<Dd> D (a, b) = f x x (a, b) f y y (a, b) − (f x y (a, b)) 2 = 2 b (b + 1) ⋅ 2 a (a + 3 b + 1) − (2 a + 2 b + 4 a b + 3 b 2) 2 . (\ displaystyle (\ begin (aligned) D (a, b) & = f_ (xx) (a, b) f_ (yy) (a, b) - \ left (f_ (xy) (a, b) \ right) ^ (2) \ \ & = 2b (b + 1) \ cdot 2a (a + 3b + 1) - (2a + 2b + 4ab + 3b ^ (2)) ^ (2). \ end (aligned))) </Dd> <P> Now we plug in all the different critical values we found to label them; we have </P> <Dl> <Dd> D (0, 0) = 0; D (0, − 1) = − 1; D (1, − 1) = − 1; D (3 8, − 3 4) = 27 128 . (\ displaystyle D (0, 0) = 0; ~ ~ D (0, - 1) = - 1; ~ ~ D (1, - 1) = - 1; ~ ~ D \ left ((\ frac (3) (8)), - (\ frac (3) (4)) \ right) = (\ frac (27) (128)).) </Dd> </Dl> <Dd> D (0, 0) = 0; D (0, − 1) = − 1; D (1, − 1) = − 1; D (3 8, − 3 4) = 27 128 . (\ displaystyle D (0, 0) = 0; ~ ~ D (0, - 1) = - 1; ~ ~ D (1, - 1) = - 1; ~ ~ D \ left ((\ frac (3) (8)), - (\ frac (3) (4)) \ right) = (\ frac (27) (128)).) </Dd>

Second derivative test for functions of two variables