Tusaale su'aal dood ah oo ku saabsan asalka shaqada aljabrada
Kala-soocidda ku jirta xisaabtu waa fikrad aasaasi ah oo loo isticmaalo in lagu qeexo sida shaqadu isu beddesho, ama janjeedhka shaqadu marka ay meel taagan tahay. Kala-soocidda waxay faa'iido u leedahay dhinacyo kala duwan sida fiisigiska, dhaqaalaha, iyo injineernimada sababtoo ah waxay bixiyaan macluumaad ku saabsan heerka isbeddelka. Maqaalkan, waxaan ka hadli doonnaa dhowr tusaale oo ka mid ah kala-soocidda shaqooyinka aljabrada iyo sida loo xalliyo.
Tusaalaha 1aad: Soo-saaridda Shaqada Polynomial
Su'aal: Marka la eego shaqada \( f(x) = 3x^3 – 5x^2 + 2x – 7 \). Go'aami nooca ka soo jeeda shaqada!
Xalka:
Annagoo adeegsanayna xeerka aasaasiga ah ee derivatives-ka ee shaqooyinka polynomial, kuwaas oo kala ah \(\frac{d}{dx} x^n = nx^{n-1} \), waxaan mid mid u xisaabin doonnaa derivative-ka erey kasta oo ka mid ah shaqada.
\[
\begin{hagaajin}
f(x) &= 3x^3 – 5x^2 + 2x – 7 \\
f'(x) &= \frac{d}{dx}(3x^3) – \frac{d}{dx}(5x^2) + \frac{d}{dx}(2x) – \frac{d}{dx}(7) \\
f'(x) &= 3 \cdot 3x^{3-1} – 5 \cdot 2x^{2-1} + 2 \cdot 1x^{1-1} – 0 \\
f'(x) &= 9x^2 – 10x + 2.
\end{align}
\]
Markaa, beddelka \( f(x) = 3x^3 – 5x^2 + 2x – 7 \) waa \( f'(x) = 9x^2 – 10x + 2 \).
Tusaalaha 2aad: Soo-saaridda shaqada oo leh jibaaranayaal jajaban
Su'aal: Go'aami kala-soocidda shaqada \( g(x) = x^{3/2} + x^{1/2} \).
Xalka:
Iyadoo la adeegsanayo isla xeerka kala soocidda, taasi waa \(\frac{d}{dx} x^n = nx^{n-1} \):
\[
\begin{hagaajin}
g(x) &= x^{3/2} + x^{1/2} \\
g'(x) &= \frac{d}{dx}(x^{3/2}) + \frac{d}{dx}(x^{1/2}) \\
g'(x) &= \frac{3}{2}x^{(3/2)-1} + \frac{1}{2}x^{(1/2)-1} \\
g'(x) &= \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2}.
\end{align}
\]
Markaa, beddelka \( g(x) = x^{3/2} + x^{1/2} \) waa \( g'(x) = \frac{3}{2}x^{1/2} + \frac{1}{2}x^{-1/2} \).
Tusaalaha 3aad: Waxyaabaha laga soo qaatay shaqooyinka jibbaaran iyo kuwa saddex-geesoodka ah
Su'aal: Go'aami kala-soocidda shaqada \( h(x) = e^x \cdot \sin(x) \).
Xalka:
Si aan u xallino noocaan wax soo saarka ah, waxaan u baahanahay Xeerka Waxsoosaarka, kaas oo sheegaya \((uv)' = u'v + uv'\). Bal qiyaas \( u(x) = e^x \) iyo \( v(x) = \sin(x) \), ka dibna:
\[
\begin{hagaajin}
u'(x) &= e^x, & \text{sababtoo ah asalka } e^x \text{ waa } e^x \\
v'(x) &= \cos(x), & \text{sababtoo ah tarjumaadda } \sin(x) \text{ waa } \cos(x).
\end{align}
\]
Isticmaalka xeerka la soo saaray ee alaabada:
\[
\begin{hagaajin}
h'(x) &= (e^x \cdot \sin(x))' \\
&= e^x \cdot (\sin(x))' + \sin(x) \cdot (e^x)' \\
&= e^x \cdot \cos(x) + \sin(x) \cdot e^x \\
&= e^x (\cos(x) + \sin(x)).
\end{align}
\]
Markaa, beddelka \( h(x) = e^x \sin(x) \) waa \( h'(x) = e^x (\cos(x) + \sin(x)) \).
Tusaalaha 4aad: Soo-saarista Shaqada iyadoo la adeegsanayo Xeerka Silsiladda
Su'aal: Go'aami kala-soocidda shaqada \( k(x) = (3x^2 – x + 4)^5 \).
Xalka:
Si aan u xallino beddelkan, waxaan u baahanahay xeerka silsiladda, kaas oo ah \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Ka soo qaad \( u(x) = 3x^2 – x + 4 \) iyo \( f(u) = u^5 \), ka dibna:
\[
\begin{hagaajin}
k(x) &= (3x^2 – x + 4)^5 \\
u(x) &= 3x^2 – x + 4, & \qoraal{so} \\
k(x) &= f(u(x)) = u^5 \\
k'(x) &= 5u^4 \cdot u'(x) \\
u'(x) &= \frac{d}{dx}(3x^2 – x + 4) \\
&= 6x – 1.
\end{align}
\]
Iyadoo la adeegsanayo xeerka silsiladda:
\[
\begin{hagaajin}
k'(x) &= 5(3x^2 – x + 4)^4 \cdot (6x – 1) \\
&= 5(3x^2 – x + 4)^4 (6x – 1).
\end{align}
\]
Markaa, beddelka \( k(x) = (3x^2 – x + 4)^5 \) waa \( k'(x) = 5 (3x^2 – x + 4)^4 (6x – 1) \).
Tusaalaha 5aad: Kala-soocidda shaqada oo leh aqoonsiyo Trigonometric ah
Su'aal: Go'aami kala-soocidda shaqada \( m(x) = \sin(x) \cdot \cos(x) \).
Xalka:
Waxaan u isticmaali doonnaa xeerka ka-soo-saarka badeecadaha. Ka soo qaad \( u(x) = \sin(x) \) iyo \( v(x) = \cos(x) \), ka dibna:
\[
\begin{hagaajin}
u'(x) &= \cos(x), \\
v'(x) &= -\sin(x).
\end{align}
\]
Isticmaalka xeerka la soo saaray ee alaabada:
\[
\begin{hagaajin}
m'(x) &= (\sin(x) \cdot \cos(x))' \\
&= (\sin(x))' \cdot \cos(x) + \sin(x) \cdot (\cos(x))' \\
&= \cos(x) \cdot \cos(x) + \sin(x) \cdot (-\sin(x)) \\
&= \cos^2(x) – \sin^2(x).
\end{align}
\]
Isticmaalka aqoonsiga trigonometric \(\cos(2x) = \cos^2(x) – \sin^2(x)\):
\[
m'(x) = \cos(2x).
\]
Markaa, beddelka \( m(x) = \sin(x) \cdot \cos(x) \) waa \( m'(x) = \cos(2x) \).
Gabagabo
Kala-soocidda shaqada aljabrada waa fikrad aasaasi ah oo ku jirta xisaabinta oo aad muhiim u ah oo waxtar u leh codsiyada kala duwan. Shuruuc kala duwan oo kala duwan oo kala duwan, sida xeerka aasaasiga ah ee kala-soocidda, xeerka badeecada, xeerka silsiladda, iyo xeerarka kala-soocidda trigonometric, dhammaantood waxay gacan ka geystaan xisaabinta kala-soocidda shaqooyinka aadka u adag. Markaan fahamno tusaalooyinka kor ku xusan iyo ku celcelinta dhibaatooyinka, waxaan horumarin karnaa fahamkeenna iyo xirfaddayada ku aaddan qaadashada kala-soocidda hawlaha aljabrada.
