9.7.7 Pembezaan, SPM Praktis (Kertas 1)

$\begin{array}{l}\frac{dA}{dr}=2\pi r\\ \\ \text{Perubahan kecil dalam luas kepada jejari,}\\ \frac{\delta A}{\delta r}\approx \frac{dA}{dr}\\ \delta A=\frac{dA}{dr}\times \delta r\end{array}$
δA = 0.08π cm²

Soalan 22 (3 markah):
Cari nilai bagi
$\begin{array}{l}\left(\text{a}\right)\underset{x\to 1}{\text{had}}\left(7-x^2\right),\\ \left(\text{b}\right)f\text{'}\text{'}\left(2\right)\text{ jika }f\text{'}\left(x\right)=2x^3-4x+3.\end{array}$

Penyelesaian:
(a)
$\begin{array}{l}\underset{x\to 1}{had}\left(7-x^2\right)\\ =7-{\left(1\right)}^2\\ =6\end{array}$

(b)
$\begin{array}{l}f\text{'}\left(x\right)=2x^3-4x+3\\ f\text{'}\text{'}\left(x\right)=6x^2-4\\ f\text{'}\text{'}\left(2\right)=6{\left(2\right)}^2-4\\ =24-4\\ =20\end{array}$

Soalan 23 (4 markah):
Diberi bahawa L = 4t – t² dan x = 3 + 6t.
(a) Ungkapkan $\frac{dL}{dx}$  dalam sebutan t.
(b) Cari perubahan kecil bagi x, apabila L berubah daripada 3 kepada 3.4 pada ketika t = 1.

Penyelesaian:
(a)
$\begin{array}{l}\text{Diberi }L=4t-t^2\text{ dan }x=3+6t\\ L=4t-t^2\\ \frac{dL}{dt}=4-2t\\ \\ x=3+6t\\ \frac{dx}{dt}=6\\ \\ \frac{dL}{dx}=\frac{dL}{dt}\times \frac{dt}{dx}\\ \frac{dL}{dx}=\left(4-2t\right)\times \frac{1}{6}\\ =\frac{4-2t}{6}\\ =\frac{2-t}{3}\end{array}$


(b)
$\begin{array}{l}\delta L=3.4-3=0.4\\ \frac{\delta L}{\delta x}\approx \frac{dL}{dx}\\ \delta x=\delta L÷\frac{\delta L}{\delta x}\\ \delta x=\delta L\times \frac{\delta x}{\delta L}\\ =0.4\times \frac{3}{2-t}\\ =\frac{2}{5}\times \frac{3}{2-t}\\ =\frac{6}{5\left(2-t\right)}\\ \\ \text{Apabila }t=1,\\ \delta x=\frac{6}{5\left(2-1\right)}=\frac{6}{5}\end{array}$