Isibonelo sombuzo wengxoxo mayelana ne-derivative yomsebenzi we-algebraic
I-derivative ku-calculus ingumqondo oyisisekelo osetshenziselwa ukuchaza ukuthi umsebenzi ushintsha kanjani, noma ukuthambekela komsebenzi endaweni ethile. Ama-derivative awusizo emikhakheni ehlukahlukene njenge-physics, ezomnotho, kanye nobunjiniyela ngoba ahlinzeka ngolwazi mayelana nesilinganiso sokushintsha. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zama-derivatives emisebenzi ye-algebraic kanye nendlela yokuwaxazulula.
Isibonelo 1: I-Derivative ye-Polynomial Function
Umbuzo: Uma ubheka umsebenzi \( f(x) = 3x^3 – 5x^2 + 2x – 7 \). Thola i-derivative yomsebenzi!
Isixazululo:
Sisebenzisa umthetho oyisisekelo wama-derivatives emisebenzini ye-polynomial, okungukuthi \(\frac{d}{dx} x^n = nx^{n-1} \), sizobala i-derivative yethemu ngayinye yomsebenzi ngamunye ngamunye.
\[
\begin{align }
f(x) &= 3x^3 – 5x^2 + 2x – 7 \\
f'(x) &= \frac{d}{dx}(3x^3) – \frac{d}{dx}(5x^2) + \frac{d}{dx}(2x) – \frac{d}{dx}(7) \\
f'(x) &= 3 \cdot 3x^{3-1} – 5 \cdot 2x^{2-1} + 2 \cdot 1x^{1-1} – 0 \\
f'(x) &= 9x^2 – 10x + 2.
\end{align }
\]
Ngakho-ke, i-derivative ye- \( f(x) = 3x^3 – 5x^2 + 2x – 7 \) ingu- \( f'(x) = 9x^2 – 10x + 2 \).
Isibonelo 2: I-Derivative ye-Function ene-Fractional Exponents
Umbuzo: Nquma i-derivative yomsebenzi \( g(x) = x^{3/2} + x^{1/2} \).
Isixazululo:
Kusetshenziswa umthetho ofanayo wokuthola, okungukuthi \(\frac{d}{dx} x^n = nx^{n-1} \):
\[
\begin{align }
g(x) &= x^{3/2} + x^{1/2} \\
g'(x) &= \frac{d}{dx}(x^{3/2}) + \frac{d}{dx}(x^{1/2}) \\
g'(x) &= \frac{3}{2}x^{(3/2)-1} + \frac{1}{2}x^{(1/2)-1} \\
g'(x) &= \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2}.
\end{align }
\]
Ngakho-ke, i-derivative ye-\( g(x) = x^{3/2} + x^{1/2} \) ingu-\( g'(x) = \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2} \).
Isibonelo 3: Izinhlobo zemisebenzi ye-Exponential kanye ne-Trigonometric
Umbuzo: Thola i-derivative yomsebenzi \( h(x) = e^x \cdot \sin(x) \).
Isixazululo:
Ukuze sixazulule le derivative, sidinga uMthetho Womkhiqizo, othi \((uv)' = u'v + uv'\). Ake sithi \( u(x) = e^x \) kanye \( v(x) = \sin(x) \), bese kuba:
\[
\begin{align }
u'(x) &= e^x, & \text{ngoba i-derivative ka } e^x \text{ ingu } e^x \\
v'(x) &= \cos(x), & \text{ngoba i-derivative ye-} \sin(x) \text{ ingu-} \cos(x).
\end{align }
\]
Ukusebenzisa umthetho osuselwe emikhiqizweni:
\[
\begin{align }
h'(x) &= (e^x \cdot \sin(x))' \\
&= e^x \cdot (\sin(x))' + \sin(x) \cdot (e^x)' \\
&= e^x \cdot \cos(x) + \sin(x) \cdot e^x \\
&= e^x (\cos(x) + \sin(x)).
\end{align }
\]
Ngakho-ke, i-derivative ka-\( h(x) = e^x \sin(x) \) ngu-\( h'(x) = e^x (\cos(x) + \sin(x)) \).
Isibonelo 4: I-Derivative of a Function Usebenzisa i-Chain Rule
Umbuzo: Thola i-derivative yomsebenzi \( k(x) = (3x^2 – x + 4)^5 \).
Isixazululo:
Ukuze sixazulule le derivative, sidinga umthetho we-chain, okungukuthi \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Ake sithi \( u(x) = 3x^2 – x + 4 \) kanye \( f(u) = u^5 \), bese kuba:
\[
\begin{align }
k(x) &= (3x^2 – x + 4)^5 \\
u(x) &= 3x^2 – x + 4, kanye no-\text{so} \\
k(x) &= f(u(x)) = u^5 \\
k'(x) &= 5u^4 \cdot u'(x) \\
u'(x) &= \frac{d}{dx}(3x^2 – x + 4) \\
&= 6x – 1.
\end{align }
\]
Ukusebenzisa umthetho weketanga:
\[
\begin{align }
k'(x) &= 5(3x^2 – x + 4)^4 \cdot (6x – 1) \\
&= 5(3x^2 – x + 4)^4 (6x – 1).
\end{align }
\]
Ngakho-ke, i-derivative ye-\( k(x) = (3x^2 – x + 4)^5 \) ingu-\( k'(x) = 5 (3x^2 – x + 4)^4 (6x – 1) \).
Isibonelo 5: I-Derivative ye-Function ene-Trigonometric Identities
Umbuzo: Thola i-derivative yomsebenzi \( m(x) = \sin(x) \cdot \cos(x) \).
Isixazululo:
Sizosebenzisa umthetho we-derivative emikhiqizweni. Ake sithi \( u(x) = \sin(x) \) kanye \( v(x) = \cos(x) \), bese:
\[
\begin{align }
u'(x) &= \cos(x), \\
v'(x) &= -\isono(x).
\end{align }
\]
Ukusebenzisa umthetho osuselwe emikhiqizweni:
\[
\begin{align }
m'(x) &= (\sin(x) \cdot \cos(x))' \\
&= (\sin(x))' \cdot \cos(x) + \sin(x) \cdot (\cos(x))' \\
&= \cos(x) \cdot \cos(x) + \sin(x) \cdot (-\sin(x)) \\
&= \cos^2(x) – \sin^2(x).
\end{align }
\]
Ukusebenzisa i-trigonometric identity \(\cos(2x) = \cos^2(x) – \sin^2(x)\):
\[
m'(x) = \cos(2x).
\]
Ngakho-ke, i-derivative ka-\( m(x) = \sin(x) \cdot \cos(x) \) ngu-\( m'(x) = \cos(2x) \).
Isiphetho
I-derivative yomsebenzi we-algebraic ingumqondo oyisisekelo ekubaleni obaluleke kakhulu futhi owusizo ekusetshenzisweni okuhlukahlukene. Imithetho ehlukahlukene yokuvela, njengomthetho oyisisekelo wokuvela, umthetho womkhiqizo, umthetho weketanga, kanye nemithetho ye-trigonometric derivatives, konke kusiza ekubaleni i-derivatives yemisebenzi eyinkimbinkimbi kakhulu. Ngokuqonda izibonelo ezingenhla kanye nokuzijwayeza izinkinga, singathuthukisa ukuqonda kwethu kanye namakhono ekuthatheni i-derivatives yemisebenzi ye-algebraic.
