Imibuzo Eyisibonelo kanye Nengxoxo Yomthetho We-Chain ku-Derivatives
Umthetho weketanga ungomunye wemibono eyisisekelo kakhulu ekubaleni okuhlukile, osetshenziselwa ukubala i-derivative yomsebenzi owakhiwe imisebenzi emibili noma ngaphezulu. Kulesi sihloko, sizoxoxa ngomqondo oyisisekelo womthetho weketanga, ukuthi ungawusebenzisa kanjani, kanye nezibonelo zokusetshenziswa kwawo ezinkingeni ze-derivative ezivame ukuvela esikoleni samabanga aphezulu nasekolishi.
1. Isingeniso kuMthetho Wezinkinobho
Ngaphambi kokuba singene enkingeni yesibonelo, ake siqale siqonde ukuthi uyini umthetho weketanga. Umthetho weketanga uthi uma sinemisebenzi emibili ehlukene \( f \) kanye \( g \), futhi sifuna ukuthola i-derivative yokwakheka kwemisebenzi \( h = f(g(x)) \), khona-ke i-derivative ka \( h \) ithi:
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
Ngamagama alula, sibala i-derivative yomsebenzi wangaphandle ku-g(x), bese siphindaphinda umphumela nge-derivative yomsebenzi wangaphakathi \( g(x) \).
2. Ukuqonda Umsebenzi Wokwakheka
Ngaphambi kokuthi singene ezinkingeni zesibonelo, kubalulekile ukuqonda imisebenzi yokwakheka. Umsebenzi wokwakheka ngumsebenzi otholakala ngokufaka umsebenzi owodwa komunye. Isibonelo, uma sine-\( f(x) = \sin(x) \) kanye ne-\( g(x) = x^2 \), khona-ke ukwakheka kwemisebenzi emibili kuzoba yi-\( h(x) = f(g(x)) = \sin(x^2) \).
Emisebenzini yokwakheka, sivame ukucabanga ngo-\( g(x) \) “njengomsebenzi wangaphakathi” kanye no-\( f(x) \) “njengomsebenzi wangaphandle”. Kulesi sibonelo, umsebenzi wangaphakathi ngu-\( x^2 \) kanti umsebenzi wangaphandle uyi-sine.
3. Imibuzo Yesibonelo Nengxoxo
Ake sibheke ezinye zezibonelo zezinkinga ezisebenzisa umthetho we-chain ukuzixazulula.
Isibonelo 1:
Njengoba unikezwe umsebenzi \( y = \cos(3x^2) \), thola i-derivative yokuqala ka-y maqondana no-x.
Ingxoxo:
Okokuqala, sithola imisebenzi yangaphakathi neyangaphandle. Lapha, umsebenzi wangaphakathi ngu-\( g(x) = 3x^2 \) kanti umsebenzi wangaphandle ngu-\( f(g) = \cos(g) \).
Siyazi:
1. \( g'(x) = 6x \)
2. \( f'(g) = -\sin(g) \)
Ngokomthetho weketanga, sithola:
\[ y' = f'(g(x)) \cdot g'(x) = -\sin(3x^2) \cdot 6x \]
Ngakho-ke, i-derivative ye-\( y = \cos(3x^2) \) ithi:
\[ y' = -6x \sin(3x^2) \]
Isibonelo 2:
Thola i-derivative yokuqala ye- \( h(x) = e^{5x^3 + 2x} \).
Ingxoxo:
Lapha umsebenzi wangaphakathi ngu-\( g(x) = 5x^3 + 2x \) kanti umsebenzi wangaphandle ngu-\( f(g) = e^g \).
Siyazi:
1. \( g'(x) = 15x^2 + 2 \)
2. \( f'(g) = e^g \)
Ngokomthetho weketanga, sithola:
\[ h'(x) = f'(g(x)) \cdot g'(x) = e^{5x^3 + 2x} \cdot (15x^2 + 2) \]
Ngakho-ke, i-derivative ye- \( h(x) = e^{5x^3 + 2x} \) ithi:
\[ h'(x) = (15x^2 + 2)e^{5x^3 + 2x} \]
Isibonelo 3:
Thola i-derivative yokuqala ye- \( y = \ln(4x^2 - 5) \).
Ingxoxo:
Umsebenzi wangaphakathi ngu-\( g(x) = 4x^2 – 5 \) kanti umsebenzi wangaphandle ngu-\( f(g) = \ln(g) \).
Siyazi:
1. \( g'(x) = 8x \)
2. \( f'(g) = \frac{1}{g} \)
Ngokomthetho weketanga, sithola:
\[ y' = f'(g(x)) \cdot g'(x) = \frac{1}{4x^2 – 5} \cdot 8x \]
Ngakho-ke, i-derivative ye-\( y = \ln(4x^2 – 5) \) ithi:
\[ y' = \frac{8x}{4x^2 – 5} \]
Isibonelo 4:
Uma ubheka umsebenzi \( y = (3x^2 + 2x + 1)^4 \), thola i-derivative yayo.
Ingxoxo:
Umsebenzi wangaphakathi ngu-\( g(x) = 3x^2 + 2x + 1 \) kanti umsebenzi wangaphandle ngu-\( f(g) = g^4 \).
Siyazi:
1. \( g'(x) = 6x + 2 \)
2. \( f'(g) = 4g^3 \)
Ngokomthetho weketanga, sithola:
\[ y' = f'(g(x)) \cdot g'(x) = 4(3x^2 + 2x + 1)^3 \cdot (6x + 2) \]
Ngakho-ke, i-derivative ye- \( y = (3x^2 + 2x + 1)^4 \) ithi:
\[ y' = 4(3x^2 + 2x + 1)^3 (6x + 2) \]
4. Amacala Akhethekile Nokuthuthukiswa Kwemithetho Yezinkinobho
Ngezinye izikhathi, umthetho weketanga awugcini ekwakhiweni kwemisebenzi emibili kuphela. Kunezikhathi lapho umsebenzi ukwakhiwa kwemisebenzi engaphezu kwemibili, isibonelo: \( h(x) = f(g(k(x))) \).
Uma kwenzeka imisebenzi emithathu, umthetho weketanga ungasetshenziswa ngezendlalelo:
\[ h'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x) \]
Singabona ukuthi kungqimba ngayinye, sibala ama-derivatives ezingqimbeni zangaphandle ngaphambi kokudlulela kuma-derivatives ezingqimbeni zangaphakathi.
Isibonelo 5:
Uma unikezwe i-\( y = \sqrt{\ln(2x^2 + 1)} \), thola i-derivative yayo.
Ingxoxo:
Umsebenzi ongaphakathi kakhulu ngu-\( k = 2x^2 + 1 \), ophakathi: \( g = \ln(k) \) kanye nowangaphandle: \( f = \sqrt{g} \).
Siyazi:
1. \( k'(x) = 4x \)
2. \( g'(k) = \frac{1}{k} \)
3. \( f'(g) = \frac{1}{2\sqrt{g}} \)
Ake sisebenzise umthetho weketanga ngezendlalelo:
\[ y' = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x) = \frac{1}{2\sqrt{\ln(2x^2 + 1)}} \cdot \frac{1}{2x^2 + 1} \cdot 4x \]
Ngakho-ke i-derivative ye-\( y = \sqrt{\ln(2x^2 + 1)} \) ithi:
\[ y' = \frac{4x}{2(2x^2 + 1)\sqrt{\ln(2x^2 + 1)}} \]
5. Isiphetho
Umthetho weketanga udlala indima ebalulekile ekubaleni okuhlukile, ikakhulukazi lapho kukhulunywa ngokwakhiwa kwemisebenzi. Ukuqonda nokuqonda kahle umthetho weketanga kunikeza isisekelo esiqinile sokubhekana nezinkinga eziyinkimbinkimbi kakhulu ekubaleni. Lesi sihloko sixoxe ngezibonelo eziningana ezibalulekile ukuze sinikeze ukuqonda okuqinile kokusetshenziswa komthetho weketanga kuma-derivatives. Sithemba ukuthi le ngxoxo iwusizo kubafundi futhi ingasetshenziswa ezimweni ezahlukahlukene zezibalo eziyinselele.
