Конспект лекций по математическому анализу. Шерстнев А.Н. - 214 стр.

   2.  pUSTX '; : [a; b] ! R NEPRERYWNY, DIFFERENCIRUEMY NA (a; b);
'(x)  (x). fUNKCIQ f ZADANA I NEPRERYWNA NA MNOVESTWE = f(x; y) j
a  x  b; '(x)  y  (x)g, A @f
                               @x OPREDELENA I NEPRERYWNA W TO^KAH
(x; y) 2 TAKIH, ^TO a < x < b. tOGDA
           Z
          d (x)f (x; y) dy = f (x; (x)) 0(x) , f (x; '(x))'0(x)
         dx '(x)             Z (x) @f
                           +          (x; y) dy (a < x < b):
                              '(x) @x
Z pUSTX fe | PRODOLVENIE f , RASSMOTRENNOE W 133.3. pOLAGAQ F (x) =
    (x)
        f (x; y) dy (a < x < b), IMEEM DLQ MALYH h
  '(x)
                                Z                           Z
 1 [F (x + h) , F (x)] = 1 [ (x+h)f (x + h; y) dy , (x)f (x; y) dy]
 h                       h Z'(x+h)                          Z'(x)(x+h)
                             '(x)
                       = h1            fe(x + h; y) dy + h1            fe(x + h; y) dy
                            '
                         Z (x)( x +h )                         ( x )
                                1 e                  e
                       +        h [f (x + h; y) , f (x; y)] dy:
                               '(x)

iSPOLXZUQ FORMULU lAGRANVA I NEPRERYWNOSTX @f             @x , IMEEM
              Z (x) 1                                 Z
          lim          [ e(x + h; y) , fe(x; y)] dy = (x) @f (x; y) dy:
                         f
          h!0 '(x) h                                    '(x) @x
s U^ETOM TEOREMY O SREDNEM (SM. 50.4) IMEEM PRI h ! 0:
     Z
   1 (x+h)fe(x + h; y) dy
   h (x)
          = fe(x + h; (x) + [ (x + h) , (x)]) 
 (x + hh) , (x)
          ! f (x; (x)) 0(x).
                      Z '(x)
aNALOGI^NO, hlim    1
                !0 h '(x+h)
                              fe(x + h; y) dy = ,f (x; '(x))'0(x): >
                                    Zb            2
   3. p R I M E R. pUSTX J ( ) =                                         d J ().
                                        expf, 2x2 g dx ( > 0). nAJDEM d
                                     a
                                                            2     d expf, x22 g =
mNOVESTWO U = fj > 0g OTKRYTO W R I expf, 2x2 g; d                    2
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