**9.2.2 Terbitan Pertama Hasil Bahagi Dua Polinomial**

**Cari hasil bahagi terbitan dengan menggunakan kaedah-kaedah yang berikut:**

*Kaedah 1: *Petua Hasil Bahagi

**Contoh 1:**

$\begin{array}{l}y=\frac{3x}{2x-1},\text{cari}\frac{dy}{dx}\\ \\ \frac{dy}{dx}=\frac{(2x-1)(3)-(3x)\left(2\right)}{{\left(2x-1\right)}^{2}}\\ \text{=}\frac{6x-3-6x}{{\left(2x-1\right)}^{2}}\\ \text{=}\frac{-3}{{\left(2x-1\right)}^{2}}\end{array}$
*Kaedah**2: (Pembezaan terus)*

$\begin{array}{l}y=\frac{u\text{}\left(pengangka\right)}{v\text{}\left(penyebut\right)}\\ \\ \frac{dy}{dx}=\frac{\left(\begin{array}{l}salin\\ penyebut\end{array}\right)\left(\begin{array}{l}bezakan\\ pengangka\end{array}\right)-\left(\begin{array}{l}salin\\ pengangka\end{array}\right)\left(\begin{array}{l}bezakan\\ penyebut\end{array}\right)}{{\left(penyebut\right)}^{2}}\end{array}$

**Contoh 2:**

$\text{Diberi}y=\frac{{x}^{2}}{2x+1},\text{cari}\frac{dy}{dx}$

*Penyelesaian:*

$\begin{array}{l}y=\frac{{x}^{2}}{2x+1}\\ \frac{dy}{dx}=\frac{(2x+1)(2x)-{x}^{2}(2)}{{(2x+1)}^{2}}\\ \text{=}\frac{4{x}^{2}+2x-2{x}^{2}}{{(2x+1)}^{2}}\\ \text{}=\frac{2{x}^{2}+2x}{{(2x+1)}^{2}}\end{array}$
**Contoh 3:**

$\text{Diberibahawa}y=\frac{4{x}^{3}}{{\left(5x+1\right)}^{3}},\text{cari}\frac{dy}{dx}$

*Penyelesaian:*

$\begin{array}{l}y=\frac{4{x}^{3}}{{\left(5x+1\right)}^{3}}\\ \frac{dy}{dx}=\frac{{(5x+1)}^{3}(12{x}^{2})-4{x}^{3}.3{(5x+1)}^{2}.5}{{\left[{\left(5x+1\right)}^{3}\right]}^{2}}\\ \text{=}\frac{{(5x+1)}^{3}(12{x}^{2})-60{x}^{3}{(5x+1)}^{2}}{{\left(5x+1\right)}^{6}}\\ \text{=}\frac{(12{x}^{2}){(5x+1)}^{2}\left[(5x+1)-5x\right]}{{\left(5x+1\right)}^{6}}\\ \text{=}\frac{(12{x}^{2}){(5x+1)}^{2}\left(1\right)}{{\left(5x+1\right)}^{6}}\\ \text{=}\frac{12{x}^{2}}{{\left(5x+1\right)}^{4}}\end{array}$