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LEMBAR KERJA PRAKTIKUM :

1. Selesaikan dalam bentuk decimal

a. 1110,112 = 10

Jawaban 507,632 = 327,799910

c. 2BA,CA316 = 10

2BA =

A / 10 x 160_{= 10} B / 11 x 161_{= 176} 2 x 162 ₌

Jawaban 2BA,CA316 = 698,789710

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LEMBAR KERJA PRAKTIKUM :

139 = 10001011

Jawaban

139,852

10

= 10001011,1101

2

 Oktal

139,85210 = 8

139 =

139 : 8 = 17 sisa 3

17 : 8 = 2 sisa 1

2 : 8 = 0 sisa 2

139 = 213

Jawaban 139,85210 = 213,66418

0,852 =

0,852 x 2 = 1,704 sisa 1 0,704 x 2 = 1,408 sisa 1 0,408 x 2 = 0,816 sisa 0 0,816 x 2 = 1,632 sisa 1

0,852 = 1101

0,852 =

0,852 x 8 = 6,816 sisa 6 0,816 x 8 = 6,528 sisa 6 0,528 x 8 = 4,224 sisa 4 0,224 x 8 = 1,792 sisa 1

0,852 = 6641

2. Konversi bilangan 139,85210 ke dalam bentuk biner, oktal dan heksa.

 Biner

139,85210 = 2

139 =

139 : 2 = 69 sisa 1 69 : 2 = 34 sisa 1 34 : 2 = 17 sisa 0 17 : 2 = 8 sisa 1 8 : 2 = 4 sisa 0 4 : 2 = 2 sisa 0 2 : 2 = 1 sisa 0 1 : 2 = 0 sisa 1

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LEMBAR KERJA PRAKTIKUM :

 Heksa

139,85210 = 16

139 : 16 = 8 sisa 11/B

8 : 16 = 0 sisa 8

139 = 8B

Jawaban 139,85210 = 8B,DA1C16

3. Konversi tiga angka terakhir dari NIM anda ke dalam bentuk biner, octal dan heksa

 NIM = 414316

 31610 = 2

316 : 2 = 158 sisa 0

158 : 2 = 79 sisa 0

79 : 2 = 39 sisa 1

39 : 2 = 19 sisa 1

19 : 2 = 9 sisa 1

9 : 2 = 4 sisa 1

4 : 2 = 2 sisa 0

2 : 2 = 1 sisa 0

1

Jawaban

316

10

= 100111100

2

0,852 x 16 = 13,632 sisa 13 / D 0,632 x 16 = 10,112 sisa 10 / A 0,112 x 16 = 1,792 sisa 1 0,792 x 16 = 12,672 sisa 12/C

0,852 = DA1C

 31610 = 8

316 : 8 = 39 sisa 4

39 : 8 = 4 sisa 7

7 : 8 = 0 sisa 1

Jawaban

316

10

= 174

8

 31610 = 16

316 : 16 = 19 sisa 12

19 : 16 = 1 sisa 3

Jawaban 31610 = 31216

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LEMBAR KERJA PRAKTIKUM :

A. Kerjakan dengan cara panjang

9. Tentukan turunan pertama dari masing-masing fungsi pada nilai x berikut ini

a. 𝒇(𝒙) = 𝟐𝒙𝟐+ 𝟒𝒙 , 𝒑𝒂𝒅𝒂 𝒙 = 𝟓

𝑓(𝑥) + Δ𝑥 = 2(𝑥 + Δ𝑥)2_{+ 4(𝑥 + Δ𝑥) − 𝑦}

𝑓(𝑥) + Δ𝑥 = 2(𝑥 + Δ𝑥)2_{+ 4(𝑥 + Δ𝑥) −}_(2𝑥2_{+ 4𝑥)} (𝑥 + Δ𝑥)2_{= 𝑥}2_{+ 2𝑥Δ𝑥 + Δ𝑥}2

2(𝑥 + Δ𝑥)2_{= 2𝑥}2_{+ 4𝑥Δ𝑥 + 2Δ𝑥}2

𝑓(𝑥) + Δ𝑥 = 2𝑥2_{+ 4𝑥Δ𝑥 + 2Δ𝑥}2_{+ 4𝑥 + 4Δ𝑥 −}_(2𝑥2_{+ 4𝑥)} 𝑓(𝑥) + Δ𝑥 = 4𝑥Δ𝑥 + 2Δ𝑥2_{+ 4Δ𝑥}

Δ𝑦 = Δ𝑥(4𝑥 + 2Δ𝑥 + 4) Δ𝑦

Δ𝑥 = limΔ𝑥→0(4𝑥 + 2Δ𝑥 + 4) Δ𝑦

Δ𝑥 = (4𝑥 + 2(0) + 4)

f′_{(x) = 4𝑥 + 4} 𝑓′_{(5) = 4(5) + 4} 𝑓′_{(5) = 24}

Jawaban adalah 24

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LEMBAR KERJA PRAKTIKUM :

b. 𝒇(𝒙) = 𝒙𝟐− 𝟓𝒙 + 𝟑 , 𝒑𝒂𝒅𝒂 𝒙 = (−𝟐)

 _{𝑓(𝑥) + Δ𝑥 = (𝑥 + 𝛥𝑥)}2_{− 5(𝑥 + 𝛥𝑥) + 3 − 𝑦}

𝑓(𝑥) + Δ𝑥 = (𝑥 + 𝛥𝑥)2_{− 5(𝑥 + 𝛥𝑥) + 3 − (𝑥}2_{− 5𝑥 + 3)} (𝑥 + Δ𝑥)2_{= 𝑥}2_{+ 2𝑥Δ𝑥 + Δ𝑥}2

𝛥𝑦 = 𝑥2_{+ 2𝑥𝛥𝑥 + 𝛥𝑥}2_{− 5𝑥 − 5𝛥𝑥 + 3 − (𝑥}2_{− 5𝑥 + 3)} 𝛥𝑦 = 2𝑥𝛥𝑥 + 𝛥𝑥2_{− 5𝛥𝑥}

𝛥𝑦 = 𝛥𝑥(2𝑥+𝛥𝑥 − 5) 𝛥𝑦

𝛥𝑥 = lim𝛥𝑥→02𝑥+𝛥𝑥 − 5 𝛥𝑦

𝛥𝑥 = 2𝑥+(0) − 5 𝑓′(𝑥) = 2𝑥 − 5

𝑓′(−2) = 2(−2) − 5 → 𝑓′_{(−2) = −4 − 5, 𝑓}′_{(−2) = (−9)}

* 𝑱𝒂𝒅𝒊 𝒕𝒖𝒓𝒖𝒏𝒂𝒏 𝒑𝒆𝒓𝒕𝒂𝒎𝒂 𝒅𝒂𝒓𝒊 𝒇(𝒙) = 𝒙𝟐− 𝟓𝒙 + 𝟑 , 𝒑𝒂𝒅𝒂 𝒙 = (−𝟐)𝒂𝒅𝒂𝒍𝒂𝒉(−𝟗)

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10.Sebuah benda bergerak melintasi garis lurus setelah t sekon menmpuh jarak yang dirumuskan dengan

𝒔 = 𝟏_𝟑 𝒕𝟑_{− 𝟐𝒕}𝟐 _{+ 𝟑𝒕 + 𝟓 (𝑠 𝑑𝑎𝑙𝑎𝑚 𝑚𝑒𝑡𝑒𝑟).}_{Tentukann waktu t pada saat v = 0.}

 𝒔 = 𝟏_𝟑 𝒕𝟑_{− 𝟐𝒕}𝟐 _{+ 𝟑𝒕 + 𝟓}

Δ𝑠 =1

3 (𝑡 + Δ𝑡)3− 2(𝑡 + Δ𝑡)2+ 3(𝑡 + Δ𝑡) + 5 − 𝑠 Δ𝑠 =1

3 (𝑡 + Δ𝑡)3− 2(𝑡 + Δ𝑡)2+ 3(𝑡 + Δ𝑡) + 5 − ( 1

3 𝑡3− 2𝑡2 + 3𝑡 + 5 ) Δ𝑠 =1_{3 (𝑡 + Δ𝑡)}3_{− 2(𝑡 + Δ𝑡)}2_{+ 3(𝑡 + Δ𝑡) + 5 − (}1

3 𝑡3− 2𝑡2 + 3𝑡 + 5 ) Δ𝑠 =1_{3 (𝑡}3_{+ 3𝑡}2_{Δ𝑡 + 3𝑡Δ𝑡}2_{+ Δ𝑡}3_{) − 2(𝑡}2_{+ 2𝑡Δ𝑡 + Δ𝑡}2₎ _{+ 3𝑡 + 3Δ𝑡 + 5 − (}1

3 𝑡3− 2𝑡2 + 3𝑡 + 5 ) Δ𝑠 =1_{3 𝑡}3_{+ 𝑡}2_{Δ𝑡 + 𝑡Δ𝑡}2₊1

3 Δ𝑡3− 2𝑡2− 4𝑡Δ𝑡 − 2Δ𝑡2 + 3𝑡 + 3Δ𝑡 + 5 − 1

3 𝑡3+ 2𝑡2 − 3𝑡 − 5 ) Δ𝑠 = 𝑡2_{Δ𝑡 + 𝑡Δ𝑡}2₊1

3 Δ𝑡3−4𝑡Δ𝑡 − 2Δ𝑡2 + 3Δ𝑡 Δ𝑠 = Δ𝑡(𝑡2_{+ 𝑡Δ𝑡 +}1

3 Δ𝑡2−4𝑡 − 2Δ𝑡 + 3) Δ𝑠

Δ𝑡 = limΔ𝑡→0(𝑡2+ 𝑡Δ𝑡 + 1

3 Δ𝑡2−4𝑡 − 2Δ𝑡 + 3) Δ𝑠

Δ𝑡 = (𝑡2+ 𝑡(0) + 1

3 (0)2−4𝑡 − 2(0) + 3) 𝑉 = 𝑡2_−4𝑡 _{+ 3}

𝑉 = 0 → 0 = 𝑡2_{−4𝑡 + 3}

0 = (𝑡 − 3)(𝑡 − 1) → 𝑡 = 3 𝑡 = 1⁄

𝑻 𝒂𝒅𝒂𝒍𝒂𝒉 (𝟑𝒔)𝒅𝒂𝒏 (𝟏𝒔)𝒑𝒂𝒅𝒂 𝒔𝒂𝒂𝒕 𝒗 = 𝟎

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LEMBAR KERJA PRAKTIKUM :

12.Sebuah partikel bergerak setelah t detik menempuh jarak yang dinyatakan dengan rumus 𝒔 = 𝟐𝟒𝒕 +

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LEMBAR KERJA PRAKTIKUM :

B. Kerjakan menggunakan cara u,v

13.Jika 𝒇(𝒙) = (𝟐𝒙 + 𝟏)𝟐+ (𝒙𝟐+ 𝟑𝒙 − 𝟏)𝟑 . 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝒇′(𝟏)

 _{𝑚𝑖𝑠𝑎𝑙 𝑢 = (2𝑥 + 1)}2_{, 𝑢}′_{= 2(2𝑥 + 1)(2)}

𝑣 = (𝑥2_{+ 3𝑥 − 1)}3_{, 𝑣}′_{= 3(𝑥}2_{+ 3𝑥 − 1)}2_{(2𝑥 + 3)} 𝑓′_{(𝑥) = 𝑢}′_{+ 𝑣}′

𝑓′_{(𝑥) = 4(2𝑥 + 1) + (𝑥}2_{+ 3𝑥 − 1)}2_{(6𝑥 + 9)} 𝑓′_{(1) = 4(2(1) + 1) + ((1)}2_{+ 3(1) − 1)}2_{(6(1) + 9)} 𝑓′_{(1) = 12 + (9)(15)}

𝑓′_{(1) = 12 + 135} 𝑓′_{(1) = 147}

14.𝐽𝑖𝑘𝑎 𝒇(𝒙) = 𝟑√𝒙𝟑 𝟐 + 𝟖√𝒙𝟒 𝟓 , 𝑚𝑎𝑘𝑎 𝑡𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝒇′(𝟏)

 _{𝑚𝑖𝑠𝑎𝑙 𝑢 = 3√𝑥}3 2

𝑎𝑡𝑎𝑢 𝑢 = 3𝑥23, 𝑢′ = 2 3 3𝑥

−1₃

𝑣 = 8√𝑥4 5 _{𝑎𝑡𝑎𝑢 𝑣 = 8𝑥}54 , 𝑣′= 5 4 8𝑥

1 4

𝑓′_{(𝑥) = 𝑢}′_{+ 𝑣}′

𝑓′_{(𝑥) = 2𝑥}−1₃_{+ 10𝑥}1₄ 𝑓′_{(𝑥) = 2(1)}−1₃_{+ 10(1)}1₄ 𝑓′_{(1) = 12}

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LEMBAR KERJA PRAKTIKUM :

15.𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑛𝑖𝑙𝑎𝑖 𝑓𝑢𝑛𝑔𝑠𝑖 𝒇(𝒙) = (𝟑𝒙+𝟏)𝟐

(𝒙−𝟑)𝟑 , 𝑢𝑛𝑡𝑢𝑘 𝑥 = (−1) 𝑢 = (3𝑥 + 1)2_{, 𝑢}′ _{= 6(3𝑥 + 1)}

𝑣 = (𝑥 − 3)3_{, 𝑣}′_{= 3(𝑥 − 3)}2

𝑓′_{(𝑥) =}𝑢′𝑣 − 𝑢𝑣′ 𝑣2

𝑓′_{(𝑥) =}6(3𝑥 + 1). ((𝑥 − 3)3) − ((3𝑥 + 1)2. (3(𝑥 − 3)2)) ((𝑥 − 3)3₎2

𝑓′_{(−1) =}6(3(−1) + 1). (((−1) − 3)3) − ((3(−1) + 1)2. (3((−1) − 3)2)) (((−1) − 3))6

𝑓′_{(−1) =}6(−2). ((−4)3) − ((−2)2. (3((−4)2)) (−4)6

𝑓′_{(−1) =}(−12). (−64) − ((4). (48)) (−4)6

𝑓

′

_{(−1) =}

768−192

4096

→ 𝑓

′

(−1) =

576 4096

𝑓′_{(−1) =} 9

64

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LEMBAR KERJA PRAKTIKUM :

16.𝐷𝑖𝑘𝑒𝑡𝑎ℎ𝑢𝑖 𝑓𝑢𝑛𝑔𝑠𝑖 𝒇(𝒙) = (𝟐𝒙𝟑+ 𝟓)(𝟑𝒙𝟒− 𝟐). 𝑁𝑖𝑙𝑎𝑖 𝑡𝑢𝑟𝑢𝑛𝑎𝑛 𝑓(𝑥)𝑢𝑛𝑡𝑢𝑘 𝑥 = 1 𝑎𝑑𝑎𝑙𝑎ℎ

 _{𝑚𝑖𝑠𝑎𝑙 𝑢 = (2𝑥}3_{+ 5), 𝑢}′_{= 6𝑥}2 𝑣 = (3𝑥4_{− 2), 𝑣}′ _{= 12𝑥}3 𝑓′_{(𝑥) = 𝑢}′_{𝑣 + 𝑢𝑣}′

𝑓′_{(𝑥) = (6𝑥}2_)(3𝑥4_{− 2) + (2𝑥}3_{+ 5)(12𝑥}3₎

𝑓′_{(1) = (6(1)}2₎₍₃₍₁₎4_{− 2) + (2(1)}3_{+ 5)(12(1)}3₎ 𝑓′_{(1) = (6)(1) + (7)(12)}

𝑓𝑢𝑜′_{(1) = (6 + 84) → 𝑓}′_{(1) = 90}

17.𝐷𝑖𝑘𝑒𝑡𝑎ℎ𝑢𝑖 𝒇(𝒙) = 𝟑𝒙𝟐− 𝟓𝒙 + 𝟐 𝑑𝑎𝑛 𝒈(𝒙) = 𝒙𝟐+ 𝟑𝒙 − 𝟑.

𝐽𝑖𝑘𝑎 ℎ(𝑥) = 𝑓(𝑥) − 2𝑔(𝑥), 𝑚𝑎𝑘𝑎 𝑏𝑒𝑠𝑎𝑟 𝑛𝑖𝑙𝑎𝑖 ℎ

′

(𝑥) 𝑎𝑑𝑎𝑙𝑎ℎ

𝑓(𝑥) = 3𝑥2_{− 5𝑥 + 2 , 𝑓}′_{(𝑥) = 6𝑥 − 5} 𝑔(𝑥) = 𝑥2_{+ 3𝑥 − 3 , 𝑔}′_{(𝑥) = 2𝑥 + 3} ℎ(𝑥) = 𝑓(𝑥) − 2𝑔(𝑥)

ℎ(𝑥) = 3𝑥2_{− 5𝑥 + 2 − 2(𝑥}2_{+ 3𝑥 − 3 )} ℎ(𝑥) = 3𝑥2_{− 5𝑥 + 2 − 2𝑥}2_{− 6𝑥 + 6)} ℎ(𝑥) = 𝑥2_{− 11𝑥 + 8}

ℎ′_{(𝑥) = 2𝑥 − 11}

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