Bab 9 Pembezaan

Posted on June 1, 2016 by user

9.7 Pembezaan, SPM Praktis (Kertas 1)

Soalan 1:

Bezakan ungkapan 2x (4x² + 2x - 5) terhadap x.

Penyelesaian:

2x (4x² + 2x – 5) = 8x³ + 4x²– 10x

d/dx (8x³ + 4x² – 10x)

= 24x + 8x –10

Soalan 2:

$Diberi y = \frac{x^{3} + 2 x^{2} + 1}{3 x}, cari \frac{d y}{d x} .$

Penyelesaian:

$\begin{array}{l} y = \frac{x^{3} + 2 x^{2} + 1}{3 x} \\ y = \frac{x^{3}}{3 x} + \frac{2 x^{2}}{3 x} + \frac{1}{3 x} \\ y = \frac{x^{2}}{3} + \frac{2 x}{3} + \frac{1}{3} x^{- 1} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3} x^{- 2} \\ \frac{d y}{d x} = \frac{2 x}{3} + \frac{2}{3} - \frac{1}{3 x^{2}} \end{array}$

Soalan 3:

$Diberi y = \frac{3}{\sqrt{5 x + 1}}, cari \frac{d y}{d x}$

Penyelesaian:

$\begin{array}{l} y = \frac{3}{\sqrt{5 x + 1}} = 3 {(5 x + 1)}^{- \frac{1}{2}} \\ \frac{d y}{d x} = - \frac{1}{2} .3 {(5 x + 1)}^{- \frac{3}{2}} (5) \\ \frac{d y}{d x} = \frac{- 15}{2 {[{(5 x + 1)}^{3}]}^{\frac{1}{2}}} \\ \frac{d y}{d x} = \frac{- 15}{2 \sqrt{{(5 x + 1)}^{3}}} \end{array}$

Soalan 4:

$\begin{array}{l} Cari \frac{d s}{d t} bagi setiap fungsi yang berikut . \\ (a) s = {(t - \frac{3}{t})}^{2} \\ (b) s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \end{array}$

Penyelesaian:

(a)

$\begin{array}{l} s = {(t - \frac{3}{t})}^{2} \\ s = (t - \frac{3}{t}) (t - \frac{3}{t}) \\ s = t^{2} - 6 + \frac{9}{t^{2}} \\ s = t^{2} - 6 + 9 t^{- 2} \\ \frac{d s}{d t} = 2 t - 18 t^{- 3} = 2 t - \frac{18}{t^{3}} \\ \frac{d s}{d t} = 2 t - \frac{18}{t^{3}} \end{array}$

(b)

$\begin{array}{l} s = \frac{(t + 1) (3 - 5 t)}{t^{2}} \\ s = \frac{3 t - 5 t^{2} + 3 - 5 t}{t^{2}} \\ s = \frac{- 5 t^{2} - 2 t + 3}{t^{2}} \\ s = - 5 - \frac{2}{t} + \frac{3}{t^{2}} \\ s = - 5 - 2 t^{- 1} + 3 t^{- 2} \\ \frac{d s}{d t} = 2 t^{- 2} - 6 t^{- 3} \\ \frac{d s}{d t} = \frac{2}{t^{2}} - \frac{6}{t^{3}} \end{array}$

Soalan 5:

$\begin{array}{l} Diberi y = \frac{3}{5} u^{5}, dengan keadaan u = 4 x + 1. \\ Cari \frac{d y}{d x} dalam sebutan x . \end{array}$

Penyelesaian:

$\begin{array}{l} y = \frac{3}{5} u^{5}, u = 4 x + 1 \\ y = \frac{3}{5} {(4 x + 1)}^{5} \\ \frac{d y}{d x} = 5. \frac{3}{5} {(4 x + 1)}^{4} .4 \\ \frac{d y}{d x} = 12 {(4 x + 1)}^{4} \end{array}$