as Chik) +000). Hence for differentiable at (0;o) (b) . Now we compute the partial derivatives of of at the points (2,4) (0,0).. f(x,y) = (az+y) sin (aszyn), Cary) +0,0) :- 4,457)=(22+ya) com Catons to cover + sin (ang) * 2x . = - 2 by cos (atryz) + 2x sin ( 2 + y2) : Now Limano la ya} does not exists. / as along the line 250 lim.co,0) {-22 y 3 = o) I and along the line ar!, ali on 1 2 3 4 = two dimit is different tim to tyr Therefore

so lim fx (npy) does not exists, Caryba(0,0) -20 does not exist as . lim Ca,y)=1(0,0) 22ty? Hast f is a c function at (0,0) lire f is continuously differentiable at (0,0) in if partial derivatives fx(x; 4); fy (ay) at the point (2,4) + (0,0) tends to fx (0,0). . and fy(0,0) respectively as (219) (0:0) (2949)+(010) * As we see above that lim : fx(my) does not exists, the above condition is not satisfied by f so if is not el at (0,0).