calc15 - (x 1 , x 2 ) = (3,1). We get _ @z 2 @x 1 _ x2 = y...

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Find _ @z 2 @x 1 _ x2 (3,−1) and simplify your answer. Solution. To get _ @z 2 @x 1 _ x2 we must multiply the second row of the first matrix above by the first column of the second matrix, and then we need to evaluate at
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Unformatted text preview: (x 1 , x 2 ) = (3,1). We get _ @z 2 @x 1 _ x2 = y 2 2 (_x 2 sin(_x 1 )) + (2y 1 y 2 + y 2 3 )(_x 2 cos(_x 1 )) + 4y 2 y 3 ( 2x 1 x 21 + x 22 )....
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