2 Frank Jone s I. Th e domai n i s one-dimensiona l This i s on e o f th e mos t basi c situation s i n al l o f calculus . I f th e function / attain s a maximu m o r a minimu m valu e a t a poin t x i n the interio r o f its domain, the n o f course the derivative is zero at tha t point: /'(*) = 0. As you know, this equation ma y be satisfied eve n if / doe s not attai n an extreme value at x. A point x which satisfies this equation is called a critical point for th e functio n / . II. Th e domai n i s n-dimensiona l We agai n suppos e tha t th e functio n / attain s a n extrem e valu e a t a point x i n M.n an d i n the interio r o f its domain. I n this case we intro- duce a vecto r h (als o i n R n ) whic h serve s a s a directio n fo r analysi s of th e function . The n th e poin t x + th, wher e £ is a rea l number , represents a poin t o n th e straigh t lin e throug h x i n th e directio n h. Thus th e functio n o f t give n b y th e expressio n f(x + th) attain s a n extreme valu e a t t = 0 . Therefor e th e situatio n give n i n part I shows that |(/(* + *fc))U=o. The quantit y o n th e lef t sid e o f thi s equatio n i s ofte n calle d th e di- rectional derivative o f / i n th e directio n h an d i s denote d D h f(x). The chai n rul e o f multivariabl e calculu s enable s u s t o rewrit e thi s equation i n th e for m n (1) ^df/dxjWh^O. Since (1 ) mus t hol d fo r ever y choic e o f th e directio n /i , w e ma y us e the n coordinat e vector s h = (0,... , 0,1,0,..., 0) t o conclud e tha t (1) i s equivalent t o (2) df/dx 3 (x) = 0 fo r al l 1 j n.
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