Bab 9 Pembezaan

Posted on June 7, 2016 by user

Bab 9 Pembezaan

9.2 Terbitan Pertama untuk Fungsi Polinomial

(A) Membezakan suatu Pemalar

Jika y = a,

Dengan keadaan a ialah suatu pemalar,

maka

\frac{d y}{d x} = 0

(B) Membezakan Pembolehubah dengan Index n

(C) Membezakan suatu Fungsi Linear

(D) Membezakan suatu Fungsi Terbitan

(E) Membezakan suatu Fungsi Pecahan

(F) Membezakan Fungsi Punca Kuasa Dua

Bab 9 Pembezaan

Posted on June 6, 2016 by user

9.2.3 Terbitan Pertama Fungsi Gubahan

(A) Membezakan fungsi gubahan dengan menggunakan Petua Rantai

Jika y = uⁿ

dengan keadaan udan v adalah fungsi dalam x

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

Contoh 1:

Bezakan y = (x² – 1)⁸

Penyelesaian:

\begin{array}{l} y = u^{8} katakan u = x^{2} - 1 \\ \frac{d y}{d u} = 8 u^{7} \frac{d u}{d x} = 2 x \\ \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} \\ \frac{d y}{d x} = 8 u^{7} \times 2 x \\ \frac{d y}{d x} = 16 x u^{7} \\ \frac{d y}{d x} = 16 x {(x^{2} - 1)}^{7} \end{array}

(B) Membezakan fungsi gubahan dengan menggunakan Kaedah Alternatif - Versi Mudah

Contoh:

Bezakan y = (x² – 1)⁸

Penyelesaian:

\begin{array}{l} y = {(x^{2} - 1)}^{8} \\ \frac{d y}{d x} = 8 {(x^{2} - 1)}^{7} \frac{d}{d x} (x^{2} - 1) \\ \frac{d y}{d x} = 8 {(x^{2} - 1)}^{7} (2 x) \\ \frac{d y}{d x} = 16 x {(x^{2} - 1)}^{7} \end{array}

Contoh 2:

Diberi y = \frac{1}{3 x - 7}, cari \frac{d y}{d x}

Penyelesaian

\begin{array}{l} y = \frac{1}{3 x - 7} = {(3 x - 7)}^{- 1} \\ \frac{d y}{d x} = - 1 {(3 x - 7)}^{- 2} .3 \\ \frac{d y}{d x} = \frac{- 3}{{(3 x - 7)}^{2}} \end{array}

Contoh 3:

Diberi y = \sqrt{2 x^{2} - 5 x + 1}, cari \frac{d y}{d x}

Penyelesaian

\begin{array}{l} y = \sqrt{2 x^{2} - 5 x + 1} = {(2 x^{2} - 5 x + 1)}^{\frac{1}{2}} \\ \frac{d y}{d x} = \frac{1}{2} {(2 x^{2} - 5 x + 1)}^{- \frac{1}{2}} (4 x - 5) \\ \frac{d y}{d x} = \frac{4 x - 5}{2 \sqrt{2 x^{2} - 5 x + 1}} \end{array}

Bab 9 Pembezaan

Posted on June 6, 2016 by user

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{3 x}{2 x - 1}, cari \frac{d y}{d x} \\ \frac{d y}{d x} = \frac{(2 x - 1) (3) - (3 x) (2)}{{(2 x - 1)}^{2}} \\ = \frac{6 x - 3 - 6 x}{{(2 x - 1)}^{2}} \\ = \frac{- 3}{{(2 x - 1)}^{2}} \end{array}

Kaedah2: (Pembezaan terus)

\begin{array}{l} y = \frac{u (p e n g a n g k a)}{v (p e n y e b u t)} \\ \frac{d y}{d x} = \frac{(\begin{array}{l} s a l i n \\ p e n y e b u t \end{array}) (\begin{array}{l} b e z a k a n \\ p e n g a n g k a \end{array}) - (\begin{array}{l} s a l i n \\ p e n g a n g k a \end{array}) (\begin{array}{l} b e z a k a n \\ p e n y e b u t \end{array})}{{(p e n y e b u t)}^{2}} \end{array}

Contoh 2:

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

Penyelesaian:

\begin{array}{l} y = \frac{x^{2}}{2 x + 1} \\ \frac{d y}{d x} = \frac{(2 x + 1) (2 x) - x^{2} (2)}{{(2 x + 1)}^{2}} \\ = \frac{4 x^{2} + 2 x - 2 x^{2}}{{(2 x + 1)}^{2}} \\ = \frac{2 x^{2} + 2 x}{{(2 x + 1)}^{2}} \end{array}

Contoh 3:

Diberi bahawa y = \frac{4 x^{3}}{{(5 x + 1)}^{3}}, cari \frac{d y}{d x}

Penyelesaian:

\begin{array}{l} y = \frac{4 x^{3}}{{(5 x + 1)}^{3}} \\ \frac{d y}{d x} = \frac{{(5 x + 1)}^{3} (12 x^{2}) - 4 x^{3} .3 {(5 x + 1)}^{2} .5}{{[{(5 x + 1)}^{3}]}^{2}} \\ = \frac{{(5 x + 1)}^{3} (12 x^{2}) - 60 x^{3} {(5 x + 1)}^{2}}{{(5 x + 1)}^{6}} \\ = \frac{(12 x^{2}) {(5 x + 1)}^{2} [(5 x + 1) - 5 x]}{{(5 x + 1)}^{6}} \\ = \frac{(12 x^{2}) {(5 x + 1)}^{2} (1)}{{(5 x + 1)}^{6}} \\ = \frac{12 x^{2}}{{(5 x + 1)}^{4}} \end{array}

Bab 9 Pembezaan

Posted on June 6, 2016 by user

9.2.1 Terbitan Pertama Hasil Darab Dua Polinomial

Cari hasil darab terbitan dengan menggunakan kaedah-kaedah yang berikut:

Kaedah 1: Petua Hasil Darab

Jika u(x) dan v(x) adalah dua fungsi x dan y = uv maka

Contoh 1:

Kaedah Alternatif: (Pembezaan terus)

\begin{array}{l} y = u v \\ \frac{d y}{d x} = (\begin{array}{l} s a l i n \\ k i r i \end{array}) (\begin{array}{l} b e z a k a n \\ k a n a n \end{array}) + (\begin{array}{l} s a l i n \\ k a n a n \end{array}) (\begin{array}{l} b e z a k a n \\ k i r i \end{array}) \end{array}

Contoh 2:

Diberi bahawa y= (2x + 3)(3x³– 2x²– x), cari dy/dx.

Penyelesaian:

y = (2x + 3)(3x³– 2x² – x)

dy/dx = (2x + 3)(9x²– 4x – 1) + (3x³– 2x²– x)(2)

dy/dx = (2x + 3)(9x²– 4x – 1) + (6x³– 4x²– 2x)

Contoh 3:

Diberi bahawa y= 4x³(3x + 1)⁵, cari dy/dx.

Penyelesaian:

y = 4x³(3x + 1)⁵

dy/dx

= 4x³. 5 (3x + 1)⁴.3 + (3x + 1)⁵.12x²

= 60x³(3x + 1)⁴ + 12x²(3x + 1)⁵

= 12x²(3x + 1)⁴ [5x+ (3x + 1)]

= 12x²(3x + 1)⁴ (8x+ 1)

Bab 9 Pembezaan

Posted on June 6, 2016 by user

9.7.4 Pembezaan, SPM Praktis (Kertas 1)

Soalan 15:

Cari koordinat bagi titik pada lengkung, y = (4x – 5)² supaya kecerunan normal lengkung itu ialah ⅛.

Penyelesaian:

y = (4x – 5)²

dy/dx = 2 (4x – 5). 4 = 32x – 40

Diberi normal ialah ⅛, maka kecerunan tangen ialah –8.

dy/dx = –8

32x – 40 = –8

32x = 32

x = 1

y = [4 (1) – 5]²= 1

Hence, the coordinates of the point on the curve, y= (4x – 5)² is (1, 1).

Soalan 16:

Suatu lengkung mempunyai fungsi kecerunan kx² – 7x, dengan keadaan k ialah pemalar. Tangen kepada lengkung di titik (1, 4) adalah selari dengan garis lurus y + 2x –1 = 0.
Cari nilai k.

Penyelesaian:

Diberi fungsi kecerunan kx² – 7x selari dengan garis lurus y + 2x –1 = 0

dy/dx = kx²– 7x

y + 2x –1 = 0, y = –2x + 1, kecerunan garis lurus = –2

Maka kx²– 7x = –2

Di titik (1, 4),

k (1)²– 7(1) = –2

k – 7 = –2

k = 5

Soalan 17:

Dalam rajah di atas, garis lurus PR adalah normal kepada lengkung

y = \frac{x^{2}}{2} + 1

at Q.
Cari nilai k.

Penyelesaian:

\begin{array}{l} y = \frac{x^{2}}{2} + 1 \\ \frac{d y}{d x} = x \\ Di titik Q, koordinat- x = 2, \\ Kecerunan lengkung, \frac{d y}{d x} = 2 \\ Maka, kecerunan normal lengkung itu, \\ P R = - \frac{1}{2} \\ \frac{3 - 0}{2 - k} = - \frac{1}{2} \\ 6 = - 2 + k \\ k = 8 \end{array}

Soalan 18:

Garis normal kepada lengkung y = x² + 3x pada titik P adalah selari dengan garis lurus y = –x + 12. Cari persamaan garis normal kepada lengkung itu pada titik P.

Penyelesaian:

Diberi normal kepada lengkung di titik P adalah selari kepada garis lurus y = –x + 12. Maka, kecerunan normal lengkung itu = –1.

Seterusnya, kecerunan tangen kepada lengkung = 1

y = x² + 3x

dydx = 2x + 3

2x + 3 = 1

2x = –2

x = –1

y = (–1)²+ 3 (–1)

y = –2

Titik P = (–1, –2).

Persamaan garis normal kepada lengkung itu pada titik P ialah,

y – (–2) = –1 (x – (–1))

y + 2 = – x – 1

y = – x – 3

2.4.1 Support and Locomotion in Humans and Animals (Structured Question 1 & 2)

Posted on June 2, 2016 by user

Question 1:
Diagram I and Diagram II show different positions of a forearm during a movement.

(a) Complete Diagram I by drawing the triceps muscle which is involved in the movement of the forearm.

(b) State one adaptive characteristic of tissue R shown in Diagram I which helps in the movement of the forearm.

(c) Explain the action of the muscles which cause the movement of the forearm in Diagram II.

(d) Diagram III shows a joint at the knee.

Explain the health problem normally faced by an old person when tissue P is impaired.

(e) An athlete must do a warming up exercise before starting an event.
Explain why.

Answer:
(a)

(b)
Strong/ tough/ non-elastic tissue

(c)
When the biceps relaxes and the triceps contracts, the radius and ulna are pulled downwards and the forearm is straightened.

(d)
- An old person will suffer from osteoarthritis.
- The wear and tear of the cartilage is due to ageing.
- The repetitive use of the joints over the years irritates and inflames the cartilage.

- Eventually, the cartilage begins to degenerate and this causes friction between the bones, leading to pain and restriction of joint mobility.

(e)
- Warming up exercise is needed so that the heart rate increases. This enables oxygen in the blood to travel faster. The capillaries dilate and let more oxygen travel in the blood.

- Warming up exercise also increases the temperature in the muscles. It loosens up the muscles and joints. This removes lactic acid, lets the muscle fibres have greater extensibility and elasticity and increases the force and contraction of the muscles.

Bab 9 Pembezaan

Posted on June 2, 2016 by user

9.7 Pembezaan, SPM Praktis (Kertas 1)

Soalan 11:

Diberi fungsi graf

f (x) = h x^{3} + \frac{k}{x^{2}}

mempunyai fungsi kecerunan

f' (x) = 12 x^{2} - \frac{258}{x^{3}}

dengan hdan k ialah pemalar. Cari nilai h dan k.

Penyelesaian:

\begin{array}{l} f (x) = h x^{3} + \frac{k}{x^{2}} = h x^{3} + k x^{- 2} \\ f' (x) = 3 h x^{2} - 2 k x^{- 3} \\ f' (x) = 3 h x^{2} - \frac{2 k}{x^{3}} \end{array}

Tetapi, diberi

f' (x) = 12 x^{2} - \frac{258}{x^{3}}

dengan perbandingan,

3h = 12 atau 2k = 258

h = 4 k = 129

Soalan 12:

\begin{array}{l} Diberi y = \frac{x^{2}}{x + 3}, tunjukkan \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} \\ Cari \frac{d^{2} y}{d x^{2}} dalam bentuk paling ringkas . \end{array}

Penyelesaian

\begin{array}{l} y = \frac{x^{2}}{x + 3} \\ \frac{d y}{d x} = \frac{(x + 3) (2 x) - x^{2} .1}{{(x + 3)}^{2}} = \frac{2 x^{2} + 6 x - x^{2}}{{(x + 3)}^{2}} \\ \frac{d y}{d x} = \frac{x^{2} + 6 x}{{(x + 3)}^{2}} (tertunjuk) \end{array}

\begin{array}{l} \frac{d^{2} y}{d x^{2}} = \frac{{(x + 3)}^{2} (2 x + 6) - (x^{2} + 6 x) .2 (x + 3)}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{(x + 3) [(x + 3) (2 x + 6) - 2 (x^{2} + 6 x)]}{{(x + 3)}^{4}} \\ \frac{d^{2} y}{d x^{2}} = \frac{[2 x^{2} + 6 x + 6 x + 18 - 2 x^{2} - 12 x]}{{(x + 3)}^{3}} \\ \frac{d^{2} y}{d x^{2}} = \frac{18}{{(x + 3)}^{3}} \end{array}

Soalan 13:

Jika y = x² + 4x, tunjukkan

x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y = 0.

Penyelesaian

\begin{array}{l} y = x^{2} + 4 x \\ \frac{d y}{d x} = 2 x + 4 \\ \frac{d^{2} y}{d x^{2}} = 2 \\ x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y \\ = x^{2} (2) - 2 x (2 x + 4) + 2 (x^{2} + 4 x) \\ = 2 x^{2} - 4 x^{2} - 8 x + 2 x^{2} + 8 x \\ = 0 (tertunjuk) \end{array}

Soalan 14:

Diberi y = x (6 – x), ungkapkan

y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18

dalam sebutan x yang paling ringkas.

Seterusnya, cari nilai xyang memuaskan persamaan

y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0

Penyelesaian:

\begin{array}{l} y = x (6 - x) = 6 x - x^{2} \\ \frac{d y}{d x} = 6 - 2 x \\ \frac{d^{2} y}{d x^{2}} = - 2 \\ ∴ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = (6 x - x^{2}) (- 2) + x (6 - 2 x) + 18 \\ = - 12 x + 2 x^{2} + 6 x - 2 x^{2} + 18 \\ = - 6 x + 18 \\ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0 \\ - 6 x + 18 = 0 \\ x = 3 \end{array}

Bab 9 Pembezaan

Posted on June 1, 2016 by user

9.7 Pembezaan, SPM Praktis (Kertas 1)

Soalan 6:

Diberi f (x) = 3x²(4x²– 1)⁷, cari f’(x).

Penyelesaian:

f (x) = 3x²(4x²– 1)⁷

f’(x) = 3x². 7(4x²– 1)⁶. 8x + (4x²– 1)⁷. 6x

f’(x) = 168x³ (4x²– 1)⁶ + 6x (4x²– 1)⁷

f’(x) = 6x (4x²– 1)⁶[28x²+ (4x²– 1)]

f’(x) = 6x (4x²– 1)⁶ (32x²– 1)

Soalan 7:

Diberi y = (1 + 4x)³(3x² – 1)⁴, cari dy/dx.

Penyelesaian:

y = (1 + 4x)³(3x²– 1)⁴

dy/dx

= (1 + 4x)³. 4(3x²– 1)³.6x + (3x²– 1)⁴. 3(1 + 4x)².4

= 24x (1 + 4x)³(3x²– 1)³ + 12 (3x² – 1)⁴(1 + 4x)²

= 12 (1 + 4x)²(3x²– 1)³ [2x (1 + 4x) + (3x²– 1)]

= 12 (1 + 4x)²(3x²– 1)³ [2x + 8x² + 3x²– 1]

= 12 (1 + 4x)²(3x²– 1)³ [11x² + 2x – 1]

Soalan 8:

Diberi f (x) = 3 x \sqrt{4 x^{2} - 1}, cari f' (x) .

Penyelesaian:

\begin{array}{l} f (x) = 3 x \sqrt{4 x^{2} - 1} = 3 x {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f' (x) = 3 x . \frac{1}{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} .8 x + {(4 x^{2} - 1)}^{\frac{1}{2}} .3 \\ f' (x) = 12 x^{2} {(4 x^{2} - 1)}^{- \frac{1}{2}} + 3 {(4 x^{2} - 1)}^{\frac{1}{2}} \\ f' (x) = 3 {(4 x^{2} - 1)}^{- \frac{1}{2}} [4 x^{2} + (4 x^{2} - 1)] \\ f' (x) = \frac{3 (8 x^{2} - 1)}{\sqrt{(4 x^{2} - 1)}} \end{array}

Soalan 9:

Diberi y = \frac{1 - 5 x^{4}}{x - 3}, cari \frac{d y}{d x} .

Penyelesaian:

\begin{array}{l} \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} \\ \frac{d y}{d x} = \frac{(x - 3) . - 20 x^{3} - (1 - 5 x^{4}) .1}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 20 x^{4} + 60 x^{3} - 1 + 5 x^{4}}{{(x - 3)}^{2}} \\ \frac{d y}{d x} = \frac{- 15 x^{4} + 60 x^{3} - 1}{{(x - 3)}^{2}} \end{array}

Soalan 10:

Diberi f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x}, cari f' (0) .

Penyelesaian:

\begin{array}{l} f (x) = \frac{{(x^{2} - 3)}^{5}}{1 - 3 x} \\ f' (x) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} \\ f' (x) = \frac{(1 - 3 x) .5 {(x^{2} - 3)}^{4} .2 x - {(x^{2} - 3)}^{5} . - 3}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{10 x (1 - 3 x) {(x^{2} - 3)}^{4} + 3 {(x^{2} - 3)}^{5}}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [10 x - 30 x^{2} + 3 (x^{2} - 3)]}{{(1 - 3 x)}^{2}} \\ f' (x) = \frac{{(x^{2} - 3)}^{4} [- 27 x^{2} + 10 x - 9]}{{(1 - 3 x)}^{2}} \\ ∴ f' (0) = \frac{{(0^{2} - 3)}^{4} [- 27 {(0)}^{2} + 10 (0) - 9]}{{(1 - 3 (0))}^{2}} \\ f' (0) = \frac{81 \times (- 9)}{1} = - 729 \end{array}

Bab 9 Pembezaan

Posted on June 1, 2016 by user

9.8 Pembezaan, SPM Praktis (Kertas 2)

Soalan 3:

Lengkung y = x³ – 6x² + 9x + 3 melalui titik P (2, 5) dan mempunyai dua titik pusingan A (3, 3) dan B.

Cari

(a) kecerunan lengkung itu pada P.

(b) persamaan normal kepada lengkung itu pada P.

Penyelesaian:

(a)

y = x³ – 6x² + 9x + 3

dy/dx = 3x² – 12x + 9

di titik P (2, 5),

dy/dx = 3(2)² – 12(2) + 9 = –3

Kecerunan lengung pada titik P = –3.

(b)

Kecerunan normal pada titik P = ⅓

persamaan normal pada P (2, 5):

y – y₁= m (x - x₁)

y – 5 = ⅓ (x– 2)

3y – 15 = x – 2

3y = x + 13

(c)

Pada titik pusingan dy/dx= 0.

3x² – 12x + 9 = 0

x² – 4x + 3 = 0

(x – 1)( x – 3) = 0

x – 1 = 0 atau x – 3 = 0

x = 1 atau x = 3 (titik A)

Pada titik B:

x = 1

y = (1)³– 6(1)² + 9(1) + 3 = 7

Maka, koordinat B = (1, 7)

\begin{array}{l} Apabila x = 1, \\ \frac{d^{2} y}{d x^{2}} = 6 x - 12 \\ \frac{d^{2} y}{d x^{2}} = 6 (1) - 12 \\ \frac{d^{2} y}{d x^{2}} = - 6 < 0 \\ \frac{d^{2} y}{d x^{2}} < 0, B ialah titik maksimum . \end{array}

Soalan 4:

Rajah di atas menunjukkan sebuah kon dengan diameter 0.8m dan tinggi 0.6m. Air dituang ke dalam kon dengan kadar tetap 0.02m³s^-1. Cari kadar perubahan tinggi paras air pada ketika tinggi parasnya ialah 0.5m.

Penyelesaian:

\begin{array}{l} Isipadu air, I = \frac{1}{3} π j^{2} h \\ I = \frac{1}{3} π {(\frac{2}{3} h)}^{2} h \\ I = \frac{4}{27} π h^{3} \\ \frac{d I}{d h} = (3) \frac{4}{27} π h^{2} \\ \frac{d I}{d h} = \frac{4}{9} π h^{2} \end{array}

Kadar perubahan tinggi paras air pada ketika tinggi parasnya ialah 0.5m =

\frac{d h}{d t} .

\begin{array}{l} \frac{d h}{d t} = \frac{d h}{d I} \times \frac{d I}{d t} \leftarrow Petua rantai \\ \frac{d h}{d t} = \frac{9}{4 π h^{2}} \times 0.02 \leftarrow Diberi \frac{d I}{d t} = 0.02 \\ \frac{d h}{d t} = \frac{9}{4 π {(0.5)}^{2}} \times 0.02 \\ \frac{d h}{d t} = 0.0572 m s^{- 1} \end{array}

Katakan,

h = tinggi paras air

j = jejari permukaan air

I = isipadu air

\begin{array}{l} \frac{j}{h} = \frac{0.4}{0.6} \leftarrow Konsep segitiga serupa \\ \frac{j}{h} = \frac{2}{3} \\ j = \frac{2}{3} h \end{array}

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}