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                                                              ‫اﻟﻜﻔﺎءات اﻟﻤﺴﺘﮭﺪﻓﺔ‬
                                                       ‫ﺣﺴﺎب اﻟﻌﺪد اﻟﻤﺸﺘﻖ ﻟﺪاﻟﺔ ﻋﻨﺪ ﻋﺪد‬
                                                                                    ‫ﺣﻘﯿﻘﻲ‬
                                                       ‫ﺗﻌﯿﯿﻦ ﻣﻌﺎدﻟﺔ ﻣﻤﺎس ﻣﻨﺤﻦ ﻓﻲ ﻧﻘﻄﺔ‬
                                                                                       ‫ﻣﻨﮫ.‬
                                                         ‫ﺣﺴﺎب ﻣﺸﺘﻘﺎت اﻟﺪوال اﻟﻤﺮﺟﻌﯿﺔ‬
                                                           ‫ﺣﺴﺎب ﻣﺸﺘﻘﺎت اﻟﺪوال ‪، f + g‬‬
                                                                             ‫‪f‬‬
                                                          ‫، ) ‪x a f ( ax + b‬‬    ‫‪، 1 ،f ×g‬‬
                                                                            ‫‪g‬‬    ‫‪g‬‬




             ‫‪∆y‬‬            ‫)‪f ( x + h ) − f ( x‬‬
      ‫‪lim ∆x = lim‬‬
       ‫0 →‪∆ x‬‬        ‫0→‪h‬‬            ‫‪h‬‬
          ‫"ﻟﻴﻭﻨﺎﺭﺩ ﺃﻭﻟﺭ" ﻤﻥ ﺃﻜﺒﺭ ﺍﻟﻌﻠﻤﺎﺀ ﺍﻟﺫﻴﻥ ﻋﺭﻓﻬﻡ‬
                                   ‫ﺭ‬
    ‫ﺍﻟﺘﺎﺭﻴﺦ ، ﺍﺴﺘﻘ ‪ ‬ﻓﻲ ﺍﻟﺒﺩﺍﻴﺔ ﺒﹻ ﺴﺎﻥ ﺒﻴﺘﺭﺴﺒﻭﺭﻕ ﺜﻡ‬
                        ‫ﻓﻲ ﺒﺭﻟﻴﻥ ﺴﻨﺔ 1471 ﺤﻴﺙ‬
              ‫ﺘﺭﺃﺱ ﺃﻜﺎﺩﻴﻤﻴﺔ ﺍﻟﻌﻠﻭﻡ ﺇﻟﻰ ﻏﺎﻴﺔ 6671‬
      ‫ﺭ‬
‫ﺘﺨﺼﺹ ﻓﻲ ﻋﻠﻡ ﺍﻟﻔﻠﻙ )ﺩﺭﺍﺴﺔ ﻤﺴﺎﺭ ﺍﻟﻤﺠ ‪‬ﺍﺕ ( ،‬
      ‫ﻋﻠﻡ ﺍﻟﻔﻴﺯﻴﺎﺀ )ﺍﻟﺤﻘل ﺍﻟﻤﻐﻨﺎﻁﻴﺴﻲ ، ﺍﻟﺒﺼﺭﻴﺎﺕ ،‬
     ‫...( ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ) ﺍﻟﺤﺴﺎﺏ، ﺍﻟﻬﻨﺩﺴﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ،‬
       ‫ﻤﺭﻭﺭﺍ ﺒﺎﻟﺘﺤﻠﻴل ﺍﻟﺭﻗﻤﻲ ﻭ ﺍﻟﻭﻅﻴﻔﻲ ، ﺤﺴﺎﺏ ،‬
    ‫ﺘﻐﻴﺭﺍﺕ ﺍﻟﺒﻴﺎﻨﺎﺕ ، ﺍﻟﻤﺴﺎﺤﺎﺕ ﺍﻟﺠﺒﺭﻴﺔ...( ، ﻤﻌﺎﺩﻟﺔ‬
                                                           ‫‪EULER Leonhard‬‬
                                                              ‫3871-7071 ,‪Suisse‬‬
                             ‫ﺃﻭﻟﺭ )ﺤﺴﺎﺏ ﺍﻟﺘﻐﻴﺭﺍﺕ(‬
            ‫ﻫﻭ ﻤﻥ ﺃﺤﺩ ﻤﺅﺴﺴﻲ ﺍﻟﺘﺤﻠﻴل ﺍﻟﻭ ﻅﻴﻔﻲ ﻭ‬
                                 ‫ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ‬

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         ‫3( 5 + ‪ − x2 + 3 = −2 x‬ﺗﻌﻨﻲ 2) 2 − ‪( x‬‬                                            ‫اﻷﻧـﺸـﻄـﺔ‬
                          ‫2‬                                                                                            ‫ﻧﺸﺎط 1:‬
       ‫وﻣﻨﮫ ) ‪ ( C f‬ﯾﻘﻄﻊ ) ‪ ( D‬ﻓﻲ ﻧﻘﻄﺔ وﺣﯿﺪة )1;2 (‪. A‬‬                          ‫اﻟﮭﺪف : إدراج ﻣﻔﮭﻮم اﻟﻌﺪد اﻟﻤﺸﺘﻖ ﺑﺎﻟﺴﺮﻋﺔ .‬
                                   ‫) (‬
                                 ‫4( ) ‪ ( D‬ﯾﻤﺲ ‪. C f‬‬                           ‫= ‪. vm‬‬
                                                                                      ‫2)2(5 − 2)‪5(2 + h‬‬
                                                                                                            ‫1( 02 + ‪= 5h‬‬
                                                                                                ‫‪h‬‬
                  ‫اﻷﻋـﻤـﺎل ﻣـﻮﺟـﮭـﺔ‬                               ‫‪h‬‬           ‫2.0-‬         ‫1.0-‬      ‫2( 100.0- 50.0-‬
         ‫ﻛﯿﻔﯿﺔ إﻧﺸﺎء ﻣﻤﺎس ﻟﻘﻄﻊ ﻣﻜﺎﻓﺊ و ﻟﻘﻄﻊ زاﺋﺪ.‬                  ‫‪vm‬‬           ‫91‬          ‫5.91‬    ‫599.91 57.91‬
                          ‫ﻣﺴﺄﻟﺔ1: ﻣﻤﺎس ﻟﻘﻄﻊ ﻣﻜﺎﻓﺊ.‬                 ‫‪h‬‬         ‫10000.0‬       ‫1000.0‬   ‫500.0‬  ‫10.0‬
                                              ‫‪2a‬‬    ‫‪a‬‬   ‫2‬          ‫‪vm‬‬       ‫50000.02‬       ‫50.02 520.02 5000.02‬
                                        ‫=‪y‬‬       ‫−‪x‬‬          ‫•‬
                                              ‫‪k‬‬     ‫‪k‬‬                                          ‫3( 1− ‪. v(2) ≈ 20ms‬‬
            ‫• ﺗﻘﺎﻃﻊ ‪ T‬ﻣﻊ ﻣﺤﻮر اﻟﻔﻮاﺻﻞ : ‪ a ‬‬                                                                          ‫ﻧﺸﺎط 2:‬
            ‫‪ ;0‬‬                      ‫) (‬
            ‫‪2 ‬‬                                                  ‫اﻟﮭﺪف : ﺗﻔﺴﯿﺮ اﻟﻌﺪد اﻟﻤﺸﺘﻖ ھﻨﺪﺳﯿﺎ وﻛﺘﺎﺑﺔ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس .‬
                      ‫‪ a2 ‬‬
      ‫• ) ‪ (T‬ﯾﻤﺮ ﺑﺎﻟﻨﻘﻄﺘﯿﻦ ‪ A a ; ‬و ‪A'  ; 0 ‬‬
           ‫‪a‬‬                                                                                             ‫3‬
         ‫‪‬‬      ‫‪‬‬                                                                                            ‫4−‬
                                                                                                ‫1‬                     ‫1‬
         ‫‪2 ‬‬         ‫‪ k ‬‬                                        ‫4 = ‪ g ( 2 ) = − (2 a‬وﻣﻨﮫ ‪g ( 2 ) = a‬‬         ‫1( − =‬
                                                                                                ‫2‬             ‫9‬       ‫2‬
                        ‫1‬
               ‫ﺗﻄﺒﯿﻖ: 2‪. k = − ، f : x a − 3x‬‬                                                            ‫+2‬
                        ‫3‬                                                                                     ‫2‬
‫• اﻟﻤﻤﺎس ) ‪ (T‬ﻟﻠﻤﻨﺤﻨﻰ ، ﻋﻨﺪ اﻟﻨﻘﻄﺔ )3 − ;1(‪ A‬ﯾﺸﻤﻞ‬
                                                                                                     ‫1‬     ‫7‬
                                                                                             ‫3( )‪. y = − x + : ( EL‬‬
                                                                                                     ‫2‬     ‫4‬
                                           ‫‪1 ‬‬                       ‫اﻟﮭﺪف : ﺗﻔﺴﯿﺮ اﻟﺴﺮﻋﺔ اﻟﻠﺤﻈﯿﺔ ھﻨﺪﺳﯿﺎ .‬      ‫ﻧﺸﺎط 3:‬
                                        ‫اﻟﻨﻘﻄﺔ ‪A'  ; 0 ‬‬
                                           ‫‪2 ‬‬
                                                                     ‫)5(5 − )‪5(5 + h‬‬
                                                                               ‫2‬         ‫2‬

         ‫• اﻟﻤﻤﺎس ) ‪ (T‬ﻟﻠﻤﻨﺤﻨﻰ ، ﻋﻨﺪ اﻟﻨﻘﻄﺔ ) 21;2− ( ‪B‬‬                                     ‫2( 05 + ‪= 5h‬‬        ‫1( اﻟﺮﺳﻢ .‬
                                                                               ‫‪h‬‬
                                   ‫ﯾﺸﻤﻞ اﻟﻨﻘﻄﺔ ) 0 ;1− ( ' ‪. B‬‬                               ‫1− ‪vm = lim 5h + 50 = 50ms‬‬
                                                                                                        ‫0→‪h‬‬
      ‫‪‬‬          ‫‪‬‬
    ‫• اﻟﻤﻤﺎس ) ‪ (T‬ﻟﻠﻤﻨﺤﻨﻰ ، ﻋﻨﺪ اﻟﻨﻘﻄﺔ ‪C  − 3 ; − 1‬‬                                       ‫3( • ﺗﺮﺗﯿﺐ اﻟﻨﻘﻄﺔ ‪ M‬ھﻮ ‪. 5t‬‬
                                                                                              ‫2‬

      ‫3 ‪‬‬        ‫‪‬‬                                                  ‫02 − ‪5t‬‬
                                                                        ‫2‬

                                  ‫‪‬‬         ‫‪‬‬                                  ‫• ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮫ ) ‪ ( AM‬ھﻮ : ) 2 + ‪= 5 ( t‬‬
                            ‫ﯾﺸﻤﻞ اﻟﻨﻘﻄﺔ ‪. C '  − 3 ; 0 ‬‬             ‫2−‪t‬‬
                                  ‫‪ 6 ‬‬                           ‫)2 ( ‪d (t ) − d‬‬
                                                                                  ‫وﻧﺴﺒﺔ ﺗﺰاﯾﺪ ‪ d‬ﻋﻨﺪ 2 = 0‪= 5 ( t + 2 ) : t‬‬
                              ‫ﻣﺴﺄﻟﺔ2: ﻣﻤﺎس ﻟﻘﻄﻊ زاﺋﺪ.‬                  ‫2−‪t‬‬
                                          ‫• ∗¡ = ‪. D f‬‬            ‫• ﻟﻠﺤﺼﻮل ﺑﯿﺎﻧﯿﺎ ﻋﻠﻰ اﻟﺴﺮﻋﺔ اﻟﻠﺤﻈﯿﺔ ﻋﻨﺪ 2 = 0‪ t‬ﻧﻘﺮب‬
                ‫‪ 1‬‬                                                                               ‫اﻟﻨﻘﻄﺔ ‪ M‬ﻧﺤﻮ اﻟﻨﻘﻄﺔ ‪. A‬‬
     ‫• ﻣﻌﺎدﻟﺔ ﻟﻠﻤﻤﺎس ) ‪ِ (T‬ـ ‪ H‬ﻋﻨﺪ ‪ M  a ; ‬ھﻲ :‬
                                 ‫ﻟ‬                                 ‫• اﻟﺴﺮﻋﺔ اﻟﻠﺤﻈﯿﺔ ھﻲ 1− ‪ lim 5 (t + 2 ) = 20ms‬و ھﺬا‬
                ‫‪ a‬‬                                                        ‫2→ ‪t‬‬

                  ‫‪ 2‬‬                       ‫1‬      ‫2‬                                                 ‫ﯾﺘﻨﺎﺳﺐ ﻣﻊ اﻟﺘﻔﺴﯿﺮ اﻟﮭﻨﺪﺳﻲ .‬
‫• ‪ A 0; ‬و ) 0 ; ‪. B ( 2a‬‬          ‫+‪. y = − 2 x‬‬
                  ‫‪ a‬‬                      ‫‪a‬‬       ‫‪a‬‬                                                                  ‫ﻧﺸﺎط 4:‬
                                                                                                  ‫اﻟﮭﺪف : إدراج ﻣﻔﮭﻮم اﻟﻤﻤﺎس‬
                     ‫‪‬‬                ‫‪2‬‬
                   ‫.‬ ‫• ‪ 0 + 2a 0 + a  =  a ; 1 ‬‬
                                           ‫‪‬‬     ‫‪‬‬
                                                                                      ‫∞− ‪x‬‬            ‫0‬           ‫1( ∞+‬
                     ‫‪‬‬          ‫;‬       ‫‪  a‬‬                                     ‫)‪f ( x‬‬             ‫3‬
                     ‫2 ‪‬‬           ‫‪2 ‬‬
  ‫• إﻧﺸﺎء ‪. H‬‬        ‫• اﻟﻤﻤﺎس ) ‪ (T‬ھﻮ اﻟﻤﺴﺘﻘﯿﻢ ) ‪( AB‬‬                                             ‫‪y‬‬                       ‫1(‬
    ‫• ) 1‪ ( T‬ھﻮ )" ‪ ( R ' R‬ﺣﯿﺚ ) 2 − ;0 ( ' ‪ R‬و ) 0 ;2− ( " ‪. R‬‬                                                         ‫2( .‬
                                                                                     ‫) ‪(C‬‬
                                                                                       ‫‪f‬‬

                        ‫‪‬‬      ‫‪2‬‬
‫• ) 2‪ ( T‬ھﻮ )" ‪ ( N ' N‬؛ ‪ N '  0; − ‬و ) 0;6− ( " ‪. N‬‬                                            ‫1‬
                        ‫‪‬‬      ‫‪3‬‬
‫• ) 3‪ ( T‬ھﻮ )" ‪ ( P ' P‬ﺣﯿﺚ ) 4 − ;0 ( ' ‪ P‬و ) 0 ;1− ( " ‪. P‬‬
                                                                                                  ‫0‬      ‫1‬              ‫‪x‬‬




                        ‫ﺗﻘﺮﯾﺒﺎت ﺗﺂﻟﻔﯿﺔ ﻣﺄﻟﻮﻓﺔ ﻋﻨﺪ0:‬                                                                  ‫)‪( D‬‬
                             ‫1( اﻟﺘﻘﺮﯾﺐ اﻟﺘﺂﻟﻔﻲ ﻋﻨﺪ 0 ھﻮ :‬
              ‫‪f ( x) ; f ( 0 ) + f ′ ( 0 ) × x‬‬

‫2‬
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                          ‫ﺗــﻤـــﺎرﯾــــﻦ‬                              ‫= )‪f ( x‬‬      ‫3)‪(1 + x)2 (1 + x‬‬          ‫‪1+ x‬‬
                                                                                                                                 ‫1‬
              ‫ﺻﺤﯿﺢ‬                                                                                                              ‫‪1+ x‬‬
                            ‫ﺧﺎﻃﺊ . 3‬      ‫2‬       ‫ﺻﺤﯿﺢ‬        ‫1‬
                                                                                                                    ‫1‬
                                                                       ‫; )‪f ( x‬‬      ‫‪1 + 2x‬‬        ‫+ 1 ‪1 + 3x‬‬         ‫‪x 1− x‬‬
              ‫6 ﺻﺤﯿﺢ‬             ‫ﺧﺎﻃﺊ‬     ‫5‬      ‫ﺻﺤﯿﺢ‬         ‫4‬                                                     ‫2‬

              ‫ﺻﺤﯿﺢ‬          ‫ﺧﺎﻃﺊ 9‬        ‫8‬       ‫ﺧﺎﻃﺊ‬        ‫7‬                            ‫1‬
                                                                          ‫= )‪f ( x‬‬                          ‫‪cos x‬‬           ‫‪sin x‬‬
                                                                                        ‫)‪(1 + x‬‬
                                                                                                   ‫2‬
             ‫ﺻﺤﯿﺢ‬           ‫11 ﺻﺤﯿﺢ 21‬            ‫ﺧﺎﻃﺊ‬       ‫01‬
                                                                          ‫; )‪f ( x‬‬     ‫‪1 − 2x‬‬         ‫1‬          ‫‪x‬‬
                                            ‫2 = )1( ' ‪f‬‬      ‫31‬                         ‫1‬
                                                                    ‫= )300 ,0 ( ‪f‬‬              ‫2( 799,0 = 300 ,0 − 1 ;‬
                           ‫)2 ( ‪f ( 2 + h) − f‬‬                                     ‫300 ,0 + 1‬
                      ‫.‬                        ‫2+‪= h‬‬         ‫41‬
                                    ‫‪h‬‬                                                            ‫1‬
                                                                              ‫= ) 20 ,0− ( ‪f‬‬           ‫20 ,1 = 20 ,0 + 1 ;‬
                          ‫اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻨﺪ 1.‬       ‫51‬                              ‫20,0 − 1‬
                                ‫اﻟﻌﺪد ) 2 ( ' ‪ f‬ھﻮ 1- .‬      ‫61‬                   ‫900 ,0 + 1 ; )300 ,0 + 1( = )300 ,0 ( ‪f‬‬
                                                                                                                    ‫3‬


                                    ‫) 0 ( ' ‪ f‬ﻏﯿﺮ ﻣﻌﺮف‬       ‫71‬                ‫60 ,0 − 1 ; ) ) 20.0− ( + 1( = ) 20 ,0− ( ‪f‬‬
                                                                                                                        ‫3‬

           ‫ﻣﻌﺎدﻟﺔ ﻣﻤﺎس اﻟﻤﻨﺤﻨﻲ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ‬           ‫81‬                       ‫400 ,1 ; ) 200 ,0 + 1( = ) 200,0 ( ‪f‬‬
                                                                                                                            ‫2‬

                    ‫)1− ;0 (‪ A‬ھﻲ 1 + ‪. y = 3x‬‬
                                                                                         ‫89 ,0 ; )10 ,0 − 1( = )10 ,0− ( ‪f‬‬
                                                                                                                                ‫2‬
                       ‫اﻟﻌﺪد )1( ' ‪ f‬ھﻮ 2 .‬                  ‫91‬
    ‫اﻟﺪاﻟﺔ اﻟﻤﺸﺘﻘﺔ ' ‪ f‬ﻟﻠﺪاﻟﺔ ‪ f‬ﻣﻌﺮﻓﺔ ﻋﻠﻰ ¡ ﺑـ:‬                                   ‫200 ,0 + 1 ; 400 ,0 + 1 = ) 400 ,0 ( ‪f‬‬
                                                             ‫02‬
                                 ‫1 + ‪. f ' ( x) = 2 x‬‬                               ‫500 ,0 − 1 ; 10 ,0 − 1 = )10 ,0 ( ‪f‬‬
                                         ‫12 0 = )1− ( ' ‪. f‬‬                ‫= )10 ,0 ( ‪f‬‬
                                                                                               ‫1‬
                                                                                                       ‫69 ,0 = 40 ,0 − 1 ;‬
                                                                                         ‫)10 ,0 + 1(‬
                                                                                                     ‫2‬
              ‫22 1( 0 = ) 0 ( ' ‪f ' ( 3) = −3 (2 ، f‬‬
                                                                                               ‫1‬
                             ‫3( 21− = ) 2− ( ' ‪f‬‬                      ‫= )10 ,0− ( ‪. f‬‬                       ‫200 ,1 = 20 ,0 + 1 ;‬
                                                                                        ‫)10 ,0 − 1(‬
                                                                                                        ‫2‬

            ‫2( 3− = )1( ' ‪f‬‬        ‫1( 1 = )1− ( ' ‪، f‬‬        ‫32‬
                                                                                                                        ‫ﺗﻄﺒﯿﻖ:‬
                         ‫1‬
                                                                                                       ‫‪. y = −x + 1 : (∆) v‬‬
                                                     ‫1‬
            ‫= )4( ' ‪f‬‬        ‫− = ) 3 − ( ' ‪(4 ، f‬‬       ‫3(‬
                          ‫8‬                         ‫81‬
             ‫‪1‬‬                                  ‫3‬                                        ‫‪.  −4, 610−7 ; 4, 610−7  v‬‬
                                                                                             ‫‪‬‬                      ‫‪‬‬
         ‫− = )1− ( ' ‪f '   = −1 (6 ، f‬‬               ‫5(‬
             ‫‪4‬‬                                                        ‫‪ 1 1‬‬                   ‫1‬                 ‫2‪x‬‬
                                               ‫3 2‬                  ‫و ﻣﻦ أﺟﻞ ‪x ∈  − ; ‬‬             ‫= )‪− (1 − x‬‬        ‫‪v‬‬
                    ‫)1− ( ‪f ( −1 + h ) − f‬‬                              ‫‪ 2 2‬‬                 ‫1+ ‪x‬‬               ‫1+ ‪x‬‬
                                            ‫42 1( 4 =‬
                              ‫‪h‬‬                                                ‫2‬      ‫1‬              ‫1‬             ‫3‬
                                                                    ‫وﯾﻌﻨﻲ أن‬      ‫≤‬       ‫وﻟﺪﯾﻨﺎ ≤ 1 + ‪ ≤ x‬وﻣﻨﮫ 2 ≤‬
                     ‫)1− ( ‪f ( −1 + h ) − f‬‬                                    ‫1+ ‪3 x‬‬                ‫2‬             ‫2‬
    ‫‪ ، lim‬ﻧﺴﺘﻨﺘﺞ‬                            ‫2( ﻟﺪﯾﻨﺎ 4 =‬                         ‫2‬                      ‫2‬       ‫2‬
               ‫0→ ‪h‬‬            ‫‪h‬‬                                                ‫‪x‬‬                    ‫‪2x‬‬        ‫‪x‬‬
                                                                          ‫≤0‬         ‫وﺑﺎﻟﺘﺎﻟﻲ 2‪≤ 2 x‬‬      ‫≤‬        ‫2‪≤ 2 x‬‬
              ‫أن اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ أﺟﻞ 1- و‬                             ‫1+ ‪x‬‬                   ‫3‬       ‫1+ ‪x‬‬
                                              ‫4 = )1− ( ' ‪f‬‬                    ‫2−01‬
                                                                        ‫=‪x‬‬           ‫‪ v‬ﺑﻮﺿﻊ 2−01 = 2‪ 2 x‬ﻧﺠﺪ 170 ,0 ≈‬
            ‫3( ﻧﻌﻢ اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ أﺟﻞ 0 .‬                              ‫2‬
                                   ‫3) ‪( 2 + h‬‬   ‫1( ﻧﻨﺸﺮ‬      ‫52‬       ‫أو 170 ,0− ≈ ‪ x‬إذن اﻟﻤﺠﺎل ھﻮ ]170 ,0 ;170 ,0− [ .‬

        ‫)‪(2 + h‬‬       ‫8−‬
                  ‫3‬

    ‫‪lim‬‬                ‫2( 21 = ) 21 + ‪= lim ( h 2 + 6h‬‬
    ‫0→ ‪h‬‬        ‫‪h‬‬        ‫0→ ‪h‬‬

                ‫ﻧﺴﺘﻨﺘﺞ أن اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻨﺪ 2 و‬
                                                   ‫21 = ) 2 ( ' ‪f‬‬
               ‫1( ‪f ( 2 + h ) − f ( 2 ) = h3 − 8h‬‬            ‫62‬

‫3‬
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                                          ‫1‬                                                    ‫)2 ( ‪f ( 2 + h ) − f‬‬
                              ‫= )7( ' ‪f‬‬           ‫و ﻧﺠﺪ‬                          ‫‪ ، lim‬إذن‬                           ‫2( 8− =‬
                                         ‫5 2‬                                            ‫0→ ‪h‬‬            ‫‪h‬‬
                       ‫)3− ( ‪f ( −3 + h ) − f‬‬                                                                  ‫8− = ) 2 ( ' ‪. f‬‬
                  ‫‪lim‬‬                           ‫73 ﻧﺤﺴﺐ‬
                  ‫0→ ‪h‬‬            ‫‪h‬‬                                                             ‫)2 ( ‪f ( 2 + h ) − f‬‬
                                             ‫1‬                  ‫‪ ، lim‬إذن 2− = ) 2 ( ' ‪. f‬‬                            ‫72 2− =‬
                              ‫و ﻧﺠﺪ − = )3− ( ' ‪f‬‬                                         ‫0→ ‪h‬‬           ‫‪h‬‬
                                             ‫4‬                                                 ‫)1− ( ‪f ( −1 + h ) − f‬‬
                           ‫83 ﺗﺼﻮﯾﺐ: اﻟﺘﺮﻗﯿﻢ ﯾﺒﺪأ ﻣﻦ 1(‬          ‫‪ ، lim‬إذن 5 = )1− ( ' ‪f‬‬                                 ‫82 5 =‬
                                                                                        ‫0→ ‪h‬‬              ‫‪h‬‬
                           ‫1( 7 − ‪ f ( x ) = 2 x‬و 3 = ‪a‬‬
                                                                                                  ‫)3 ( ‪f ( 3 + h ) − f‬‬
                              ‫)3 ( ‪f ( 3 + h ) − f‬‬                     ‫‪ ، lim‬إذن 1 = )3 ( ' ‪f‬‬                           ‫92 1 =‬
     ‫‪ lim‬و ﻧﺠﺪ 2 = )3 ( ' ‪f‬‬                        ‫ﻧﺤﺴﺐ‬                                     ‫0→ ‪h‬‬           ‫‪h‬‬
                         ‫0→ ‪h‬‬          ‫‪h‬‬                                                       ‫) 2− ( ‪f ( −2 + h ) − f‬‬
                 ‫) ‪f ( a + h) − f ( a‬‬                                            ‫‪ ، lim‬إذن‬                               ‫03 1 =‬
                                                                                       ‫0→ ‪h‬‬
          ‫‪lim‬‬                         ‫ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻧﺤﺴﺐ‬                                                   ‫‪h‬‬
                                                                                                                    ‫1 = ) 2− ( ' ‪f‬‬
           ‫0→ ‪h‬‬           ‫‪h‬‬
                ‫و ﻧﺠﺪ ) ‪ f ' ( a‬ﻓﻲ ﺑﺎﻗﻲ اﻟﺤﺎﻻت اﻷﺧﺮى.‬
                                                                                             ‫)2( ‪f ( 2 + h) − f‬‬              ‫13 1‬
      ‫93 ﻧﻔﺲ اﻟﻄﺮﯾﻘﺔ اﻟﻤﺘﺒﻌﺔ ﻓﻲ ﺣﻞ اﻟﺘﻤﺮﯾﻦ رﻗﻢ 83 .‬                                 ‫‪، lim‬‬                               ‫−=‬
                                                                                      ‫0→ ‪h‬‬            ‫‪h‬‬                      ‫9‬
                                                                                                                     ‫1‬
      ‫ﻧﻔﺲ اﻟﻄﺮﯾﻘﺔ اﻟﻤﺘﺒﻌﺔ ﻓﻲ ﺣﻞ اﻟﺘﻤﺮﯾﻦ رﻗﻢ 83 .‬          ‫04‬                                          ‫− = )2( ' ‪f‬‬      ‫إذن‬
               ‫1( 2 + 2‪a = 6 ، f : x a x‬‬                  ‫14‬
                                                                                                                     ‫9‬
                                                                                              ‫)2( ‪f ( 2 + h) − f‬‬         ‫1‬
             ‫1+ )‪f ( a + h ) = f (6 + h ) = (6 + h‬‬                                    ‫‪، lim‬‬                          ‫23 − =‬
                                                   ‫2‬
                                                                                        ‫0→ ‪h‬‬            ‫‪h‬‬                ‫9‬
               ‫)6( ‪f (6 + h) − f‬‬
‫‪f ' ( 6 ) = lim‬‬                  ‫21 = ) 21 + ‪= lim ( h‬‬                                                               ‫1‬
                                                                                                       ‫إذن − = ) 2 ( ' ‪f‬‬
           ‫0→ ‪h‬‬        ‫‪h‬‬           ‫0→ ‪h‬‬
                                                                                                                     ‫9‬
            ‫إذن اﻟﻌﺪد اﻟﻤﺸﺘﻖ ﻟﻠﺪاﻟﺔ ‪ f‬ﻣﻦ اﺟﻞ 6 = ‪ a‬ھﻮ‬                                      ‫)3 ( ‪f ( 3 + h ) − f‬‬
                                                                                  ‫‪، lim‬‬                            ‫33 1( 2− =‬
                                               ‫21 = ) 6 ( ' ‪f‬‬                       ‫0→ ‪h‬‬            ‫‪h‬‬
        ‫و ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻧﺤﺴﺐ ) ‪ f ' ( a‬ﻓﻲ اﻟﺤﺎﻻت اﻷﺧﺮى‬                                                 ‫إذن 2− = )3 ( ' ‪f‬‬
                                                    ‫اﻟﻤﺘﺒﻘﯿﺔ.‬           ‫‪ 1‬‬          ‫8‬                           ‫2‬
                                                                     ‫− = ‪f ' − ‬‬          ‫− = ) 5 ( ' ‪(3 ، f‬‬         ‫2(‬
                                   ‫1( ‪f : x a x‬‬
                                              ‫2‬
                                                         ‫24‬             ‫‪ 2‬‬         ‫52‬                           ‫9‬
                   ‫)3 ( ‪f ( 3 + h ) − f‬‬
    ‫‪f ' ( 3) = lim‬‬
               ‫0→‪h‬‬          ‫‪h‬‬
                                        ‫6 = ) 6 + ‪= lim ( h‬‬
                                          ‫0→ ‪h‬‬
                                                                               ‫1‬
                                                                 ‫− = 3 ' ‪f ' ( 0 ) = − (5 ، f‬‬
                                                                               ‫2‬
                                                                                               ‫) (‬        ‫3 4−7‬
                                                                                                               ‫2‬
                                                                                                                      ‫4(‬

       ‫2( أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد ) ‪ f ( 3 + h‬ﻣﻦ اﺟﻞ اﻟﻘﯿﻢ‬                                    ‫)3 ( ‪f ( 3 + h ) − f‬‬
                                                                  ‫‪ ، lim‬إذن 2 = )3 ( ' ‪f‬‬                          ‫43 2 =‬
     ‫اﻟﺼﻐﯿﺮة ﻟﻠﻌﺪد ‪ h‬ھﻮ )3 ( ' ‪ f ( 3 ) + h f‬أي ‪9 + 6 h‬‬                                ‫0→ ‪h‬‬           ‫‪h‬‬
                                  ‫34 1( 2 + 2‪f : x a x‬‬                      ‫)1( ‪f (1 + h ) − f‬‬        ‫2−4+‪h‬‬
                                                                                                 ‫=‬                  ‫53 1(‬
                           ‫)1− ( ‪f ( h − 1) − f‬‬                                     ‫‪h‬‬                     ‫‪h‬‬
          ‫‪f ' ( −1) = lim‬‬                       ‫2− =‬                                        ‫2( ﻣﻦ أﺟﻞ 4− > ‪ h‬و 0 ≠ ‪: h‬‬
                                                                                                  ‫(‬                ‫()‬            ‫)‬
                      ‫0→ ‪h‬‬           ‫‪h‬‬
       ‫= 2 − 4 + ‪ (2 f (1+ h) − f (1) = h‬أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد )1 − ‪ f ( h‬ﻣﻦ اﺟﻞ اﻟﻘﯿﻢ‬
                                                                                                      ‫2− 4+‪h‬‬            ‫2 + 4 +‪h‬‬

                  ‫) (‬          ‫) (‬
            ‫اﻟﺼﻐﯿﺮة ﻟﻠﻌﺪد ‪ h‬ھﻮ 1− ' ‪ f −1 + h f‬أي‬               ‫‪h‬‬              ‫‪h‬‬                          ‫‪h‬‬   ‫(‬   ‫2+ 4+‪h‬‬     ‫)‬
                                                 ‫‪.3− 2 h‬‬                                                 ‫1‬
                                                                                                 ‫=‬
                         ‫44 1( ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ ‪f : x a x‬‬
                                   ‫2‬
                                                                                                      ‫2+ 4+‪h‬‬
    ‫اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ اﺟﻞ اﻟﻘﯿﻤﺔ 2 و ﻟﺪﯾﻨﺎ :‬                                               ‫1 )1( ‪f (1 + h ) − f‬‬
                                                                                             ‫‪lim‬‬                    ‫3( =‬
               ‫)2 ( ‪f ( 2 + h ) − f‬‬       ‫4 = 4 − )‪( 2 + h‬‬
                                                   ‫2‬
                                                                                             ‫0→ ‪h‬‬         ‫‪h‬‬          ‫4‬
‫‪f ' ( 2) = lim‬‬                      ‫‪= lim‬‬                                       ‫1‬
           ‫0→‪h‬‬          ‫‪h‬‬             ‫0→‪h‬‬       ‫‪h‬‬                   ‫= )1( ' ‪f‬‬     ‫إذن اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻨﺪ 1 و ﻟﺪﯾﻨﺎ‬
‫أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد ) ‪ ( 2 + h‬ﻋﻨﺪﻣﺎ ﯾﻨﺘﮭﻲ ‪ h‬إﻟﻰ 0‬                            ‫4‬
                                  ‫2‬

                                                                                        ‫)7( ‪f (7 + h ) − f‬‬
                     ‫ھﻮ ) 2 ( ' ‪ f ( 2 ) + h f‬أي ‪. 4 + 4 h‬‬                        ‫‪lim‬‬                       ‫63 ﻧﺤﺴﺐ‬
                                                                                  ‫0→ ‪h‬‬          ‫‪h‬‬
                                   ‫2( 40 ,0 + 2 = 40 ,2‬

‫4‬
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        ‫ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻧﺠﺪ ﻗﯿﻤﺔ ﻣﻘﺮﺑﺔ ﻟـ 79,4 و 38,4‬                ‫61,4 ≅ 2) 40 ,2 (‬       ‫إذن ) 40 ,0 ( 4 + 4 ≅ ) 40 ,2 ( أي‬
                                                                                                           ‫2‬

‫)ﺑﻤﻼﺣﻈﺔ أن 30 ,0 − 5 = 79,4 و 71,0 − 5 = 38,4 (‬                                                           ‫20 ,0 − 2 = 89 ,1‬
                  ‫84 1( )1,0 + 2 ( ‪f ( 2,1) = f‬‬                   ‫) 89,1(‬
                                                                            ‫2‬
                                                                                ‫إذن ) 20 ,0− ( 4 + 4 ≅ ) 89,1( أي 29 ,3 ≅‬
                                                                                                       ‫2‬


                 ‫) 2 ( ‪ f ( 2,1) ≅ 0,1× f ' ( 2 ) + f‬أي‬                                            ‫100 ,0 + 2 = 100 ,2‬
                                            ‫4 ,3 ≅ )1,2 ( ‪f‬‬                          ‫أي‬    ‫)100 ,2 (‬
                                                                                                       ‫2‬
                                                                                                           ‫إذن )100 ,0 ( 4 + 4 ≅‬
                               ‫2( )1,0 + 2 ( ‪f ( 2,1) = f‬‬                                                      ‫400 ,4 ≅ 2)100 ,2 (‬
                     ‫)1,2 ( ‪f ( 2, 2 ) ≅ 0,1× f ' ( 2,1) + f‬‬                                     ‫1‬
                                                                                          ‫‪f :xa‬‬    ‫54 1( ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ‬
    ‫أي 4 ,3 + ) 2 ×1,0 ( ≅ ) 2 ,2 ( ‪ f‬أي 6 ,3 ≅ ) 2 ,2 ( ‪f‬‬                                       ‫‪x‬‬
           ‫94 1( ﻣﻌﺎدﻟﺔ ﻣﻤﺎس اﻟﻤﻨﺤﻨﻲ ) ‪ ( C‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ‬                     ‫اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ اﺟﻞ اﻟﻘﯿﻤﺔ 3 و ﻟﺪﯾﻨﺎ :‬
                                                                                                                       ‫2‬
                     ‫) 0 ;2 (‪ A‬و اﻟﺬي ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮭﮫ 1 = ‪a‬‬                                                   ‫1 ‪ 1 ‬‬
                                                                                ‫)3 ( ‪f ( 3 + h ) − f‬‬       ‫‪‬‬      ‫− ‪‬‬
                                                                 ‫‪f ' ( 3) = lim‬‬                            ‫1 − = 3 ‪ 3+ h ‬‬
    ‫ھﻲ: ) 0‪ y = a ( x − x0 ) + f ( x‬ﺣﯿﺚ 0‪ x‬ھﻲ ﻓﺎﺻﻠﺔ ‪A‬‬                       ‫0→‪h‬‬
                                                                                                     ‫‪= lim‬‬
                                                                                                       ‫0→‪h‬‬
                                                                                         ‫‪h‬‬                       ‫‪h‬‬        ‫9‬
                               ‫أي ) 2 ( ‪y = 1( x − 2 ) + f‬‬                                   ‫1‬
                                                                   ‫ﻋﻨﺪﻣﺎ ﯾﻨﺘﮭﻲ ‪ h‬إﻟﻰ 0‬            ‫أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد‬
                        ‫أي ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ھﻲ 2 − ‪y = x‬‬                                      ‫‪3+ h‬‬
  ‫‪ v‬و ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻧﻌﯿﻦ اﻟﻤﻤﺎس ﻓﻲ اﻟﺤﺎﻻت اﻷﺧﺮى.‬                                                               ‫1‬    ‫1‬
                                                                                                          ‫+‪.− h‬‬        ‫ھﻮ‬
                             ‫2‪2 x‬‬                                                                             ‫9‬    ‫3‬
        ‫= ‪ y‬و 3 = 0‪x‬‬                ‫05 ﻣﻌﺎدﻟﺔ ) ‪ ( C‬ھﻲ‬              ‫1‬         ‫1‬             ‫1‬
                              ‫5‬                                           ‫2( 20 ,0 + 3 = 20 ,3 إذن + ) 20 ,0 ( − ≅‬
    ‫ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮫ اﻟﻤﻤﺎس ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ 3 = 0‪x‬‬              ‫20 ,3‬       ‫9‬             ‫3‬
                            ‫21 )3 ( ‪f ( 3 + h ) − f‬‬                                         ‫1‬
                                                                                                 ‫أي 111111133,0 ≅‬
           ‫‪، f ' ( 3) = lim‬‬                      ‫=‬     ‫ھﻮ:‬
                        ‫0→‪h‬‬          ‫‪h‬‬             ‫5‬                                      ‫20 ,3‬
                                               ‫2‪2x‬‬                      ‫1‬         ‫1‬
                                   ‫= )‪( f ( x‬‬      ‫)ﺑﻮﺿﻊ‬                    ‫و‬         ‫و ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻧﺠﺪ ﻗﯿﻤﺔ ﺗﻘﺮﯾﺒﯿﺔ ﻟـ‬
                                                ‫5‬                      ‫1,3‬      ‫99,2‬
‫ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ھﻲ : )3 ( ‪ y = f ' ( 3)( x − 3) + f‬و ﻧﺠﺪ‬                                  ‫64 1( ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ 3‪f : x a x‬‬
                                                ‫21‬    ‫81‬          ‫اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ اﺟﻞ اﻟﻘﯿﻤﺔ 1و ﻟﺪﯾﻨﺎ :‬
                    ‫، ) 81 = ) 3 ( ‪( f‬‬     ‫=‪y‬‬      ‫−‪x‬‬
                                                                             ‫)1( ‪f (1 + h ) − f‬‬        ‫3 = 1 − ) ‪(1 + h‬‬
                                                                                                                           ‫3‬
                                ‫5‬                ‫5‬     ‫5‬
‫و ﺑﻨﻔﺲ اﻟﻄﺮﯾﻘﺔ ﯾﺘﻢ ﺗﻌﯿﯿﻦ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻲ ) ‪ ( C‬ﻓﻲ‬        ‫‪f ' (1) = lim‬‬                    ‫‪= lim‬‬
                                                                         ‫0→‪h‬‬         ‫‪h‬‬            ‫0→ ‪h‬‬      ‫‪h‬‬
                                    ‫اﻟﺤﺎﻻت اﻷﺧﺮى اﻟﻤﺘﺒﻘﯿﺔ .‬
                                                                ‫أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد ) ‪ (1 + h‬ﻋﻨﺪﻣﺎ ﯾﻘﺘﺮب ‪ h‬ﻣﻦ 0‬
                                                                                              ‫3‬

  ‫15 ﺑﻮﺿﻊ: ‪ . f ( x) = x2 − 2 x‬ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮫ اﻟﻤﻤﺎس‬
                             ‫ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ 1− ھﻮ:‬                              ‫ھﻮ )1( ' ‪ f (1) + h f‬أي ‪.1 + 3h‬‬
                               ‫)1− ( ‪f ( −1 + h ) − f‬‬                                                  ‫2( 40 ,0 + 1 = 40 ,1‬
          ‫‪f ' ( −1) = lim‬‬                               ‫4− =‬
                         ‫0→ ‪h‬‬              ‫‪h‬‬                          ‫إذن ) 40 ,0 ( 3 + 1 ≅ ) 40 ,1( أي 21,1 ≅ ) 40,1(‬
                                                                              ‫3‬                      ‫3‬


  ‫ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ھﻲ : )1− ( ‪ y = f ' ( −1)( x + 1) + f‬و‬                                                    ‫88,0 ≅ 3) 69,0 (‬
                 ‫ﻧﺠﺪ 1 − ‪( f ( − 1 ) = 3 ) ، y = − 4 x‬‬
                                                                                      ‫1( ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ ‪f : x a x‬‬
                                                ‫4‬                                                                        ‫74‬
    ‫25 ﺑﻮﺿﻊ: − = )‪ f ( x‬ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮫ اﻟﻤﻤﺎس ﻋﻨﺪ‬                     ‫اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ اﺟﻞ اﻟﻘﯿﻤﺔ 5 و ﻟﺪﯾﻨﺎ :‬

                                   ‫( ‪ f ' ( 5) = lim‬اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ 2 ھﻮ:‬
                                                ‫‪x‬‬                            ‫)5 ( ‪f 5 + h) − f‬‬           ‫5 − ‪5+ h‬‬        ‫1‬
                                                                                                ‫‪= lim‬‬                 ‫=‬
                                      ‫)2( ‪f ( 2 + h) − f‬‬                 ‫0→‪h‬‬         ‫‪h‬‬            ‫0→‪h‬‬        ‫‪h‬‬          ‫5 2‬
                   ‫‪f ' ( 2 ) = lim‬‬                        ‫1=‬
                                ‫0→ ‪h‬‬           ‫‪h‬‬                  ‫أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﻌﺪد ‪ 5 + h‬ﻋﻨﺪﻣﺎ ﯾﻨﺘﮭﻲ ‪ h‬إﻟﻰ 0‬
‫ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ھﻲ : ) 2 ( ‪ y = f ' ( 2 )( x − 2 ) + f‬و ﻧﺠﺪ‬                       ‫+5 .‬
                                                                                          ‫‪h‬‬
                                                                                                ‫ھﻮ ) 5 ( ' ‪ f ( 5 ) + h f‬أي‬
                           ‫4 − ‪( f ( 2 ) = −2 ) ، y = x‬‬                                 ‫5 2‬
                                                                                                       ‫2( 10 ,0 + 5 = 10 ,5‬
                                             ‫2 1‬
 ‫ﺑﻮﺿﻊ: ‪ ، f ( x) = 2 − x‬ﻣﻌﺎﻣﻞ ﺗﻮﺟﯿﮫ اﻟﻤﻤﺎس‬                                                                        ‫10 ,0‬
                                             ‫2‬             ‫35‬                                    ‫+ 5 ≅ 10 ,5‬             ‫إذن‬
                              ‫ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ 1 ھﻮ:‬                                                        ‫5 2‬
                                      ‫)1( ‪f (1 + h ) − f‬‬                                     ‫أي 540403832 ,2 ≅ 10 ,5‬
                    ‫‪f ' (1) = lim‬‬                        ‫1− =‬
                                ‫0→ ‪h‬‬           ‫‪h‬‬

‫5‬
                                             ‫‪http://assil.yoo7.com‬‬
    ‫1 ) ‪ϕ ( a + h) − ϕ ( a ) f ( a + h ) − f ( a‬‬                     ‫ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ھﻲ : )1( ‪ y = f ' (1)( x − 1) + f‬و ﻧﺠﺪ‬
                        ‫=‬                     ‫2− ‪+ h+a‬‬
                                  ‫‪h‬‬            ‫2‬                                                                                 ‫5‬
             ‫‪h‬‬                                                                               ‫، ) 3 = ) 1( ‪( f‬‬       ‫+‪y = −x‬‬
                               ‫ب( ‪ ϕ‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ ¡‬                                                   ‫2‬                       ‫2‬
                             ‫) ‪f ( a + h) − f ( a‬‬                     ‫ﻣﻦ اﻟﻮاﺿﺢ أن اﻟﻤﻤﺎس ﯾﻘﻄﻊ ﻣﺤﻮر اﻟﻔﻮاﺻﻞ ﻓﻲ اﻟﻨﻘﻄﺔ‬
                                                  ‫ﺟـ( ‪= 2 − a‬‬                                               ‫5‬
                                      ‫‪h‬‬                                                                   ‫.‬   ‫اﻟﺘﻲ ﻓﺎﺻﻠﺘﮭﺎ‬
                                     ‫1 ) ‪ϕ ( a + h) −ϕ ( a‬‬                                                  ‫2‬
                                                        ‫و‪= h‬‬        ‫45 1( ﻧﺤﻞ اﻟﻤﻌﺎدﻟﺔ ذات اﻟﻤﺠﮭﻮل ‪x2 = −4 x − 4 : x‬‬
                                                                           ‫و ﻧﺠﺪ 2− = ‪ ، x‬إذن ) ‪ ( C‬و ) ‪ ( D‬ﯾﺘﻘﺎﻃﻌﺎن ﻓﻲ‬
                                              ‫‪h‬‬          ‫2‬
                              ‫) ‪ϕ ( a + h) −ϕ ( a‬‬
   ‫‪ ، lim‬إذن اﻟﺪاﻟﺔ ‪ ϕ‬ﺛﺎﺑﺘﺔ‬                        ‫و ﻣﻨﮫ 0 =‬                                                  ‫اﻟﻨﻘﻄﺔ ) 4 ;2 − ( ‪. A‬‬
                          ‫0→‪h‬‬           ‫‪h‬‬
                ‫16 1( ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪ a‬ﻟﺪﯾﻨﺎ:‬                                   ‫2( ﻧﺴﺘﻨﺘﺞ أن ) ‪ ( D‬ھﻮ اﻟﻤﻤﺎس ﻟـ ) ‪ ( C‬ﻓﻲ‬
                              ‫) ‪f ( a + h) − f ( a‬‬                                                    ‫اﻟﻨﻘﻄﺔ ) 4 ;2 − ( ‪. A‬‬
‫‪ ، lim‬إذن اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ‬                             ‫5 − ‪= 2a‬‬
                        ‫0→ ‪h‬‬           ‫‪h‬‬                           ‫55 ﺗﺼﺤﯿﺢ : ﻣﻌﺎدﻟﺔ ) ‪ y = −2 x − 2 : ( D‬وﻓﻲ اﻟﺴﺆال1(‬
       ‫ﻟﻼﺷﺘﻘﺎق ﻋﻨﺪ ﻛﻞ ‪ x‬ﻣﻦ ¡ و 5 − ‪. f ' ( x) = 2 x‬‬
                                                                   ‫) ‪ 3 x3 + 2 x2 − 5 x − 4 = ( x + 1) ( ax2 + bx + c‬وﻧﺠﺪ‬
       ‫2( ﻣﻌﺎدﻟﺔ ﻣﻤﺎس اﻟﻤﻨﺤﻨﻲ ) ‪ (P‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ) 4 ;0 ( ‪E‬‬
                                                                         ‫) 4 − ‪3 x3 + 2 x2 − 5 x − 4 = ( x + 1) ( 3 x2 − x‬‬
                                         ‫ھﻲ: 4 + ‪y = −5 x‬‬
      ‫3( ﻧﻌﻢ ﺗﻮﺟﺪ ﻧﻘﻄﺔ ‪ M‬ﻣﻦ ) ‪ (P‬ﯾﻜﻮن ﻣﻤﺎﺳﮫ ﻋﻨﺪھﺎ‬                         ‫2( ﻧﺤﻞ اﻟﻤﻌﺎدﻟﺔ 2 − ‪3 x3 + 2 x2 − 7 x − 6 = −2 x‬‬
                                                                                   ‫4‬
                               ‫1‬
   ‫ﻣﻮازﯾﺎ ﻟﻠﻤﺴﺘﻘﯿﻢ اﻟﺬي ﻣﻌﺎدﻟﺘﮫ ‪ y = x‬ﺣﯿﺚ ﻓﺎﺻﻠﺔ ‪M‬‬                            ‫وﻧﺴﺘﻌﻤﻞ اﻟﺴﺆال اﻟﺴﺎﺑﻖ وﻧﺠﺪ 1− = ‪ x‬أو = ‪x‬‬
                               ‫2‬                                                   ‫3‬
                                                          ‫11‬            ‫إذن اﻟﻨﻘﻄﺔ اﻟﻤﺸﺘﺮﻛﺔ ذات اﻟﺘﺮﺗﯿﺐ ﻣﻌﺪوم ھﻲ ) 0 ;1− (‪A‬‬
                                                        ‫.‬    ‫ھﻲ‬                      ‫3( 1− = ‪ x‬ھﻮ ﺣﻞ ﻣﻀﺎﻋﻒ ﻟﻠﻤﻌﺎدﻟﺔ :‬
                                                           ‫4‬
                            ‫1‬                                         ‫2 − ‪ 3 x3 + 2 x2 − 7 x − 6 = −2 x‬إذن ) ‪ ( D‬ﻣﻤﺎس ِـ‬
                                                                       ‫ﻟ‬
            ‫)ﻹﯾﺠﺎد ھﺬه اﻟﻔﺎﺻﻠﺔ ﻧﺤﻞ اﻟﻤﻌﺎدﻟﺔ = )‪( f ' ( x‬‬
                            ‫2‬                                                                   ‫) ‪ ( C‬ﻓﻲ اﻟﻨﻘﻄﺔ ) 0 ;1− (‪. A‬‬
 ‫4( ﻣﻌﺎدﻟﺔ ﻣﻤﺎس اﻟﻤﻨﺤﻨﻲ ) ‪ (P‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ ‪a‬‬                 ‫65 ﻣﻌﺎدﻟﺔ ﻣﻤﺎس اﻟﻤﻨﺤﻨﻲ ) ‪ ( C‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ) 4 ;2 (‪A‬‬
                            ‫ھﻲ: 4 + 2 ‪y = ( 2a − 5 ) x − a‬‬                                ‫ھﻲ: ) 2 ( ‪y = f ' ( 2 )( x − 2 ) + f‬‬
 ‫5( اﻟﻤﻨﺤﻨﻲ ) ‪ (P‬ﯾﺸﻤﻞ ﻣﻤﺎﺳﯿﻦ ﻛﻞ ﻣﻨﮭﻤﺎ ﯾﺸﻤﻞ اﻟﻤﺒﺪأ إذا‬
                                                                                  ‫ﺑﻤﺎ أن اﻟﻤﻤﺎس ﯾﻮازي ) ∆ ( ﻓﺈن 3 = ) 2 ( ' ‪f‬‬
           ‫ﻛﺎن 0 = 4 + 2 ‪ − a‬أي ) 2− = ‪ ( a‬أو ) 2 = ‪. ( a‬‬
                                                                                              ‫إذن ﻣﻌﺎدﻟﺔ ﻣﻤﺎس ھﻲ 2 − ‪y = 3 x‬‬
  ‫26 1( اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ ¡ و ﻟﺪﯾﻨﺎ ﻣﻦ أﺟﻞ‬                                           ‫‪r‬‬
                                                                       ‫75 ﺑﻤﺎ أن ﺷﻌﺎع ﺗﻮﺟﯿﮫ اﻟﻤﻤﺎس ‪ i‬ﻓﺈﻧﮫ ﯾﻮازي ﺣﺎﻣﻞ‬
                   ‫ﻛﻞ ‪ x‬ﻣﻦ ¡ : 1 − ‪f ' ( x) = 6 x2 + 10 x‬‬                               ‫ﻣﺤﻮر اﻟﻔﻮاﺻﻞ و ﺑﺎﻟﺘﺎﻟﻲ ﻣﻌﺎدﻟﺘﮫ 3− = ‪y‬‬
‫2( اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ ¡ و ﻟﺪﯾﻨﺎ ﻣﻦ أﺟﻞ ﻛﻞ ‪x‬‬                                             ‫) ﺗﺮﺗﯿﺐ اﻟﻨﻘﻄﺔ ‪ A‬ھﻮ 3− (‬
                                                  ‫‪π‬‬                                 ‫85 1( ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪ a‬ﻟﺪﯾﻨﺎ:‬
                              ‫ﻣﻦ ¡ : 1 − ‪f ' ( x) = 2 x cos‬‬
                                                   ‫3‬                                                       ‫) ‪f ( a + h) − f ( a‬‬
                                                                           ‫‪ ، lim‬إذن اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ‬                              ‫3=‬
‫3( اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ [∞+ ;1] و ﻟﺪﯾﻨﺎ ﻣﻦ أﺟﻞ ﻛﻞ‬                                              ‫0→ ‪h‬‬              ‫‪h‬‬
                                        ‫1‬                                                        ‫ﻟﻼﺷﺘﻘﺎق ﻋﻨﺪ ‪ a‬و 3 = ) ‪. f ' ( a‬‬
                       ‫= )‪f ' ( x‬‬             ‫‪ x‬ﻣﻦ [∞+ ;1] :‬
                                    ‫1− ‪2 x‬‬                                                                      ‫2( ‪f ' : x a m‬‬
 ‫4( اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ ¡ و ﻟﺪﯾﻨﺎ ﻣﻦ أﺟﻞ ﻛﻞ ‪x‬‬                                 ‫95 1( ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪ a‬ﻟﺪﯾﻨﺎ:‬
                                     ‫‪2 x4 + 6 x2 + 10 x‬‬                                                 ‫) ‪f ( a + h) − f (a‬‬
                        ‫= )‪f ' ( x‬‬                        ‫ﻣﻦ ¡ :‬      ‫‪ ، lim‬إذن اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ‬                               ‫2 ‪= 3a‬‬
                                                                                                 ‫0→ ‪h‬‬
                                         ‫)1 + ‪( x‬‬                                                                 ‫‪h‬‬
                                                    ‫2‬

                                                                              ‫ﻟﻼﺷﺘﻘﺎق ﻋﻨﺪ ﻛﻞ ‪ x‬ﻣﻦ ¡ و ‪. f ' ( x) = 3x‬‬
                                                                                                ‫2‬
             ‫اﻟﺪاﻟﺔ ‪ x a x‬ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ¡‬  ‫36‬                  ‫2( ﻣﻌﺎدﻟﺔ ﻣﻤﺎس ﻣﻨﺤﻨﻲ اﻟﺪاﻟﺔ ‪ f‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات اﻟﻔﺎﺻﻠﺔ 1‬
         ‫]‬‫و اﻟﺪاﻟﺔ ‪ x a x‬ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ [∞+ ;0‬
                                                                                                               ‫ھﻲ: 2 − ‪y = 3 x‬‬
 ‫ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ [∞+ ;0]‬           ‫و ﺑﺎﻟﺘﺎﻟﻲ اﻟﺪاﻟﺔ ‪x a x x‬‬          ‫06 1( اﻟﺪاﻟﺔ ‪ g‬ھﻲ ﻣﺠﻤﻮع داﻟﺘﯿﻦ ﻗﺎﺑﻠﺘﯿﻦ ﻟﻼﺷﺘﻘﺎق‬
           ‫= )‪f ' ( x‬‬
                        ‫3‬
                          ‫وﻣﻦ أﺟﻞ ﻛﻞ ‪ x‬ﻣﻦ [∞+ ;0] : ‪x‬‬                  ‫ﻋﻠﻰ ¡ ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪f ' ( x) = − x + 2 : x‬‬
                        ‫2‬                                                                                                       ‫2( أ(‬
    ‫−‪f ': x a x‬‬
                   ‫1‬
                      ‫46 1( 4 − ‪(2 ، f ' : x a 6 x‬‬                  ‫) ‪ϕ ( a + h) − ϕ ( a ) f ( a + h) − f ( a ) g ( a + h ) − g ( a‬‬
                                                                                        ‫=‬                    ‫−‬
                   ‫2‬                                                         ‫‪h‬‬                    ‫‪h‬‬                    ‫‪h‬‬
                     ‫3( 2 + ‪f ' : x a x‬‬

‫6‬
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                                         ‫1‬          ‫1‬                       ‫4( 3 + ‪f ' : x a 4 3x3 − 3 2 x2 − 2 6 x‬‬
                        ‫= )‪f ' ( x‬‬            ‫−‬               ‫ﻟﺪﯾﻨﺎ:‬
                                                 ‫)1 + ‪( x‬‬
                                                        ‫2‬
                                     ‫‪2 x‬‬                                                 ‫3‬                     ‫2‬
                                                                        ‫− ‪f ': x a‬‬              ‫56 1( 2 ‪(2 ، f ' : x a‬‬
           ‫، 94 ,1 ≅ ) 69,0 ( ‪f‬‬              ‫505 ,1 ≅ ) 20 ,1( ‪f‬‬                     ‫)2 + ‪( x‬‬
                                                                                              ‫2‬
                                                                                                              ‫‪x‬‬
           ‫1( 4 + ‪، y = 11x + 5 ( 2 ، y = 3 x‬‬                  ‫17‬                 ‫9 − ‪−3x2 −10x‬‬                                   ‫4‬
                              ‫3( 11 + ‪y = −7 x‬‬                         ‫‪f ': x a‬‬                     ‫+ 2 ‪(4 f ' : x a‬‬                      ‫3(‬
                                                                                    ‫)3− ‪( x‬‬                              ‫)3 − ‪( x‬‬
                                                                                              ‫2‬                                       ‫2‬
                                                                                       ‫2‬

         ‫1( ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ) 1‪ (T‬ﻟـ ) 1‪ ( C‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ‬           ‫27‬
                                                                                                          ‫66 1( 2‪f ' : x a 3x‬‬
              ‫) ) 0‪ A( x0 , f ( x‬ھﻲ 3 + 0‪y = −2 x0 x + x‬‬
                                     ‫2‬
                                                                                                      ‫‪g ( x) = f ( x − 3) v‬‬
                ‫و ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ) 2‪ (T‬ﻟـ ) 2‪ ( C‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ‬
                                                                                     ‫و ﻣﻨﮫ ) 3 − ‪g ' ( x) = f ' ( x − 3 ) = 3 ( x‬‬
                                                                                                                              ‫2‬


                   ‫) ) 0‪ A( x0 , g ( x‬ھﻲ + 2 − = ‪y‬‬
                           ‫4 ‪2x‬‬
                                                                                                     ‫‪g ( x) = f ( 2 x + 5) v‬‬
                           ‫0‪x‬‬     ‫0‪x‬‬
                                                                         ‫و ﻣﻨﮫ ) 5 + ‪g ' ( x) = 2 f ' ( 2 x + 5 ) = 2 × 3 ( 2 x‬‬
                                                                                                                          ‫2‬
                         ‫4‬                           ‫2−‬
         ‫ﻓﯿﻜﻮن 1 = 0‪x‬‬       ‫2 = 0‪ −2x‬و 3 + 2 0‪= x‬‬
                        ‫0‪x‬‬                           ‫0‪x‬‬                                    ‫‪ g ( x) = f ( −3 x + 2 ) v‬و ﻣﻨﮫ‬
    ‫إذن ﯾﻮﺟﺪ ﻣﺴﺘﻘﯿﻢ ) ∆ ( ﯾﻤﺲ اﻟﻤﻨﺤﻨﯿﯿﻦ ) 1‪ ( C‬و ) 2‪ ( C‬ﻓﻲ‬             ‫) 2 + ‪g ' ( x) = −3 f ' ( −3 x + 2 ) = −3 × 3 ( −3x‬‬
                                                                                                                                  ‫2‬


                                                  ‫اﻟﻨﻘﻄﺔ ) 2 ;1(‪. A‬‬                                      ‫1( ‪ f ( x ) = x‬و‬             ‫76‬
                        ‫2( ﻣﻌﺎدﻟﺔ ) ∆ ( ھﻲ : 4 + ‪y = −2 x‬‬
                                                                                                  ‫1 − ‪g ( x) = f ( x − 1) = x‬‬
    ‫، ) ∆ ( أﻋﻠﻰ ) 2‪ ( C‬ﻓﻲ [0 ;∞−]‬            ‫3( ) ∆ ( أﻋﻠﻰ ) 1‪( C‬‬           ‫اﻟﺪاﻟﺔ ‪ f‬ﻣﻌﺮﻓﺔ ﻋﻠﻰ [∞+ ;0 [ و اﻟﺪاﻟﺔ ‪ g‬ﻣﻌﺮﻓﺔ ﻋﻠﻰ‬
                       ‫و ) ∆ ( أﺳﻔﻞ ) 2‪ ( C‬ﻓﻲ [∞+ ;0] .‬
                                                                                                                       ‫[∞+ ;1[ .‬
                       ‫‪−α x2 + ( 6 − 2 β ) x + α‬‬                       ‫2( اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻋﻠﻰ [∞+ ;0] و اﻟﺪاﻟﺔ ‪ g‬ﺗﻘﺒﻞ‬
          ‫= )‪f ' ( x‬‬                                    ‫1(‬     ‫37‬
                              ‫‪(x‬‬         ‫)1 +‬
                                             ‫2‬
                                                                                                                ‫اﻻﺷﺘﻘﺎق ﻋﻠﻰ ]∞+ ;1] .‬
                                     ‫2‬


                                            ‫2( 4 = ‪ α‬و 3 = ‪β‬‬                                                                      ‫1‬
                                         ‫47 1− = ‪ α‬و 2 = ‪β‬‬                                                           ‫= )‪f ' ( x‬‬       ‫3(‬
                                                                                                                                ‫‪2 x‬‬
                                                                                                                                  ‫1‬
‫57 ﻧﻨﺎﻗﺶ ﺣﺴﺐ ﻗﯿﻢ ‪ m‬ﻋﺪد ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ 0 = )‪f ' ( x‬‬                                               ‫= )1 − ‪g ' ( x) = f ' ( x‬‬              ‫و‬
                                                                                                                               ‫1− ‪2 x‬‬
                   ‫إذا ﻛﺎن 0 = ‪ m‬ﻓﺈﻧﮫ ﯾﻮﺟﺪ ﻣﻤﺎس واﺣﺪ.‬                            ‫‪ v‬ﻧﺘﺒﻊ ﻧﻔﺲ اﻟﻄﺮﯾﻘﺔ ﻓﻲ اﻟﺤﺎﻟﺘﯿﻦ اﻟﻤﺘﺒﻘﯿﺘﯿﻦ.‬
                      ‫و إذ ﻛﺎن 0 ≠ ‪ m‬ﻓﺈﻧﮫ ﯾﻮﺟﺪ ﻣﻤﺎﺳﺎن.‬
                                                                                             ‫1( ) 2 − ‪، f ' ( x ) = 6 ( 3 x‬‬
    ‫67 1( ‪DE = x tan 60° = x 3 ، DG = m − 2 x‬‬                                                                                        ‫86‬
                                                                                                                           ‫3‬
‫وﻣﻨﮫ ﻣﺴﺎﺣﺔ اﻟﻤﺴﺘﻄﯿﻞ ھﻲ : ‪R ( x) = −2 3x2 + m 3 x‬‬                                                          ‫= )‪f ' ( x‬‬           ‫2(‬
                                                                                                                       ‫‪2 x‬‬
       ‫‪m‬‬                                                                               ‫3−‬
‫=‪x‬‬        ‫3 ‪ R ' ( x) = −4 3 x + m‬؛ 0 = )‪ R ' ( x‬ﻣﻌﻨﺎه‬                 ‫= )‪f ' ( x‬‬                 ‫= ) ‪(4 ، f ' ( x‬‬
                                                                                                                                ‫1‬
                                                                                                                                     ‫3(‬
       ‫4‬                                                                          ‫‪2 2 − 3x‬‬                                  ‫3−‪2 x‬‬
         ‫ﺑﻤﺎ أن )‪ R ( x‬ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﯿﺔ و 0 < 3 2− ﻓﺈن‬                                                                       ‫3 − ‪6x‬‬
                                                                                                            ‫= )‪، f ' ( x‬‬             ‫5(‬
                    ‫‪ R  ‬ھﻲ اﻟﻘﯿﻤﺔ اﻟﺤﺪﯾﺔ اﻟﻜﺒﺮى وﻣﻨﮫ‬
               ‫‪m‬‬                                       ‫‪m‬‬
  ‫= ‪ x‬وﻟﺪﯾﻨﺎ‬                                         ‫‪ ‬‬
                                                                                                                               ‫‪2 x‬‬
               ‫4‬                                     ‫‪4‬‬                                                            ‫51 + ‪−5 x2 + 6 x‬‬
                                                                                                      ‫= )‪f ' ( x‬‬                      ‫6(‬
                                        ‫= ‪. R ‬‬                                                                       ‫3 + ‪2 −x‬‬
                                            ‫‪m‬‬         ‫2 3‬
                                           ‫‪ ‬‬          ‫‪m‬‬
                                           ‫8 ‪4‬‬                                          ‫1( ) 2 − ‪f ' ( x) = −3sin ( 3x‬‬
                                        ‫2( ﻣﺴﺎﺣﺔ اﻟﻤﺜﻠﺚ ھﻲ‬                                                                           ‫96‬
                                                                                                      ‫2( ) 2 − ‪f ' ( x) = 3cos ( 3x‬‬
                             ‫1‬       ‫‪m‬‬
                  ‫= °06 ‪T ( m) = m × tan‬‬
                                                      ‫2 3‬
                                                        ‫‪m‬‬                                            ‫3( ‪f ' ( x) = cos 2 x − sin 2 x‬‬
                             ‫2‬        ‫2‬              ‫4‬
                                                                       ‫4( ) ‪f '( x) =cos( x−2π ) cos( x+π ) −sin( x+π ) sin( x−2π‬‬
                                                ‫) (‪T‬‬
                                     ‫وﻣﻨﮫ ‪R   = m‬‬
                                          ‫‪m‬‬
                                       ‫‪ ‬‬
                                       ‫‪4‬‬         ‫2‬                                           ‫5( ‪f ' ( x) = − 2cos 3 x sin 3x‬‬
                    ‫) 4 ( ' ‪T ( 4, 002 ) ; T ( 4 ) + 0, 002T‬‬              ‫أﻛﺒﺮ ﻣﺠﻤﻮﻋﺔ ﺑﺤﯿﺚ ﺗﻜﻮن اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق‬
                                                                                                                            ‫07‬
                          ‫وﻣﻨﮫ 3 × 400 ,4 ; ) 200 ,4 ( ‪T‬‬                 ‫ھﻲ [∞+ ;0] ، وﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪ x‬ﻣﻦ [∞+ ;0]‬
                              ‫و 3 × 200 ,2 ; )100 ,2 ( ‪R‬‬

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                  ‫) ‪ ( c f‬أﺳﻔﻞ ) ‪(Ta‬‬   ‫إذا ﻛﺎن ‪ x < −2a‬ﻓﺈن‬                                            ‫‪ −8 ‬‬
                                                                                  ‫77 1( ‪ A 0; ‬و ) 0 ; ‪B ( 2m‬‬
                                                                                                      ‫‪ m‬‬
                  ‫) ‪ ( c f‬ﯾﻘﻄﻊ ) ‪(Ta‬‬   ‫إذا ﻛﺎن ‪ x = −2a‬ﻓﺈن‬                            ‫4‬        ‫8‬
                                                                                ‫2( ﻣﻌﺎدﻟﺔ ) ‪ ( AB‬ھﻲ − ‪y = 2 x‬‬
              ‫‪IT‬‬                                                                     ‫‪m‬‬         ‫‪m‬‬
                  ‫28 1( ﻓﻲ اﻟﻤﺜﻠﺚ اﻟﻘﺎﺋﻢ ‪ OIT‬؛ ‪= sin x‬‬                                         ‫4 4−‬           ‫8‬
             ‫‪OT‬‬                                                      ‫ﺗﻘﺒﻞ ﺣﻼ ﻣﻀﺎﻋﻔﺎ ‪x = m‬‬         ‫−‪= 2 x‬‬          ‫اﻟﻤﻌﺎدﻟﺔ‬
                                         ‫‪OI‬‬                                                     ‫‪x m‬‬           ‫‪m‬‬
           ‫، ﺑﻤﺎ أن 1 = ‪ OI‬ﻧﺤﺼﻞ ﻋﻠﻰ‬            ‫و ‪= cos x‬‬                ‫و ﺑﺎﻟﺘﺎﻟﻲ اﻟﻤﺴﺘﻘﯿﻢ ) ‪ ( AB‬ﻣﻤﺎس ﻟﻠﻤﻨﺤﻨﻲ ) ‪ ( H‬ﻓﻲ‬
                                         ‫‪OT‬‬
                                                                                                               ‫اﻟﻨﻘﻄﺔ ‪. M‬‬
                                                  ‫‪sin x‬‬
       ‫= ‪) IT‬ﻻﻧﺪرج ‪ tan‬ﻟﻜﻮﻧﮭﺎ ﻏﯿﺮ ﻣﻮﺟﻮدة ﻓﻲ‬                                                ‫‪−12 − 4h‬‬
                                                  ‫‪cos x‬‬                    ‫= )‪T (h‬‬                            ‫87 1( أ(‬
                                               ‫اﻟﺒﺮﻧﺎﻣﺞ(.‬                            ‫4 + 2‪16 − 12h − 4h‬‬
                                                                                                                   ‫3‬
            ‫1‬
  ‫= ‪. A2 = IT × OI‬‬
                          ‫‪1 sin x‬‬            ‫1‬
                                   ‫2( ‪ A = sin x‬و‬                  ‫ب( − = ) ‪ lim T ( h‬و ﻣﻨﮫ اﻟﺪاﻟﺔ ‪ f‬ﺗﻘﺒﻞ اﻻﺷﺘﻘﺎق ﻣﻦ‬
                                       ‫1‬                                                          ‫0→ ‪h‬‬             ‫2‬
            ‫2‬              ‫‪2 cos x‬‬           ‫2‬
‫‪ A‬ﻣﺴﺎﺣﺔ اﻟﺠﺰء ﻣﻦ اﻟﻘﺮص اﻟﻤﺮﻓﻖ ﻟﻠﺰاوﯾﺔ ‪ ، x‬وﻣﺴﺎﺣﺔ‬                                             ‫‪‬‬ ‫‪3‬‬      ‫3 3‬
                                                                                          ‫و − = ‪f ' ‬‬           ‫أﺟﻞ اﻟﻘﯿﻤﺔ‬
     ‫اﻟﻘﺮص ھﻲ ‪ π R2 = π‬وھﻲ ﻣﺮﻓﻘﺔ ﻟﻠﺰاوﯾﺔ ‪ 2π‬إذن :‬                                            ‫‪2‬‬       ‫2 2‬
                                             ‫1 ‪πx‬‬                                                   ‫‪x‬‬              ‫‪x‬‬
                                       ‫=‪A‬‬         ‫‪= x‬‬                             ‫2( = ‪ v‬و ﻣﻨﮫ = 2 أي ‪x = 2 t‬‬
                                             ‫2 ‪2π‬‬                                                   ‫‪t‬‬              ‫‪t‬‬
                                                                     ‫2‪ OB = AB − OA‬و ﻣﻨﮫ ) ‪OB = 25 − ( 2t‬‬
                                                                         ‫2‬              ‫2‬              ‫2‬        ‫2‬
                        ‫3( ﺑﻤﺎ أن : ‪ A ≤ A ≤ A‬ﻓﺈن :‬
                                ‫1‬          ‫2‬

               ‫‪sin x‬‬         ‫1‬         ‫1‬        ‫‪1 sin x‬‬                                          ‫أي 2 ‪OB = 25 − 4t‬‬
‫≤ ‪sin x ≤ x‬‬           ‫≤ ‪ sin x ≤ x‬أي :‬
                                                                                                     ‫3‬
                                                                            ‫إذا ﻛﺎن 3 = ‪ x‬ﻓﺈن = ‪f ( t ) = 25 − t 2 ، t‬‬
               ‫‪cos x‬‬         ‫2‬         ‫2‬        ‫‪2 cos x‬‬
              ‫‪‬‬    ‫‪π‬‬                           ‫‪sin x‬‬                                                ‫2‬
‫≤ ‪ x‬وﺑﻤﺎ أن ﻓﻲ اﻟﻤﺠﺎل ‪cos x > 0 :  0 ; ‬‬             ‫إذن‬
              ‫‪‬‬    ‫‪2‬‬                           ‫‪cos x‬‬                                            ‫‪3  3‬‬
                                                                                               ‫‪f  + h − f  ‬‬
     ‫ﻓﺈن: ‪ x cos x ≤ sin x‬ﺧﻼﺻﺔ ‪x cos x ≤ sin x ≤ x‬‬                                        ‫‪lim ‬‬         ‫3 − = ‪ 2‬‬
                                                                                                   ‫2‬
      ‫‪ π‬‬                  ‫‪sin x‬‬                                                         ‫0→ ‪h‬‬          ‫‪h‬‬            ‫2‬
‫≤ ‪ cos x‬ﻷن ‪x ∈  0 ; ‬‬            ‫• ﻧﺴﺘﻨﺘﺞ ﻣﻦ ھﺬا أن 1 ≤‬
                                                                                             ‫1( ﻧﻨﺸﺮ )‪ ( R + x‬ﻓﯿﻜﻮن‬
                                                                                                           ‫2‬
      ‫‪ 2‬‬                     ‫‪x‬‬                                                                                    ‫97‬
   ‫4( ﻣﻦ اﻟﺮﺳﻢ ﻧﺨﻤﻦ اﻟﻨﺘﯿﺠﺔ 1 = )‪lim g ( x) = lim h( x‬‬                                                               ‫2‪R‬‬
    ‫0 →‪x‬‬          ‫0→ ‪x‬‬                                                                          ‫× 0‪g = g‬‬
          ‫5( ‪ f ' ( x) = cos x‬وﻣﻨﮫ 1 = 0 ‪f ' ( 0 ) = cos‬‬                                                     ‫‪ 2x  x 2 ‬‬
                                                                                                           ‫‪R 1+ +   ‬‬
                                                                                                             ‫2‬

                          ‫(‪f‬‬        ‫)0( ‪) − f‬‬                                                                ‫‪ R  R ‬‬
           ‫‪ lim 0 + h‬وﻣﻨﮫ :‬                       ‫1 = )0(' ‪= f‬‬                                               ‫‪‬‬           ‫‪‬‬
                     ‫0→ ‪h‬‬         ‫0−‪h‬‬                                                                            ‫1‬
                                                                                          ‫× 0‪g = g‬‬                 ‫2‬
                                                                                                                     ‫و ﻣﻨﮫ‬
                                ‫‪sin x‬‬                  ‫‪sinh‬‬                                             ‫‪2x  x ‬‬
‫‪ lim‬وﻣﻨﮫ : 1 = )‪lim h( x‬‬              ‫‪ lim‬أي 1 =‬             ‫1=‬                                    ‫+1‬      ‫+‬
‫0 →‪x‬‬                       ‫0 →‪x‬‬   ‫‪x‬‬              ‫‪h →0 h‬‬
                                                                                                        ‫‪R  R‬‬
                                                                                                             ‫‪ ‬‬
                                         ‫38 اﻟﻄﺮﯾﻘﺔ اﻷوﻟﻰ :‬                               ‫2‬
                                                                                   ‫‪‬‬  ‫‪x‬‬                 ‫‪2x‬‬         ‫‪2x‬‬
‫1( ) ‪ d (t ) = −5(t 2 − 12t‬؛ 081 + )6 − ‪d (t ) = −5(t‬‬                              ‫+1 و 0 ; ‪ ‬‬              ‫−1 ≅‬       ‫2(‬
                        ‫2‬

                                                                                   ‫‪ R‬‬                   ‫‪R‬‬          ‫‪R‬‬
  ‫وﻣﻨﮫ اﻟﻘﯿﻤﺔ اﻟﺤﺪﯾﺔ اﻟﻌﻈﻤﻰ ﻟﻼرﺗﻔﺎع ھﻲ 081 = )6( ‪. d‬‬
                                                                                                                ‫‪ 2x ‬‬
                      ‫2( اﻟﺴﺮﻋﺔ ﻓﻲ اﻟﻠﺤﻈﺔ 6 ﺗﻜﻮن ﻣﻌﺪوﻣﺔ .‬                                             ‫‪g ; g0 × 1 − ‬‬
                                         ‫اﻟﻄﺮﯾﻘﺔ اﻟﺜﺎﻧﯿﺔ :‬                                                      ‫‪‬‬     ‫‪R‬‬
                                                                                                          ‫3( 587 ,9 ; ‪g‬‬
                                    ‫1( 06 + ‪d ' ( t ) = −10t‬‬
             ‫0 ‪t‬‬                      ‫6‬                   ‫∞+‬               ‫1( ﻧﻨﺸﺮ ) 2 ‪( x − a ) ( x2 + ax − 2a‬‬        ‫08‬
      ‫) ‪d ' (t‬‬              ‫+‬         ‫0‬         ‫-‬                   ‫2( ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ) ‪ (Ta‬ﻟﻠﻤﻨﺤﻨﻲ ) ‪ ( c f‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ذات‬
       ‫) ‪d (t‬‬                         ‫081‬                                                  ‫اﻟﻔﺎﺻﻠﺔ ‪ a‬ھﻲ: 3 ‪y = 3a 2 x − 2a‬‬
                                                                               ‫ﻟﺪﯾﻨﺎ ) ‪( x2 + ax − 2a 2 ) = ( x − a )( x + 2a‬‬
                   ‫0‬
              ‫وﻣﻨﮫ 081 = )6( ‪ d‬ھﻲ اﻟﻘﯿﻤﺔ اﻟﺤﺪﯾﺔ اﻟﻌﻈﻤﻰ .‬
                                                ‫2( 0 = ) 6 ( ' ‪d‬‬   ‫- ﻟﺪراﺳﺔ اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟـ ) ‪ ( c f‬و ) ‪ (Ta‬ﻧﺪرس إﺷﺎرة اﻟﻌﺪد‬
          ‫48 1( ﻟﺪﯾﻨﺎ ) ‪DC = f ( x0 + h ) − f ( x0 − h‬‬                                                   ‫) ‪( x − a )2 ( x + 2 a‬‬
‫و ‪ BD = 2h‬وﻣﻨﮫ ‪S = h  f ( x0 + h ) − f ( x0 − h ) ‬‬
      ‫‪‬‬                             ‫‪‬‬                                               ‫) ‪ ( c f‬أﻋﻠﻰ ) ‪(Ta‬‬   ‫إذا ﻛﺎن ‪ x > −2a‬ﻓﺈن‬


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                  ‫‪S = h  f ( x0 ) + hf ' ( x0 ) − f ( x0 ) + hf ' ( x0 ) ‬‬
                        ‫‪‬‬                                                 ‫‪‬‬
                                                 ‫أي : ) 0‪S = 2h f ' ( x‬‬
                                                          ‫2‬


                               ‫2( 2610 ,0 = 9 × )30 ,0 ( × 2 = ‪. S‬‬
                                                   ‫2‬


                   ‫58 1( أﺣﺴﻦ ﺗﻘﺮﯾﺐ ﺗﺂﻟﻔﻲ ﻟﻠﺪاﻟﺔ ‪ f‬ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد‬
                 ‫ﺣﻘﯿﻘﻲ ‪ x‬ھﻮ )‪ f ( x + h ) = f ( x) + hf ' ( x‬وﻣﻦ أﺟﻞ‬
                      ‫)‪ ( − x‬ﻟﺪﯾﻨﺎ )‪f ( − x − h ) = f ( − x) − hf ' ( − x‬‬
                   ‫ﺑﻤﺎ أن ‪ f‬زوﺟﯿﺔ ﻧﺤﺼﻞ ﻋﻠﻰ ) ‪. f ' ( x) = − f ' ( − x‬‬
                                                              ‫‪4x‬‬
                             ‫= )‪ g ' ( x‬وﻣﻨﮫ 1 = )1( ' ‪g‬‬               ‫2(‬
                                                          ‫2)1 + ‪( x‬‬
                                                             ‫2‬

                                ‫وﻟﺪﯾﻨﺎ 0 = )1( ‪ g‬إذن اﻟﻤﻌﺎدﻟﺔ 1 − ‪y = x‬‬
                                ‫اﻻﺳﺘﻨﺘﺎج ‪ g‬زوﺟﯿﺔ وﻣﻨﮫ ' ‪ g‬ﻓﺮدﯾﺔ إذن :‬
                    ‫1− = )1( ' ‪ g ' ( −1) = − g‬و 0 = )1( ‪g ( −1) = g‬‬
                                              ‫واﻟﻤﻌﺎدﻟﺔ ھﻲ : 1 − ‪y = − x‬‬




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posted:4/7/2011
language:Arabic
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