Page 42 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(b) y = 3 – x 2 (c) y = x(x – 4)

y + δy = 3 – (x + δx) 2 y = x – 4x
2
3 – x + δy = 3 – x – 2xδx – (δx) 2 y + δy = (x + δx) – 4(x + δx)
2
2
2
δy = [–2x + δx] δx x – 4x + δy = x + 2xδx + (δx) – 4x – 4 δx
2
2
2
δy = – 2x + δx δy = (2x – 4 + δx)δx
δx δy = 2x – 4 + δx
Maka/Hence dy = had/lim δy δx
dx δx → 0 δx dy δy
= had/lim (–2x + δx) Maka/Hence = had/lim
δx → 0 dx δx → 0 δx
= –2x = had/lim (2x – 4 + δx)
δx → 0
= 2x – 4
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1

x
x
x
had/lim  – 2 = had/lim ( – 2)( + 2)
x → 4 (x – 4) x → 4 (x – 4)(x + 2)
x
(x – 4) Gunakan penggantian secara langsung, had  – 2 =
= had/lim 0 x → 4 (x – 4)
x
x → 4 (x – 4)( + 2) tidak tertakrif.
1 0
= Perlu gunakan konjugat bagi x – 2, iaitu x + 2.
4  – 2 0
x
Use direct substitution, lim = is not defined.
x → 4 (x – 4) 0
Need to use the conjugate of x – 2, which is x + 2.
Kriteria Kejayaan: ............................................................................................ .
Saya berjaya
• Mencari had fungsi dengan menggunakan gantian terus.
• Mencari had fungsi dengan menggunakan pemfaktoran dan merasionalkan pengangka atau penyebut.
• Menentukan sama ada wujudnya had bagi suatu fungsi yang diberi.
• Menggunakan prinsip pertama.

2.2 Pembezaan Peringkat Pertama Buku Teks
PBD
PBD The First Derivative ms. 38 – 48
PBD
FOKUS TOPIK

Rumus terbitan pertama
First derivative formula
y = f(x) = ax , a dan n ialah pemalar, maka Fungsi ungkapan algebra
n
Dari prinsip pertama y = f(x) = ax , a and n are constants, hence Function of algebraic expression
n
From the first principles dy = df(x) = anx n–1 y = f(x) = h(x) ± g(x),
dy df(x) had/lim f(x) dx dx maka/hence
dx = dx = δx → 0 dy = df(x) = d (h(x) ± g(x))
dx dx dx
Pembezaan/Differentiation
dy , f'(x), df(x)
dx dx
Fungsi hasil bahagi Fungsi hasil darab
Function involving quotient Function involving product
u y = uv, u = h(x), v = g(x)
y = , u = h(x), v = g(x). Fungsi gubahan/Composite function maka/hence
v
Maka/Hence y = g(u) dan/and u = h(x), dy d dv du
du dv f'(x) = g'(u) × h'(x), dx = dx (uv) = u dx + v dx
dy d u v dx – u dx iaitu/which is dy = dy × du

dx = dx  v  = v 2 dx du dx
(Petua rantai/Chain rule)

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