f(x,y)=x2ey z=x4+x2y,x=s+2t−u,y=stu2 ∂s∂z(4,2,1),∂t∂z(4,2,1),∂u∂z(4,2,1) - Answer Happy

f(x,y)=x2ey z=x4+x2y,x=s+2t−u,y=stu2 ∂s∂z(4,2,1),∂t∂z(4,2,1),∂u∂z(4,2,1)

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answerhappygod: Site Admin; Posts: 899604; Joined: Mon Aug 02, 2021 8:13 am

f(x,y)=x2ey z=x4+x2y,x=s+2t−u,y=stu2 ∂s∂z(4,2,1),∂t∂z(4,2,1),∂u∂z(4,2,1)

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Post by answerhappygod » Thu Jul 14, 2022 4:46 pm

: F X Y X2ey Z X4 X2y X S 2t U Y Stu2 S Z 4 2 1 T Z 4 2 1 U Z 4 2 1 1 (125.63 KiB) Viewed 35 times

explain why function is differentially at given point. Then find linearization L(x,y) of function at that point

: F X Y X2ey Z X4 X2y X S 2t U Y Stu2 S Z 4 2 1 T Z 4 2 1 U Z 4 2 1 2 (197.63 KiB) Viewed 35 times

use chain rule to find indicated partial derivatives
f(x,y)=x2ey
z=x4+x2y,x=s+2t−u,y=stu2 ∂s∂z(4,2,1),∂t∂z(4,2,1),∂u∂z(4,2,1)

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