Ngā tauira pātai me te matapakinga mō ngā āhuatanga o ngā mahi taupū
Ko te pānga o tētahi mahi he ariā taketake i roto i te tātaitai e tino whai hua ana mō te tātari i te whanonga o ētahi mahi. I roto i tēnei tuhinga, ka matapakihia e mātou ētahi tauira raruraru me ngā āhuatanga o te pānga o tētahi mahi.
Kupu Whakataki ki ngā Pānga Taurite
Ko te pānga o tētahi mahi \( f \) ka whakaaturia ko \( f'(x) \). Ko te pānga tuatahi o tētahi mahi ka hoatu i te tere o te huringa o te mahi e pā ana ki tōna taurangi motuhake. Ko tētahi atu kupu e whakamahia ana ko te rerekētanga. Mena ko \( y = f(x) \), ko te pānga o \( f \) e pā ana ki \( x \) ko:
\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) – f(x)}{h} \]
Ngā Āhuatanga o ngā Pānga Taurite
Ko ētahi āhuatanga nui o te taupū o tētahi mahi ko:
1. Te Raina: Mena he mahi rerekētanga a \( f(x) \) me \( g(x) \) , ā, he pūmau a \( c \), kāti:
\[
\frac{d}{dx} [cf(x) + g(x)] = c f'(x) + g'(x)
\]
2. Ture Mekameka: Mō te mahi hiato \( g(f(x)) \):
\[
\frac{d}{dx} g(f(x)) = g'(f(x)) \cdot f'(x)
\]
3. Hua: Mō ngā mahi \( u(x) \) me \( v(x) \):
\[
\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
4. Tauwehenga: Mō ngā mahi \( u(x) \) me \( v(x) \) kei reira \( v(x) \neq 0 \):
\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) – u(x)v'(x)}{(v(x))^2}
\]
Ngā Pātai Tauira me te Kōrero
Tauira 1: Te Whakatau i te Pānga o tētahi Mahi Māmā
Me kī \( f(x) = 3x^2 + 5x – 4 \). Tātaihia te pānga o te mahi.
Otinga:
Ka whakamahia e mātou ngā ture taketake mō te wehewehe.
\[
f(x) = 3x^2 + 5x – 4
\]
Te taupū tuatahi:
\[
f'(x) = \frac{d}{dx} (3x^2) + \frac{d}{dx} (5x) – \frac{d}{dx} (4)
\]
Te tatau i ia pārōnaki:
\[
\frac{d}{dx} (3x^2) = 6x
\]
\[
\frac{d}{dx} (5x) = 5
\]
\[
\frac{d}{dx} (4) = 0
\]
Nō reira:
\[
f'(x) = 6x + 5
\]
Tauira 2: Te Whakamahi i te Ture Mekameka
I hoatu te mahi \( y = (2x^3 – x^2 + 1)^5 \). Whakatauhia te pānga o te mahi.
Otinga:
Whakamahia te ture mekameka. Mēnā ko \( u = 2x^3 – x^2 + 1 \), ka taea te tuhi anō i te mahi kia \( y = u^5 \).
Tuatahi, kimihia te taupū o \( y \) e pā ana ki \( u \):
\[
\frac{dy}{du} = 5u^4
\]
Muri iho, kimihia te pānga o \( u \) e pā ana ki \( x \):
\[
u = 2x^3 – x^2 + 1
\]
\[
\frac{du}{dx} = 6x^2 – 2x
\]
Whakakotahitia ngā pānga e rua ki te ture mekameka:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 5u^4 \cdot (6x^2 – 2x)
\]
Whakakapia anō \( u = 2x^3 – x^2 + 1 \):
\[
\frac{dy}{dx} = 5(2x^3 – x^2 + 1)^4 \cdot (6x^2 – 2x)
\]
Tauira 3: Te Whakamahi i ngā Ture Hua
Homai \( f(x) = x^2 e^x \). Whakatauhia te pānga o te mahi.
Otinga:
Whakamahia te ture hua, arā, mēnā ko \( u(x) = x^2 \) me \( v(x) = e^x \), kātahi:
\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]
Tuatahi, tatauhia ngā pānga o \( u(x) \) me \( v(x) \):
\[
u(x) = x^2 \e tohu ana u'(x) = 2x
\]
\[
v(x) = e^x \e kīia ana ko v'(x) = e^x
\]
Mā te whakamahi i ngā ture mō te hua:
\[
f'(x) = 2x \cdot e^x + x^2 \cdot e^x = e^x (2x + x^2)
\]
Tauira 4: Te Whakamahi i te Ture Huarahi
Homai \( f(x) = \frac{x^2 + 1}{x + 2} \). Kimihia te pānga o te mahi.
Otinga:
Whakamahia te ture haurua, arā, mēnā ko \( u(x) = x^2 + 1 \) me \( v(x) = x + 2 \), kātahi:
\[
f'(x) = \frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}
\]
Tuatahi, tatauhia ngā pānga o \( u(x) \) me \( v(x) \):
\[
u(x) = x^2 + 1 \e kīia ana u'(x) = 2x
\]
\[
v(x) = x + 2 \e kīia ana ko v'(x) = 1
\]
Mā te whakamahi i te ture haurua:
\[
f'(x) = \frac{2x(x + 2) – (x^2 + 1)(1)}{(x + 2)^2}
\]
\[
f'(x) = \frac{2x^2 + 4x – x^2 – 1}{(x + 2)^2}
\]
\[
f'(x) = \frac{x^2 + 4x – 1}{(x + 2)^2}
\]
Whakamutunga
I roto i te tātaitai, he mea nui te mārama ki te ariā taketake o ngā pāngarau me ō rātou āhuatanga hei whakaoti rapanga pāngarau. E whakarāpopoto ana tēnei tuhinga i ētahi tikanga mō te whakaputa i ngā mahi mā te whakaatu i te whakamahinga o ngā ture taketake pēnei i te rārangi, ngā mekameka, ngā hua, me ngā tauwehenga mā roto i ētahi tauira me ngā kōrero taipitopito. Mā te mārama me te mahi auau i ngā pāngarau, ka taea e tātou te matatau ake ki te tātari i ngā huringa o ngā mahi i roto i ngā horopaki rerekē.
