Exercise 9.11

Exercise 11: If f and g are differentiable real functions in n , prove that

( fg ) = fg + gf

and that ( 1 f ) = f 2 f wherever f 0 .

Answers

(analambanomenos)

( fg ) = i = 1 n ( D i ( fg ) ) ě i = i = 1 n ( f ( D i g ) + g ( D i f ) ) ě i = f i = 1 n ( D i g ) ě i + g i = 1 n ( D i f ) ě i = fg + gf 0 = ( 1 ) = ( f f 1 ) = f ( f 1 ) + f 1 f f ( f 1 ) = f 1 f ( f 1 ) = f 2 f
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2023-08-07 00:00
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