ბიზნესის მათემატიკა

63 rogorc vxedavT, sruli diferenciali warmoadgens Δx da Δy argumentebis wrfiv funqcias. davadginoT A da B ricxvebis mniSvnelobani. amisaTvis (1) tolobis orive mxare gavyoT jer Δx -ze, Semdeg miRebuli tolobis orive mxares gadavideT zRvarze, rodesac Δx → 0 , miviRebT: f x′ ( x0 , y 0 ) = A . analogiurad, (1) tolobis orive mxare gavyoT jer Δy -ze, Semdeg miRebuli tolobis orive mxares gadavideT zRvarze, rodesac Δy → 0 , miviRebT: f y′ ( x 0 , y 0 ) = B . am dazustebebis Semdeg, z = f ( x, y ) funqcis sruli diferenciali Δx da N 0 ( x0 , y 0 ) wertilSi, SeiZleba warmovadginoT, rogorc Δy argumentebis konkretuli wrfivi funqcia

dz

x = x0 y = y0

′ = f x′ ( x0 , y 0 )Δx + f y′ ( x 0 , y 0 )Δy .

x da y cvladebis diferencialebi dx, dy Tu CavTvliT, rom rac am cvladebis nazrdebi Δx da Δy , SeiZleba davweroT dz

x = x0 y = y0

igivea,

′ = f x′ ( x0 , y 0 )dx + f y′ ( x0 , y 0 )dy .

Tu funqcia diferencirebadia D aris nebismier wertilSi, maSin mas uwodeben diferencirebads D areze. magaliTi 1. vipovoT π z = ax 2 + sin 3 ( x + 2 y 2 ) funqciis sruli diferenciali N (π , ) wertilSi. 2 amoxsna:

dz

x =π y=

π 2

= (2ax + 3 sin 2 ( x + 2 y 2 ))′x

x =π y=

π 0

dx + (6 y sin 2 ( x + 2 y 2 ))′y

x =π y=

dy = (2aπ + 3)dx + 3πdy . π 2

sruli diferenciali saSualebas gvaZlevs gamovTvaloT funqciis miaxloebiTi mniSvneloba, marTlac f ( x 0 + Δx, y 0 + Δy ) − f ( x0 , y 0 ) = f x′ ( x0 , y 0 )Δx + f x′ ( x0 , y 0 )Δy + α (Δx)Δx + β (Δy )Δy . Δx da Δy nazrdebis sakmaod mcire mniSvnelobebisTvis gveqneba miaxloebiTi toloba: f ( x0 + Δx, y 0 + Δy ) − f ( x0 , y 0 ) ≈ f x′ ( x0 , y 0 )Δx + f x′ ( x0 , y 0 )Δy . am miaxloebiTi tolobidan SegviZlia davweroT: f ( x0 + Δx, y 0 + Δy ) ≈ f ( x0 , y 0 ) + f x′ ( x0 , y 0 )Δx + f x′ ( x0 , y 0 )Δy .

Tu x0 ≤ x ≤ x0 + Δx da y 0 ≤ y ≤ y 0 + Δy , sakmaod mcire Δx da Δy nazrdebis mniSvnelo- bebisTvis gveqneba: f ( x0 , y ) ≈ f ( x 0 , y 0 ) + f x′ ( x 0 , y 0 )Δx + f x′ ( x0 , y 0 )Δy .

(2) magaliTi 2. gamovTvaloT z = e xy funqciis mniSvneloba, rodesac x = 0,35, y = −0,5 . amoxsna: (2) formuliT: z = e 0,35⋅( −0,5) = e 0⋅0 + (−0,5)e 0 0,35 + 0,35e 0 (−0,5) = 1 − 0,35 = 0,65 .