He tauira pātai kōrero mō te pānga o tētahi mahi taurangi
Ko te tātaitanga i roto i te tātaitai he ariā taketake e whakamahia ana hei whakaahua i te huringa o tētahi mahi, i te pikinga rānei o tētahi mahi i tētahi pūwāhi. He whai hua ngā tātaitanga i roto i ngā momo mara pērā i te ahupūngao, te ōhanga, me te hangarau nā te mea ka whakaratohia he mōhiohio mō te tere o te huringa. I roto i tēnei tuhinga, ka matapakihia e mātou ētahi tauira o ngā tātaitanga o ngā mahi taurangi me pēhea te whakaoti i aua mea.
Tauira 1: Te Pūtake o te Mahi Pūrau
Pātai: Ki te hoatu te mahi \( f(x) = 3x^3 – 5x^2 + 2x – 7 \). Whakatauhia te pānga o te mahi!
Otinga:
Mā te whakamahi i te ture taketake o ngā tātaitanga mō ngā mahi pūrau, arā, \(\frac{d}{dx} x^n = nx^{n-1} \), ka tatauhia e tātou te tātaitanga o ia wāhanga o te mahi takitahi.
\[
\begin{align}
f(x) &= 3x^3 – 5x^2 + 2x – 7 \\
f'(x) &= \frac{d}{dx}(3x^3) – \frac{d}{dx}(5x^2) + \frac{d}{dx}(2x) – \frac{d}{dx}(7) \\
f'(x) &= 3 \cdot 3x^{3-1} – 5 \cdot 2x^{2-1} + 2 \cdot 1x^{1-1} – 0 \\
f'(x) &= 9x^2 – 10x + 2.
\end{whakahāngai}
\]
Nō reira, ko te pānga o \( f(x) = 3x^3 – 5x^2 + 2x – 7 \) ko \( f'(x) = 9x^2 – 10x + 2 \).
Tauira 2: Te Pānga o tētahi Mahi me ngā Taupū Hautau
Pātai: Whakatauhia te pānga o te mahi \( g(x) = x^{3/2} + x^{1/2} \).
Otinga:
Mā te whakamahi i te ture taua mō te whakaputanga, arā, \(\frac{d}{dx} x^n = nx^{n-1} \):
\[
\begin{align}
g(x) &= x^{3/2} + x^{1/2} \\
g'(x) &= \frac{d}{dx}(x^{3/2}) + \frac{d}{dx}(x^{1/2}) \\
g'(x) &= \frac{3}{2}x^{(3/2)-1} + \frac{1}{2}x^{(1/2)-1} \\
g'(x) &= \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2}.
\end{whakahāngai}
\]
Nō reira, ko te pānga o \( g(x) = x^{3/2} + x^{1/2} \) ko \( g'(x) = \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2} \).
Tauira 3: Ngā Pānga o ngā Mahi Taupū me te Pānga-toru
Pātai: Whakatauhia te pānga o te mahi \( h(x) = e^x \cdot \sin(x) \).
Otinga:
Hei whakaoti i tēnei taupū, me whakamahi tātou i te Ture Hua, e kī ana \((uv)' = u'v + uv'\). Me kī \( u(x) = e^x \) me \( v(x) = \sin(x) \), kātahi:
\[
\begin{align}
u'(x) &= e^x, & \text{nā te mea ko te pānga o } e^x \text{ ko } e^x \\
v'(x) &= \cos(x), & \text{nā te mea ko te pānga o } \sin(x) \text{ ko } \cos(x).
\end{whakahāngai}
\]
Mā te whakamahi i te ture i ahu mai mō ngā hua:
\[
\begin{align}
h'(x) &= (e^x \cdot \sin(x))' \\
&= e^x \cdot (\sin(x))' + \sin(x) \cdot (e^x)' \\
&= e^x \cdot \cos(x) + \sin(x) \cdot e^x \\
&= e^x (\cos(x) + \sin(x)).
\end{whakahāngai}
\]
Nō reira, ko te pānga o \( h(x) = e^x \sin(x) \) ko \( h'(x) = e^x (\cos(x) + \sin(x)) \).
Tauira 4: Te Whakapūtanga o tētahi Mahi mā te Whakamahi i te Ture Mekameka
Pātai: Whakatauhia te pānga o te mahi \( k(x) = (3x^2 – x + 4)^5 \).
Otinga:
Hei whakaoti i tēnei tauwehenga, me whakamahi tātou i te ture mekameka, arā, \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Me kī ko \( u(x) = 3x^2 – x + 4 \) me \( f(u) = u^5 \), kātahi:
\[
\begin{align}
k(x) &= (3x^2 – x + 4)^5
u(x) &= 3x^2 – x + 4, & \text{so} \\
k(x) &= f(u(x)) = u^5
k'(x) &= 5u^4 \cdot u'(x) \\
u'(x) &= \frac{d}{dx}(3x^2 – x + 4) \\
&= 6x – 1.
\end{whakahāngai}
\]
Mā te whakamahi i te ture mekameka:
\[
\begin{align}
k'(x) &= 5(3x^2 – x + 4)^4 \cdot (6x – 1) \\
&= 5(3x^2 – x + 4)^4 (6x – 1).
\end{whakahāngai}
\]
Nō reira, ko te pānga o \( k(x) = (3x^2 – x + 4)^5 \) ko \( k'(x) = 5 (3x^2 – x + 4)^4 (6x – 1) \).
Tauira 5: Te Pūtake o tētahi Mahi me ngā Tuakiri Pātoru
Pātai: Whakatauhia te pānga o te mahi \( m(x) = \sin(x) \cdot \cos(x) \).
Otinga:
Ka whakamahia e tātou te ture tātai mō ngā hua. Me kī ko \( u(x) = \sin(x) \) me \( v(x) = \cos(x) \), kātahi:
\[
\begin{align}
u'(x) &= \cos(x), \\
v'(x) &= -\sin(x).
\end{whakahāngai}
\]
Mā te whakamahi i te ture i ahu mai mō ngā hua:
\[
\begin{align}
m'(x) &= (\sin(x) \cdot \cos(x))' \\
&= (\sin(x))' \cdot \cos(x) + \sin(x) \cdot (\cos(x))' \\
&= \cos(x) \cdot \cos(x) + \sin(x) \cdot (-\sin(x)) \\
&= \cos^2(x) – \sin^2(x).
\end{whakahāngai}
\]
Mā te whakamahi i te tuakiri pātoru \(\cos(2x) = \cos^2(x) – \sin^2(x)\):
\[
m'(x) = \cos(2x).
\]
Nō reira, ko te tauwehenga o \( m(x) = \sin(x) \cdot \cos(x) \) ko \( m'(x) = \cos(2x) \).
Whakamutunga
Ko te pānga o tētahi mahi taurangi he ariā taketake i roto i te tātaitai he mea tino nui, he mea whai hua hoki i roto i ngā momo tono. He maha ngā ture pānga, pērā i te ture pānga taketake, te ture hua, te ture mekameka, me ngā ture mō ngā pānga pārōnaki, e āwhina katoa ana ki te tatau i ngā pānga o ngā mahi uaua ake. Mā te mārama ki ngā tauira i runga ake nei me te whakaharatau i ngā rapanga, ka taea e tātou te whakapai ake i tō tātou māramatanga me ō tātou pūkenga ki te tango pānga o ngā mahi taurangi.
