Tambayoyi da Tattaunawa game da Dokar Sarka a cikin Abubuwan da aka samo
Dokar sarka tana ɗaya daga cikin mahimman ra'ayoyi a cikin lissafin bambancin lissafi, wanda ake amfani da shi don ƙididdige asalin aikin da ya ƙunshi ayyuka biyu ko fiye. A cikin wannan labarin, za mu tattauna ainihin manufar dokar sarka, yadda ake amfani da ita, da misalan amfani da ita a cikin matsalolin da suka samo asali waɗanda galibi ke tasowa a makarantar sakandare da kwaleji.
1. Gabatarwa ga Dokar Sarka
Kafin mu shiga cikin matsalar misali, bari mu fara fahimtar menene ƙa'idar sarkar. Ƙa'idar sarkar ta bayyana cewa idan muna da ayyuka biyu masu bambancewa \( f \) da \( g \), kuma muna son nemo asalin abun da ke cikin ayyukan \( h = f(g(x)) \), to asalin \( h \) shine:
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
A taƙaice, muna ƙididdige abin da aka samo daga aikin waje akan g(x), sannan mu ninka sakamakon da abin da aka samo daga aikin ciki \( g(x) \).
2. Fahimtar Aikin Tsarin
Kafin mu shiga cikin matsalolin misalan, yana da mahimmanci a fahimci ayyukan haɗin. Aikin haɗin aiki aiki ne da aka samu ta hanyar saka aiki ɗaya a cikin wani. Misali, idan muna da \( f(x) = \sin(x) \) da \( g(x) = x^2 \), to tsarin ayyukan biyu zai zama \( h(x) = f(g(x)) = \sin(x^2) \).
A cikin ayyukan haɗin kai, sau da yawa muna tunanin \( g(x) \) a matsayin "aikin ciki" da \( f(x) \) a matsayin "aikin waje". A cikin wannan misalin, aikin ciki shine \( x^2 \) kuma aikin waje shine sine.
3. Tambayoyi da Tattaunawa Misali
Bari mu duba wasu misalan matsaloli da ke amfani da ƙa'idar sarka don magance su.
Misali na 1:
Idan aka yi la'akari da aikin \( y = \cos(3x^2) \), nemo farkon abin da aka samo daga y dangane da x.
Tattaunawa:
Da farko, mun gano ayyukan ciki da na waje. A nan, aikin ciki shine \( g(x) = 3x^2 \) kuma aikin waje shine \( f(g) = \cos(g) \).
Mun sani:
1. \( g'(x) = 6x \)
2. \( f'(g) = -\sin(g) \)
Ta hanyar tsarin sarkar, mun sami:
\[ y' = f'(g(x)) \cdot g'(x) = -\sin(3x^2) \cdot 6x \]
Don haka, abin da aka samo daga \( y = \cos(3x^2) \) shine:
\[y' = -6x \sin(3x^2) \]
Misali na 2:
Nemo farkon abin da aka samo daga \( h(x) = e^{5x^3 + 2x} \).
Tattaunawa:
A nan aikin ciki shine \( g(x) = 5x^3 + 2x \) kuma aikin waje shine \( f(g) = e^g \).
Mun sani:
1. \( g'(x) = 15x^2 + 2 \)
2. \( f'(g) = e^g \)
Ta hanyar tsarin sarkar, mun sami:
\[ h'(x) = f'(g(x)) \cdot g'(x) = e^{5x^3 + 2x} \cdot (15x^2 + 2) \]
Don haka, abin da aka samo daga \( h(x) = e^{5x^3 + 2x} \) shine:
[h'(x) = (15x^2 + 2)e^{5x^3 + 2x} \]
Misali na 3:
Nemo farkon abin da aka samo daga \( y = \ln(4x^2 – 5) \).
Tattaunawa:
Aikin ciki shine \( g(x) = 4x^2 – 5 \) kuma aikin waje shine \( f(g) = \ln(g) \).
Mun sani:
1. \( g'(x) = 8x \)
2. \( f'(g) = \frac{1}{g} \)
Ta hanyar tsarin sarkar, mun sami:
\[y' = f'(g(x)) \cdot g'(x) = \frac{1}{4x^2 – 5} \cdot 8x \]
Don haka, abin da aka samo daga \( y = \ln(4x^2 – 5) \) shine:
\[y' = \frac{8x}{4x^2 – 5} \]
Misali na 4:
Idan aka yi la'akari da aikin \( y = (3x^2 + 2x + 1)^4 \), nemo asalinsa.
Tattaunawa:
Aikin ciki shine \( g(x) = 3x^2 + 2x + 1 \) kuma aikin waje shine \( f(g) = g^4 \).
Mun sani:
1. \( g'(x) = 6x + 2 \)
2. \( f'(g) = 4g^3 \)
Ta hanyar tsarin sarkar, mun sami:
\[ y' = f'(g(x)) \cdot g'(x) = 4(3x^2 + 2x + 1)^3 \cdot (6x + 2) \]
Don haka, abin da aka samo daga \( y = (3x^2 + 2x + 1)^4 \) shine:
\[ y' = 4(3x^2 + 2x + 1)^3 (6x + 2) \]
4. Shari'o'i na Musamman da Ci gaban Dokokin Sarka
Wani lokaci, ƙa'idar sarkar ba ta tsaya a kan haɗakar ayyuka biyu kawai ba. Akwai lokutan da aiki ya zama haɗin ayyuka fiye da biyu, misali: \( h(x) = f(g(k(x))) \).
Dangane da ayyuka uku, ana iya amfani da ƙa'idar sarkar a cikin yadudduka:
\[ h'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x) \]
Za mu iya ganin cewa a cikin kowane layi, muna ƙididdige abubuwan da suka samo asali daga cikin yadudduka na waje kafin mu ci gaba zuwa abubuwan da suka samo asali daga cikin yadudduka na ciki.
Misali na 5:
Idan aka ba da \( y = \sqrt{\ln(2x^2 + 1)} \), sami asalinsa.
Tattaunawa:
Aikin ciki shine \( k = 2x^2 + 1 \), tsakiya: \( g = \ln(k) \) da waje: \( f = \sqrt{g} \).
Mun sani:
1. \( k'(x) = 4x \)
2. \( g'(k) = \frac{1}{k} \)
3. \( f'(g) = \frac{1}{2\sqrt{g}} \)
Bari mu yi amfani da ƙa'idar sarkar a cikin yadudduka:
\[ y' = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x) = \frac{1}{2\sqrt{\ln(2x^2 + 1)}} \cdot \frac{1}{2x^2 + 1} \cdot 4x \]
Don haka abin da aka samo daga \( y = \sqrt{\ln(2x^2 + 1)} \) shine:
\[ y' = \frac{4x}{2(2x^2 + 1)\sqrt{\ln(2x^2 + 1)}} \]
5. Kesimpulan
Dokar sarka tana taka muhimmiyar rawa a cikin lissafin bambanci, musamman lokacin da ake mu'amala da tsarin ayyuka. Fahimtar da kuma sanin ƙa'idar sarka tana ba da tushe mai ƙarfi don magance matsaloli masu rikitarwa a cikin lissafin. Wannan labarin ya tattauna misalai da yawa masu mahimmanci don samar da fahimtar aiki da ƙa'idar sarka ga abubuwan da aka samo asali. Muna fatan wannan tattaunawar za ta taimaka wa ɗalibai kuma za a iya amfani da ita ga yanayi daban-daban na lissafi masu ƙalubale.
