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lect12_5

# lect12_5 - Chain Rule(12.5 1 Chain Rule Let z f!x y and x...

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Chain Rule - (12.5) 1. Chain Rule Let z ! f ! x , y " and x ! x ! t " and y ! y ! t " . So implicitly z is a function of t , z ! g ! t " . What is dz dt ? lim h " 0 g ! t # h " ! g ! t " h ! lim h " 0 f ! x ! t # h " , y ! t # h "" ! f ! x ! t " , y ! t "" h ! lim h " 0 f ! x ! t # h " , y ! t # h "" ! z ! x ! t # h " , y ! t "" # f ! x ! t # h " , y ! t "" ! z ! x ! t " , y ! t "" h ! lim h " 0 f ! x ! t # h " , y ! t # h "" ! f ! x ! t # h " , y ! t "" h # lim h " 0 f ! x ! t # h " , y ! t "" ! f ! x ! t " , y ! t "" h By the Mean Value Theorem, we know there exist x 1 between x ! t # h " and x ! t " ; and y 1 between y ! t # h " and y ! t " such that f ! x ! t # h " , y ! t # h "" ! f ! x ! t # h " , y " ! f y ! x ! t # h " , y 1 "! y ! t # h " ! y ! t "" f ! x ! t # h " , y ! t "" ! f ! x ! t " , y ! t "" ! f x x 1 , y ! t " ! x ! t # h " ! x ! t "" So lim h " 0 g ! t # h " ! g ! t " h ! lim h " 0 f y ! x ! t # h " , y 1 " y ! t # h " ! y ! t " h # lim h " 0 f x x 1 , y ! t " x ! t # h " ! x ! t " h as h " 0, y 1 " y ! t " so f y ! x ! t # h " , y 1 " " f y ! x ! t " , y ! t "" as h " 0, x 1 " x ! t " so f x x 1 , y ! t " " f x ! x ! t " , y ! t "" as h " 0, y ! t # h " ! y ! t " h " dy dt ! y \$ ! t " as h " 0, x ! t # h " ! x ! t " h " dx dt ! x \$ ! t " lim h " 0 g ! t # h " ! g ! t " h ! f x dx dt # f y dy dt When z ! f ! x , y " and x ! s , t " y ! s , t " , z is a function of s and t . In a same way, we can derive f s ! f x x s # f y y s , f t ! f x x t # f y y t In the same way, we can derive the second partial derivatives w.r.t. s and t : Note that both f x and f y

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lect12_5 - Chain Rule(12.5 1 Chain Rule Let z f!x y and x...

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