Misalan tambayoyi da tattaunawa kan halayen ayyukan da aka samo asali
Asalin aikin wani muhimmin ra'ayi ne a cikin lissafi wanda yake da matuƙar amfani wajen nazarin halayen wasu ayyuka. A cikin wannan labarin, za mu tattauna misalai da dama na matsaloli kuma mu tattauna halayen asalin aikin.
Gabatarwa ga Abubuwan da aka samo daga Aiki
An bayyana asalin aikin \( f \) a matsayin \( f'(x) \). Asalin farko na aikin yana ba da ƙimar canjin aikin dangane da canjinsa mai zaman kansa. Wani kalma da ake amfani da shi akai-akai shine bambanci. Idan \( y = f(x) \), to asalin \( f \) dangane da \( x \) shine:
\[f'(x) = \lim_{{h \to 0}} \frac{f(x+h) – f(x)}{h} \]
Halayen Abubuwan da Aka Samu na Aiki
Wasu muhimman halaye na asali na aikin sune:
1. Layi: Idan \( f(x) \) da \( g(x) \) ayyuka ne masu bambancewa, kuma \( c \) madaidaci ne, to:
\[
\frac{d}{dx} [cf(x) + g(x)] = c f'(x) + g'(x)
\]
2. Dokar Sarka: Don aikin haɗaka \( g(f(x)) \):
\[
\frac{d}{dx} g(f(x)) = g'(f(x)) \cdot f'(x)
\]
3. Samfura: Don ayyukan \( u(x) \) da \( v(x) \):
\[
\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
4. Adadin kuɗi: Don ayyukan \( u(x) \) da \( v(x) \) inda \( v(x) \neq 0 \):
\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) – u(x)v'(x)}{(v(x))^2}
\]
Tambayoyi da Tattaunawa Samfura
Misali na 1: Tantance Ma'anar Aiki Mai Sauƙi
A ce \( f(x) = 3x^2 + 5x – 4 \). Kayyade abin da aka samo daga aikin.
Mafita:
Za mu yi amfani da ƙa'idodin asali na bambance-bambance.
\[
f(x) = 3x^2 + 5x – 4
\]
Asalin farko:
\[
f'(x) = \frac{d}{dx} (3x^2) + \frac{d}{dx} (5x) – \frac{d}{dx} (4)
\]
Lissafin kowane abin da aka samo asali:
\[
\frac{d}{dx} (3x^2) = 6x
\]
\[
\frac{d}{dx} (5x) = 5
\]
\[
\frac{d}{dx} (4) = 0
\]
Don haka:
\[
f'(x) = 6x + 5
\]
Misali na 2: Amfani da Dokar Sarka
Idan aka ba da aikin \( y = (2x^3 – x^2 + 1)^5 \). Kayyade abin da aka samo daga aikin.
Mafita:
Yi amfani da ƙa'idar sarkar. A ce \( u = 2x^3 – x^2 + 1 \), to za a iya sake rubuta aikin kamar \( y = u^5 \).
Da farko, nemo wanda aka samo daga \( y \) dangane da \( u \):
\[
\frac{dy}{du} = 5u^4
\]
Na gaba, nemo wanda aka samo daga \(u \) dangane da \( x \):
\[
u = 2x^3 – x^2 + 1
\]
\[
\frac{du}{dx} = 6x^2 – 2x
\]
Haɗa nau'ikan biyu tare da ƙa'idar sarkar:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 5u^4 \cdot (6x^2 – 2x)
\]
Sake maye gurbin \( u = 2x^3 – x^2 + 1 \):
\[
\frac{dy}{dx} = 5(2x^3 – x^2 + 1)^4 \cdot (6x^2 – 2x)
\]
Misali na 3: Amfani da Dokokin Samfura
An bayar da \( f(x) = x^2 e^x \). Kayyade abin da aka samo daga aikin.
Mafita:
Yi amfani da ƙa'idar samfurin, wato, idan \( u(x) = x^2 \) da \( v(x) = e^x \), to:
\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]
Da farko, ƙididdige abubuwan da suka samo asali daga \( u(x) \) da \( v(x) \):
\[
u(x) = x^2 \yana nufin u'(x) = 2x
\]
\[
v(x) = e^x \yana nufin v'(x) = e^x
\]
Ta hanyar amfani da ƙa'idodin samfurin:
\[
f'(x) = 2x \cdot e^x + x^2 \cdot e^x = e^x (2x + x^2)
\]
Misali na 4: Amfani da Dokar Kuɗi
An bayar da \( f(x) = \frac{x^2 + 1}{x + 2} \). Nemo abin da aka samo daga aikin.
Mafita:
Yi amfani da ƙa'idar rabo, wato idan \( u(x) = x^2 + 1 \) da kuma \( v(x) = x + 2 \), to:
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\]
Da farko, ƙididdige abubuwan da suka samo asali daga \( u(x) \) da \( v(x) \):
\[
u(x) = x^2 + 1 \yana nufin u'(x) = 2x
\]
\[
v(x) = x + 2 \yana nufin v'(x) = 1
\]
Ta hanyar amfani da ƙa'idar rabo:
\[
f'(x) = \frac{2x(x + 2) – (x^2 + 1)(1)}{(x + 2)^2}
\]
\[
f'(x) = \frac{2x^2 + 4x – x^2 – 1}{(x + 2)^2}
\]
\[
f'(x) = \frac{x^2 + 4x – 1}{(x + 2)^2}
\]
Kammalawa
A cikin lissafi, fahimtar ainihin ra'ayin abubuwan da suka samo asali da halayensu yana da mahimmanci don magance matsalolin lissafi daban-daban. Wannan labarin ya taƙaita hanyoyi da dama don samar da ayyuka ta hanyar nuna amfani da ƙa'idodi na asali kamar layi, sarƙoƙi, samfura, da kwatance ta hanyar misalai da tattaunawa dalla-dalla. Ta hanyar fahimtar da kuma yin amfani da abubuwan da suka samo asali akai-akai, za mu iya zama ƙwararru wajen nazarin canje-canje a cikin ayyuka a cikin yanayi daban-daban.
