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Suppose ๐‘“ = ๐‘“(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘” = ๐‘”(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘• = ๐‘•(๐‘ข. ๐‘ฃ. ๐‘ค), and ๐‘ข = ๐‘ข(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ฃ =๐‘ฃ(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ค = ๐‘ค(๐‘Ÿ, ๐‘ , ๐‘ก). Then, write the formula for the Jacobian ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก)

Question

Suppose ๐‘“ = ๐‘“(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘” = ๐‘”(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘• = ๐‘•(๐‘ข. ๐‘ฃ. ๐‘ค), and ๐‘ข = ๐‘ข(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ฃ =๐‘ฃ(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ค = ๐‘ค(๐‘Ÿ, ๐‘ , ๐‘ก). Then, write the formula for the Jacobian ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก)

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Solution 1

Given ๐‘“ = ๐‘“(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘” = ๐‘”(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘• = ๐‘•(๐‘ข. ๐‘ฃ. ๐‘ค), and ๐‘ข = ๐‘ข(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ฃ =๐‘ฃ(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ค = ๐‘ค(๐‘Ÿ, ๐‘ , ๐‘ก), the formula for the Jacobian ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) is as follows:

๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) = ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘ข,๐‘ฃ,๐‘ค) * ๐œ•(๐‘ข,๐‘ฃ,๐‘ค)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก)

This can be further expanded as:

๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) = ๐œ•๐‘“๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘Ÿ + ๐œ•๐‘“๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘Ÿ + ๐œ•๐‘“๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘Ÿ + ๐œ•๐‘”๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘  + ๐œ•๐‘”๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘  + ๐œ•๐‘”๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘  + ๐œ•๐‘•๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘ก + ๐œ•๐‘•๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘ก + ๐œ•๐‘•๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘ก

This formula represents the partial derivatives of ๐‘“, ๐‘”, and ๐‘• with respect to ๐‘Ÿ, ๐‘ , and ๐‘ก, respectively.

This problem has been solved

Solution 2

Given ๐‘“ = ๐‘“(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘” = ๐‘”(๐‘ข, ๐‘ฃ, ๐‘ค), ๐‘• = ๐‘•(๐‘ข. ๐‘ฃ. ๐‘ค), and ๐‘ข = ๐‘ข(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ฃ =๐‘ฃ(๐‘Ÿ, ๐‘ , ๐‘ก), ๐‘ค = ๐‘ค(๐‘Ÿ, ๐‘ , ๐‘ก), the formula for the Jacobian ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) is as follows:

๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) = ๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘ข,๐‘ฃ,๐‘ค) * ๐œ•(๐‘ข,๐‘ฃ,๐‘ค)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก)

This can be further expanded as:

๐œ•(๐‘“,๐‘”,๐‘•)๐œ•(๐‘Ÿ,๐‘ ,๐‘ก) = ๐œ•๐‘“๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘Ÿ + ๐œ•๐‘“๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘Ÿ + ๐œ•๐‘“๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘Ÿ + ๐œ•๐‘”๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘  + ๐œ•๐‘”๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘  + ๐œ•๐‘”๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘  + ๐œ•๐‘•๐œ•๐‘ข * ๐œ•๐‘ข๐œ•๐‘ก + ๐œ•๐‘•๐œ•๐‘ฃ * ๐œ•๐‘ฃ๐œ•๐‘ก + ๐œ•๐‘•๐œ•๐‘ค * ๐œ•๐‘ค๐œ•๐‘ก

where ๐œ•๐‘“๐œ•๐‘ข, ๐œ•๐‘“๐œ•๐‘ฃ, ๐œ•๐‘“๐œ•๐‘ค, ๐œ•๐‘”๐œ•๐‘ข, ๐œ•๐‘”๐œ•๐‘ฃ, ๐œ•๐‘”๐œ•๐‘ค, ๐œ•๐‘•๐œ•๐‘ข, ๐œ•๐‘•๐œ•๐‘ฃ, ๐œ•๐‘•๐œ•๐‘ค represent the partial derivatives of ๐‘“, ๐‘”, ๐‘• with respect to ๐‘ข, ๐‘ฃ, ๐‘ค respectively, and ๐œ•๐‘ข๐œ•๐‘Ÿ, ๐œ•๐‘ข๐œ•๐‘ , ๐œ•๐‘ข๐œ•๐‘ก, ๐œ•๐‘ฃ๐œ•๐‘Ÿ, ๐œ•๐‘ฃ๐œ•๐‘ , ๐œ•๐‘ฃ๐œ•๐‘ก, ๐œ•๐‘ค๐œ•๐‘Ÿ, ๐œ•๐‘ค๐œ•๐‘ , ๐œ•๐‘ค๐œ•๐‘ก represent the partial derivatives of ๐‘ข, ๐‘ฃ, ๐‘ค with respect to ๐‘Ÿ, ๐‘ , ๐‘ก respectively.

This problem has been solved

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