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15.6.Ex14-15 - = 1 x = e cos 1 = cos 1 y = e sin 1 = sin 1...

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S E C T I O N 15.6 The Chain Rule (ET Section 14.6) 711 14. g s at s = 4, where g ( x , y ) = x 2 y 2 , x = s 2 + 1, y = 1 2 s . SOLUTION We find the primary derivatives of g ( x , y ) = x 2 y 2 : g x = 2 x , g y = − 2 y Applying the Chain Rule gives g s = g x · dx ds + g y · dy ds = 2 x dx ds 2 y dy ds (1) We compute dx ds and dy ds : dx ds = 2 s , dy ds = − 2 Substituting in (1) we obtain g s = 4 xs + 4 y (2) We now determine ( x , y ) for s = 4: x = 4 2 + 1 = 17 , y = 1 2 · 4 = − 7 Substituting ( x , y ) = ( 17 , 7 ) and s = 4 in (2) gives the following derivative: g s s = 4 = 4 · 17 · 4 4 · 7 = 244 15. g u at ( u , v) = ( 0 , 1 ) , where g ( x , y ) = x 2 y 2 , x = e u cos v , y = e u sin v . SOLUTION The primary derivatives of g ( x , y ) = x 2 y 2 are g x = 2 x , g y = − 2 y By the Chain Rule we have g u = g x · x u + g y · y u = 2 x x u 2 y y u (1) We find x u and y u : x u = e u cos v, y u = e u sin v Substituting in (1) gives g u = 2 xe u cos v 2 ye u sin v = 2 e u ( x cos v y sin v) (2) We determine ( x , y ) for
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Unformatted text preview: = ( , 1 ) : x = e cos 1 = cos 1 , y = e sin 1 = sin 1 ±inally, we substitute ( u , v) = ( , 1 ) and ( x , y ) = ( cos 1 , sin 1 ) in (2) and use the identity cos 2 α − sin 2 = cos 2 , to obtain the Following derivative: ∂ g ∂ u ¯ ¯ ¯ ¯ ( u ,v) = ( , 1 ) = 2 e ³ cos 2 1 − sin 2 1 ´ = 2 · cos 2 · 1 = 2 cos 2 16. ∂ h ∂ q at ( q , r ) = ( 3 , 2 ) , where h ( u , v) = ue v , u = q 3 , v = qr 2 . SOLUTION We frst fnd the primary derivatives oF h ( u , v) = ue v : ∂ h ∂ u = e v , ∂ h ∂v = ue v...
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