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Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: Sistem Bilangan
Asisten :
LEMBAR KERJA PRAKTIKUM :
1. Selesaikan dalam bentuk decimal
a. 1110,112 =
10
1110 =
0 x 20 =
1 x 21 =
1 x 22 =
1 x 23 =
0,11 =
0
2
4
8
1 x 2-1 = 1/2 = 0.5
1 x 2-2 = 1/4 = 0.25
—
+
0,75
— +
14
Jawaban 111,112 = 14,7510
b. 507,6328 =
10
507 =
0,632 =
7 x 80 = 7
0 x 81 = 0
5 x 82 = 320
—+
327
6 x 8-1 = 0,75
3 x 8-2 = 0,046
2 x 8-3 = 0,0039
+
0,7999
Jawaban 507,632 = 327,799910
c. 2BA,CA316 =
10
2BA =
A / 10 x 160 = 10
B / 11 x 161 = 176
2 x 162 = 512
698
Jawaban 2BA,CA316 = 698,789710
0,CA3
C / 12 x 16-1 = 0,75
A / 10 x 16-2 = 0,0390
3 x 16-3 = 0,0007
0,7897
Halaman : 1
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
2. Konversi bilangan 139,85210 ke dalam bentuk biner, oktal dan heksa.
Biner
139,85210 =
2
139 =
0,852 =
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
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
139 = 10001011
Jawaban 139,85210 = 10001011,11012
Oktal
139,85210 =
8
139 =
139 : 8 = 17 sisa 3
17 : 8 = 2 sisa 1
2 : 8 = 0 sisa 2
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
139 = 213
Jawaban 139,85210 = 213,66418
Halaman : 2
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
Heksa
139,85210 =
16
139 : 16 = 8 sisa 11/B
8 : 16 = 0 sisa 8
139 = 8B
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
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
31610 =
8
316 : 8 = 39 sisa 4
39 : 8 =
4 sisa 7
7:8=
0 sisa 1
Jawaban 31610 = 1748
39 : 2 = 19 sisa 1
19 : 2 = 9 sisa 1
31610 =
16
9:2=
4 sisa 1
4:2=
2 sisa 0
19 : 16 = 1 sisa 3
2:2=
1 sisa 0
Jawaban 31610 = 31216
316 : 16 = 19 sisa 12
1
Jawaban 31610 = 1001111002
Halaman : 3
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: LIMIT
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
4. Hitunglah 𝐥𝐢𝐦
𝒙→(−𝟐)
𝐥𝐢𝐦
𝒙→(−𝟐)
𝒙𝟑 − 𝒙𝟐 −𝒙+𝟏𝟎
𝒙𝟑 − 𝒙𝟐 −𝒙+𝟏𝟎
𝒙𝟐 +𝟑𝒙+𝟐
𝒙𝟐 +𝟑𝒙+𝟐
= 𝐥𝐢𝐦
𝒙→(−𝟐)
(𝒙+𝟐)(𝒙𝟐 −𝟑𝒙+𝟓)
(𝒙+𝟐)(𝒙+𝟏)
=
((−2)2 (−3(−2))+5)
=
4+6+5
=
/
15
−1
−1
= −15
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝒙→ ∞
5. Buktikan bahwa 𝐥𝐢𝐦
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝒙→ ∞
𝐥𝐢𝐦
⇔ 𝐥𝐢𝐦
𝒙→ ∞
𝒙𝟐 + 𝟒𝒙𝟐 −𝟖
𝟐𝒙𝟐 −𝟑
𝟓𝒙𝟐 −𝟖
=
𝟓
𝟓−𝟎
𝒙→ ∞
𝟓
⇔ 𝟐−𝟎 =
𝟐
𝟓
𝟐
⇔ =
𝟓
⇔ 𝐥𝐢𝐦 𝟐𝒙𝟐⁄𝒙𝟐−𝟑⁄𝒙𝟐 =
𝟓
⇔ 𝟐−𝟑⁄∞𝟐 = 𝟐
=𝟐
𝟓𝒙𝟐⁄𝒙𝟐 −𝟖⁄𝒙𝟐
𝟓
=𝟐
𝟓−𝟖⁄∞𝟐
𝟓
𝟐
⇔ 𝐥𝐢𝐦 𝟐𝒙𝟐−𝟑 =
𝟐
𝒙→ ∞
/
((−2)+1 )
𝟓
𝟐
𝟓
𝟐
maka,
𝑻𝒆𝒓𝒃𝒖𝒌𝒕𝒊 𝐥𝐢𝐦
𝒙→ ∞
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝟓
=𝟐
Halaman : 4
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
LEMBAR KERJA PRAKTIKUM :
𝒙
6. Hitunglah 𝐥𝐢𝐦
𝒙
𝒙→𝟎 𝟐−√𝟒−𝒙
𝒙
𝐥𝐢𝐦 𝟐−√𝟒−𝒙 = 𝐥𝐢𝐦 𝟐−√𝟒−𝒙 𝒙
𝒙→𝟎
𝒙→𝟎
𝟐+√𝟒−𝒙
𝟐+√𝟒−𝒙
𝟐𝒙+𝒙√𝟒−𝒙
= 𝐥𝐢𝐦 𝟒+𝟐√𝟒−𝒙−𝟐√𝟒−𝒙−(𝟒−𝒙)
𝒙→𝟎
𝟐𝒙+𝒙√𝟒−𝒙
𝒙→𝟎 𝟒−(𝟒−𝒙)
= 𝐥𝐢𝐦
= 𝐥𝐢𝐦
𝒙→𝟎
= 𝐥𝐢𝐦
𝒙→𝟎
𝟐𝒙+𝒙√𝟒−𝒙
𝒙
𝒙(𝟐+√𝟒−𝒙 )
/𝒙
/
= 𝟐 + √𝟒 − (𝟎) = 2 + 2 = 4
𝑱𝒂𝒘𝒂𝒃𝒂𝒏 𝐥𝐢𝐦
𝒙→𝟎
𝒙𝟐 +𝟒𝒙−𝟓 5
)
𝒙𝟐 −𝟏
𝒙→𝟏
7. Hitunglah 𝐥𝐢𝐦(
𝒙𝟐 +𝟒𝒙−𝟓 5
)
𝒙𝟐 −𝟏
𝒙→𝟏
𝐥𝐢𝐦(
𝒙
=𝟒
𝟐 − √𝟒 − 𝒙
(𝒙−𝟏)(𝒙+𝟓)
= 𝐥𝐢𝐦 ( (𝒙−𝟏)(𝒙+𝟏) )5
𝒙→𝟏
(𝒙+𝟓)
= 𝐥𝐢𝐦 ( (𝒙+𝟏) )5
𝒙→𝟏
=(
((𝟏)+𝟓) 5
)
((𝟏)+𝟏)
𝟔
=( )5 = (3)5 = 243
𝟐
𝒙𝟐 + 𝟒𝒙 − 𝟓 𝟓
𝑱𝒂𝒘𝒂𝒃𝒂𝒏 𝒅𝒂𝒓𝒊 𝐥𝐢𝐦(
) = 𝟐𝟒𝟑
𝒙→𝟏
𝒙𝟐 − 𝟏
8. Hitunglah 𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐)
𝒙→(−𝟏)
𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐) = 𝟓( 𝟑(−𝟏)𝟐 + 𝟐)
𝒙→(−𝟏)
=5(5) = 25
𝒉𝒂𝒔𝒊𝒍 𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐) = 𝟐𝟓
𝒙→(−𝟏)
Halaman : 5
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: Differensial
Asisten : .........................................................
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) − (2𝑥 2 + 4𝑥)
(𝑥 + Δ𝑥)2 = 𝑥 2 + 2𝑥Δ𝑥 + Δ𝑥 2
2(𝑥 + Δ𝑥)2 = 2𝑥 2 + 4𝑥Δ𝑥 + 2Δ𝑥 2
𝑓(𝑥) + Δ𝑥 = 2𝑥 2 + 4𝑥Δ𝑥 + 2Δ𝑥 2 + 4𝑥 + 4Δ𝑥 − (2𝑥 2 + 4𝑥)
𝑓(𝑥) + Δ𝑥 = 4𝑥Δ𝑥 + 2Δ𝑥 2 + 4Δ𝑥
Δ𝑦 = Δ𝑥(4𝑥 + 2Δ𝑥 + 4)
Δ𝑦
= lim (4𝑥 + 2Δ𝑥 + 4)
Δ𝑥 Δ𝑥→0
Δ𝑦
= (4𝑥 + 2(0) + 4)
Δ𝑥
f ′ (x) = 4𝑥 + 4
𝑓 ′ (5) = 4(5) + 4
𝑓 ′ (5) = 24
Jawaban adalah 24
Halaman : 6
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
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 = 𝑥 2 + 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝑥 − 5𝛥𝑥 + 3 − (𝑥 2 − 5𝑥 + 3)
𝛥𝑦 = 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝛥𝑥
𝛥𝑦 = 𝛥𝑥(2𝑥+𝛥𝑥 − 5)
𝛥𝑦
= lim 2𝑥+𝛥𝑥 − 5
𝛥𝑥 𝛥𝑥→0
𝛥𝑦
= 2𝑥+(0) − 5
𝛥𝑥
𝑓′(𝑥) =
𝑓′(−2) =
2𝑥 − 5
2(−2) − 5 → 𝑓 ′ (−2) = −4 − 5, 𝑓 ′ (−2) = (−9)
* 𝑱𝒂𝒅𝒊 𝒕𝒖𝒓𝒖𝒏𝒂𝒏 𝒑𝒆𝒓𝒕𝒂𝒎𝒂 𝒅𝒂𝒓𝒊 𝒇 (𝒙) = 𝒙𝟐 − 𝟓𝒙 + 𝟑 , 𝒑𝒂𝒅𝒂 𝒙 = (−𝟐)𝒂𝒅𝒂𝒍𝒂𝒉(−𝟗)
Halaman : 7
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
10. Sebuah benda bergerak melintasi garis lurus setelah t sekon menmpuh jarak yang dirumuskan dengan
𝒔=
𝒔=
𝟏
𝟑
𝟏
𝟑
𝒕𝟑 − 𝟐𝒕𝟐 + 𝟑𝒕 + 𝟓 (𝑠 𝑑𝑎𝑙𝑎𝑚 𝑚𝑒𝑡𝑒𝑟). Tentukann waktu t pada saat v = 0.
𝒕𝟑 − 𝟐𝒕𝟐 + 𝟑𝒕 + 𝟓
1
(𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − 𝑠
3
1
1
Δ𝑠 = (𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
Δ𝑠 = (𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
Δ𝑠 = (𝑡 3 + 3𝑡 2 Δ𝑡 + 3𝑡Δ𝑡 2 + Δ𝑡 3 ) − 2(𝑡 2 + 2𝑡Δ𝑡 + Δ𝑡 2 ) + 3𝑡 + 3Δ𝑡 + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
1
Δ𝑠 = 𝑡 3 + 𝑡 2 Δ𝑡 + 𝑡Δ𝑡 2 + Δ𝑡 3 − 2𝑡 2 − 4𝑡Δ𝑡 − 2Δ𝑡 2 + 3𝑡 + 3Δ𝑡 + 5 − 𝑡 3 + 2𝑡 2 − 3𝑡 − 5 )
3
3
3
1
Δ𝑠 = 𝑡 2 Δ𝑡 + 𝑡Δ𝑡 2 + Δ𝑡 3 −4𝑡Δ𝑡 − 2Δ𝑡 2 + 3Δ𝑡
3
1
Δ𝑠 = Δ𝑡(𝑡 2 + 𝑡Δ𝑡 + Δ𝑡 2 −4𝑡 − 2Δ𝑡 + 3)
3
Δ𝑠
1
= lim (𝑡 2 + 𝑡Δ𝑡 + Δ𝑡 2 −4𝑡 − 2Δ𝑡 + 3)
Δ𝑡 Δ𝑡→0
3
Δ𝑠
1
= (𝑡 2 + 𝑡(0) + (0)2 −4𝑡 − 2(0) + 3)
Δ𝑡
3
Δ𝑠 =
𝑉 = 𝑡 2 −4𝑡 + 3
𝑉 = 0 → 0 = 𝑡 2 −4𝑡 + 3
0 = (𝑡 − 3)(𝑡 − 1) → 𝑡 = 3⁄𝑡 = 1
𝑻 𝒂𝒅𝒂𝒍𝒂𝒉 (𝟑𝒔)𝒅𝒂𝒏 (𝟏𝒔)𝒑𝒂𝒅𝒂 𝒔𝒂𝒂𝒕 𝒗 = 𝟎
Halaman : 8
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
11. Tentukan turunan dari fungsi 𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟑 𝒑𝒂𝒅𝒂 𝒙 = 𝟐
𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟑
𝛥𝑦 = (𝑥 + 𝛥𝑥)2 − 5(𝑥 + 𝛥𝑥) + 3 − 𝑦
𝛥𝑦 = (𝑥 + 𝛥𝑥)2 − 5(𝑥 + 𝛥𝑥) + 3 − 𝑥 2 + 5𝑥 − 3
𝛥𝑦 = 𝑥 2 + 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝑥 − 5𝛥𝑥 + 3 − 𝑥 2 + 5𝑥 − 3
𝛥𝑦 = 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝛥𝑥
𝛥𝑦 = 𝛥𝑥(2𝑥 + 𝛥𝑥 − 5)
𝛥𝑦
= lim (2𝑥 + 𝛥𝑥 − 5)
𝛥𝑥 𝛥𝑥→0
𝛥𝑦
= (2𝑥 + (0) − 5)
𝛥𝑥
𝑓 ′ (𝑥) = (2𝑥 − 5)
𝑓 ′ (2) = 2(2) − 5
𝑓 ′ (2) = (−1)
12. Sebuah partikel bergerak setelah t detik menempuh jarak yang dinyatakan dengan rumus 𝒔 = 𝟐𝟒𝒕 +
𝟏
𝟔𝒕𝟐 − 𝟑 𝒕𝟑 . Tentukan besarnya kecepatan pada saat t = 3 sekon.
𝟏
𝟑
𝒔 = 𝟐𝟒𝒕 + 𝟔𝒕𝟐 − 𝒕𝟑
1
𝛥𝑠 = 24(𝑡 + 𝛥𝑡) + 6(𝑡 + 𝛥𝑡 )2 − (𝑡 + 𝛥𝑡)3 − 𝑠
3
1
1
𝛥𝑠 = 24(𝑡 + 𝛥𝑡) + 6(𝑡 + 𝛥𝑡 )2 − (𝑡 + 𝛥𝑡)3 − 24𝑡 − 6𝑡 2 + 𝑡 3
3
3
1
1
1
𝛥𝑠 = 24𝑡 + 24𝛥𝑡 + 6𝑡 2 + 12𝑡𝛥𝑡 + 6𝛥𝑡 2 − 𝑡 3 − 𝑡 2 𝛥𝑡 − 𝑡𝛥𝑡 2 − 𝛥𝑡 3 − 24𝑡 − 6𝑡 2 + 𝑡 3
3
3
3
1
𝛥𝑠 = 24𝛥𝑡+12𝑡𝛥𝑡 + 6𝛥𝑡 2 − 𝑡 2 𝛥𝑡 − 𝑡𝛥𝑡 2 − 𝛥𝑡 3
3
𝟏
𝒔 = 𝟐𝟒𝒕 + 𝟔𝒕𝟐 − 𝒕𝟑 pada saat
𝛥𝑠
1 2
𝟑
2
= lim 24+12𝑡 + 6𝛥𝑡 − 𝑡 − 𝑡𝛥𝑡 − 𝛥𝑡
t= 3 sekon yaitu
𝛥𝑡 𝛥𝑡→0
3
(-51𝒎⁄𝒔𝟐 )
1
𝛥𝑠
2
2
= 24+12𝑡 + 6(0) − 𝑡 − 𝑡(0) − (0)
3
𝛥𝑡
𝛥𝑠
= 24 + 12𝑡 − 𝑡 2
𝛥𝑡
𝑉(𝑡) = 24 + 12𝑡 − 𝑡 2 → 𝑉(3) = 24 + 12(3) − (3)2 , 𝑉(3) = 51 𝑚⁄𝑠 2
Halaman : 9
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
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 = (𝑥 2 + 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
𝟒
𝟑
14. 𝐽𝑖𝑘𝑎 𝒇(𝒙) = 𝟑√𝒙𝟐 + 𝟖√𝒙𝟓 , 𝑚𝑎𝑘𝑎 𝑡𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝒇′(𝟏)
2
3
𝑚𝑖𝑠𝑎𝑙 𝑢 = 3√𝑥 2 𝑎𝑡𝑎𝑢 𝑢 = 3𝑥 3 , 𝑢′ =
5
4
𝑣 = 8√𝑥 5 𝑎𝑡𝑎𝑢 𝑣 = 8𝑥 4 , 𝑣 ′ =
𝑓 ′ (𝑥) = 𝑢′ + 𝑣 ′
1
5
4
1
3𝑥 −3
1
8𝑥 4
1
𝑓 ′ (𝑥) = 2𝑥 −3 + 10𝑥 4
1
2
3
1
𝑓 ′ (𝑥) = 2(1)−3 + 10(1)4
𝑓 ′ (1) = 12
Halaman : 10
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Paraf :
Pertemuan Ke :
Materi
: ...................................................................
Asisten : .........................................................
............................................................................................................
LEMBAR KERJA PRAKTIKUM :
15. 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑛𝑖𝑙𝑎𝑖 𝑓𝑢𝑛𝑔𝑠𝑖 𝒇(𝒙) =
𝑢 = (3𝑥 + 1)2 , 𝑢′ = 6(3𝑥 + 1)
(𝟑𝒙+𝟏)𝟐
(𝒙−𝟑)𝟑
, 𝑢𝑛𝑡𝑢𝑘 𝑥 = (−1)
𝑣 = (𝑥 − 3)3 , 𝑣 ′ = 3(𝑥 − 3)2
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) =
𝑢′ 𝑣 − 𝑢𝑣′
𝑣2
6(3𝑥 + 1). ((𝑥 − 3)3 ) − ((3𝑥 + 1)2 . (3(𝑥 − 3)2 ))
((𝑥 − 3)3 )2
𝑓 ′ (−1) =
𝑓 ′ (−1) =
𝑓 ′ (−1) =
6(3(−1) + 1). (((−1) − 3)3 ) − ((3(−1) + 1)2 . (3((−1) − 3)2 ))
(((−1) − 3))6
6(−2). ((−4)3 ) − ((−2)2 . (3((−4)2 ))
(−4)6
(−12). (−64) − ((4). (48))
(−4)6
𝑓 ′ (−1) =
𝑓 ′ (−1) =
9
64
768−192
4096
→ 𝑓 ′ (−1) =
576
4096
Halaman : 11
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
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 = (3𝑥 4 − 2), 𝑣 ′ = 12𝑥 3
𝑓 ′ (𝑥) = 𝑢′ 𝑣 + 𝑢𝑣 ′
𝑓 ′ (𝑥) = (6𝑥 2 )(3𝑥 4 − 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)
𝑓 𝑢𝑜′ (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) = 3𝑥 2 − 5𝑥 + 2 , 𝑓 ′ (𝑥) = 6𝑥 − 5
𝑔(𝑥) = 𝑥 2 + 3𝑥 − 3 , 𝑔′ (𝑥) = 2𝑥 + 3
ℎ(𝑥) = 𝑓(𝑥) − 2𝑔(𝑥)
ℎ(𝑥) = 3𝑥 2 − 5𝑥 + 2 − 2(𝑥 2 + 3𝑥 − 3 )
ℎ(𝑥) = 3𝑥 2 − 5𝑥 + 2 − 2𝑥 2 − 6𝑥 + 6)
ℎ(𝑥) = 𝑥 2 − 11𝑥 + 8
ℎ′ (𝑥) = 2𝑥 − 11
Halaman : 12
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: Sistem Bilangan
Asisten :
LEMBAR KERJA PRAKTIKUM :
1. Selesaikan dalam bentuk decimal
a. 1110,112 =
10
1110 =
0 x 20 =
1 x 21 =
1 x 22 =
1 x 23 =
0,11 =
0
2
4
8
1 x 2-1 = 1/2 = 0.5
1 x 2-2 = 1/4 = 0.25
—
+
0,75
— +
14
Jawaban 111,112 = 14,7510
b. 507,6328 =
10
507 =
0,632 =
7 x 80 = 7
0 x 81 = 0
5 x 82 = 320
—+
327
6 x 8-1 = 0,75
3 x 8-2 = 0,046
2 x 8-3 = 0,0039
+
0,7999
Jawaban 507,632 = 327,799910
c. 2BA,CA316 =
10
2BA =
A / 10 x 160 = 10
B / 11 x 161 = 176
2 x 162 = 512
698
Jawaban 2BA,CA316 = 698,789710
0,CA3
C / 12 x 16-1 = 0,75
A / 10 x 16-2 = 0,0390
3 x 16-3 = 0,0007
0,7897
Halaman : 1
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
2. Konversi bilangan 139,85210 ke dalam bentuk biner, oktal dan heksa.
Biner
139,85210 =
2
139 =
0,852 =
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
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
139 = 10001011
Jawaban 139,85210 = 10001011,11012
Oktal
139,85210 =
8
139 =
139 : 8 = 17 sisa 3
17 : 8 = 2 sisa 1
2 : 8 = 0 sisa 2
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
139 = 213
Jawaban 139,85210 = 213,66418
Halaman : 2
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
Heksa
139,85210 =
16
139 : 16 = 8 sisa 11/B
8 : 16 = 0 sisa 8
139 = 8B
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
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
31610 =
8
316 : 8 = 39 sisa 4
39 : 8 =
4 sisa 7
7:8=
0 sisa 1
Jawaban 31610 = 1748
39 : 2 = 19 sisa 1
19 : 2 = 9 sisa 1
31610 =
16
9:2=
4 sisa 1
4:2=
2 sisa 0
19 : 16 = 1 sisa 3
2:2=
1 sisa 0
Jawaban 31610 = 31216
316 : 16 = 19 sisa 12
1
Jawaban 31610 = 1001111002
Halaman : 3
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: LIMIT
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
4. Hitunglah 𝐥𝐢𝐦
𝒙→(−𝟐)
𝐥𝐢𝐦
𝒙→(−𝟐)
𝒙𝟑 − 𝒙𝟐 −𝒙+𝟏𝟎
𝒙𝟑 − 𝒙𝟐 −𝒙+𝟏𝟎
𝒙𝟐 +𝟑𝒙+𝟐
𝒙𝟐 +𝟑𝒙+𝟐
= 𝐥𝐢𝐦
𝒙→(−𝟐)
(𝒙+𝟐)(𝒙𝟐 −𝟑𝒙+𝟓)
(𝒙+𝟐)(𝒙+𝟏)
=
((−2)2 (−3(−2))+5)
=
4+6+5
=
/
15
−1
−1
= −15
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝒙→ ∞
5. Buktikan bahwa 𝐥𝐢𝐦
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝒙→ ∞
𝐥𝐢𝐦
⇔ 𝐥𝐢𝐦
𝒙→ ∞
𝒙𝟐 + 𝟒𝒙𝟐 −𝟖
𝟐𝒙𝟐 −𝟑
𝟓𝒙𝟐 −𝟖
=
𝟓
𝟓−𝟎
𝒙→ ∞
𝟓
⇔ 𝟐−𝟎 =
𝟐
𝟓
𝟐
⇔ =
𝟓
⇔ 𝐥𝐢𝐦 𝟐𝒙𝟐⁄𝒙𝟐−𝟑⁄𝒙𝟐 =
𝟓
⇔ 𝟐−𝟑⁄∞𝟐 = 𝟐
=𝟐
𝟓𝒙𝟐⁄𝒙𝟐 −𝟖⁄𝒙𝟐
𝟓
=𝟐
𝟓−𝟖⁄∞𝟐
𝟓
𝟐
⇔ 𝐥𝐢𝐦 𝟐𝒙𝟐−𝟑 =
𝟐
𝒙→ ∞
/
((−2)+1 )
𝟓
𝟐
𝟓
𝟐
maka,
𝑻𝒆𝒓𝒃𝒖𝒌𝒕𝒊 𝐥𝐢𝐦
𝒙→ ∞
𝒙𝟐 + √𝟏𝟔𝒙𝟒 −𝟔𝟒
𝟐𝒙𝟐 −𝟑
𝟓
=𝟐
Halaman : 4
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
LEMBAR KERJA PRAKTIKUM :
𝒙
6. Hitunglah 𝐥𝐢𝐦
𝒙
𝒙→𝟎 𝟐−√𝟒−𝒙
𝒙
𝐥𝐢𝐦 𝟐−√𝟒−𝒙 = 𝐥𝐢𝐦 𝟐−√𝟒−𝒙 𝒙
𝒙→𝟎
𝒙→𝟎
𝟐+√𝟒−𝒙
𝟐+√𝟒−𝒙
𝟐𝒙+𝒙√𝟒−𝒙
= 𝐥𝐢𝐦 𝟒+𝟐√𝟒−𝒙−𝟐√𝟒−𝒙−(𝟒−𝒙)
𝒙→𝟎
𝟐𝒙+𝒙√𝟒−𝒙
𝒙→𝟎 𝟒−(𝟒−𝒙)
= 𝐥𝐢𝐦
= 𝐥𝐢𝐦
𝒙→𝟎
= 𝐥𝐢𝐦
𝒙→𝟎
𝟐𝒙+𝒙√𝟒−𝒙
𝒙
𝒙(𝟐+√𝟒−𝒙 )
/𝒙
/
= 𝟐 + √𝟒 − (𝟎) = 2 + 2 = 4
𝑱𝒂𝒘𝒂𝒃𝒂𝒏 𝐥𝐢𝐦
𝒙→𝟎
𝒙𝟐 +𝟒𝒙−𝟓 5
)
𝒙𝟐 −𝟏
𝒙→𝟏
7. Hitunglah 𝐥𝐢𝐦(
𝒙𝟐 +𝟒𝒙−𝟓 5
)
𝒙𝟐 −𝟏
𝒙→𝟏
𝐥𝐢𝐦(
𝒙
=𝟒
𝟐 − √𝟒 − 𝒙
(𝒙−𝟏)(𝒙+𝟓)
= 𝐥𝐢𝐦 ( (𝒙−𝟏)(𝒙+𝟏) )5
𝒙→𝟏
(𝒙+𝟓)
= 𝐥𝐢𝐦 ( (𝒙+𝟏) )5
𝒙→𝟏
=(
((𝟏)+𝟓) 5
)
((𝟏)+𝟏)
𝟔
=( )5 = (3)5 = 243
𝟐
𝒙𝟐 + 𝟒𝒙 − 𝟓 𝟓
𝑱𝒂𝒘𝒂𝒃𝒂𝒏 𝒅𝒂𝒓𝒊 𝐥𝐢𝐦(
) = 𝟐𝟒𝟑
𝒙→𝟏
𝒙𝟐 − 𝟏
8. Hitunglah 𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐)
𝒙→(−𝟏)
𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐) = 𝟓( 𝟑(−𝟏)𝟐 + 𝟐)
𝒙→(−𝟏)
=5(5) = 25
𝒉𝒂𝒔𝒊𝒍 𝐥𝐢𝐦 𝟓( 𝟑𝒙𝟐 + 𝟐) = 𝟐𝟓
𝒙→(−𝟏)
Halaman : 5
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: Differensial
Asisten : .........................................................
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) − (2𝑥 2 + 4𝑥)
(𝑥 + Δ𝑥)2 = 𝑥 2 + 2𝑥Δ𝑥 + Δ𝑥 2
2(𝑥 + Δ𝑥)2 = 2𝑥 2 + 4𝑥Δ𝑥 + 2Δ𝑥 2
𝑓(𝑥) + Δ𝑥 = 2𝑥 2 + 4𝑥Δ𝑥 + 2Δ𝑥 2 + 4𝑥 + 4Δ𝑥 − (2𝑥 2 + 4𝑥)
𝑓(𝑥) + Δ𝑥 = 4𝑥Δ𝑥 + 2Δ𝑥 2 + 4Δ𝑥
Δ𝑦 = Δ𝑥(4𝑥 + 2Δ𝑥 + 4)
Δ𝑦
= lim (4𝑥 + 2Δ𝑥 + 4)
Δ𝑥 Δ𝑥→0
Δ𝑦
= (4𝑥 + 2(0) + 4)
Δ𝑥
f ′ (x) = 4𝑥 + 4
𝑓 ′ (5) = 4(5) + 4
𝑓 ′ (5) = 24
Jawaban adalah 24
Halaman : 6
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
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 = 𝑥 2 + 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝑥 − 5𝛥𝑥 + 3 − (𝑥 2 − 5𝑥 + 3)
𝛥𝑦 = 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝛥𝑥
𝛥𝑦 = 𝛥𝑥(2𝑥+𝛥𝑥 − 5)
𝛥𝑦
= lim 2𝑥+𝛥𝑥 − 5
𝛥𝑥 𝛥𝑥→0
𝛥𝑦
= 2𝑥+(0) − 5
𝛥𝑥
𝑓′(𝑥) =
𝑓′(−2) =
2𝑥 − 5
2(−2) − 5 → 𝑓 ′ (−2) = −4 − 5, 𝑓 ′ (−2) = (−9)
* 𝑱𝒂𝒅𝒊 𝒕𝒖𝒓𝒖𝒏𝒂𝒏 𝒑𝒆𝒓𝒕𝒂𝒎𝒂 𝒅𝒂𝒓𝒊 𝒇 (𝒙) = 𝒙𝟐 − 𝟓𝒙 + 𝟑 , 𝒑𝒂𝒅𝒂 𝒙 = (−𝟐)𝒂𝒅𝒂𝒍𝒂𝒉(−𝟗)
Halaman : 7
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
10. Sebuah benda bergerak melintasi garis lurus setelah t sekon menmpuh jarak yang dirumuskan dengan
𝒔=
𝒔=
𝟏
𝟑
𝟏
𝟑
𝒕𝟑 − 𝟐𝒕𝟐 + 𝟑𝒕 + 𝟓 (𝑠 𝑑𝑎𝑙𝑎𝑚 𝑚𝑒𝑡𝑒𝑟). Tentukann waktu t pada saat v = 0.
𝒕𝟑 − 𝟐𝒕𝟐 + 𝟑𝒕 + 𝟓
1
(𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − 𝑠
3
1
1
Δ𝑠 = (𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
Δ𝑠 = (𝑡 + Δ𝑡)3 − 2(𝑡 + Δ𝑡)2 + 3(𝑡 + Δ𝑡) + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
Δ𝑠 = (𝑡 3 + 3𝑡 2 Δ𝑡 + 3𝑡Δ𝑡 2 + Δ𝑡 3 ) − 2(𝑡 2 + 2𝑡Δ𝑡 + Δ𝑡 2 ) + 3𝑡 + 3Δ𝑡 + 5 − ( 𝑡 3 − 2𝑡 2 + 3𝑡 + 5 )
3
3
1
1
1
Δ𝑠 = 𝑡 3 + 𝑡 2 Δ𝑡 + 𝑡Δ𝑡 2 + Δ𝑡 3 − 2𝑡 2 − 4𝑡Δ𝑡 − 2Δ𝑡 2 + 3𝑡 + 3Δ𝑡 + 5 − 𝑡 3 + 2𝑡 2 − 3𝑡 − 5 )
3
3
3
1
Δ𝑠 = 𝑡 2 Δ𝑡 + 𝑡Δ𝑡 2 + Δ𝑡 3 −4𝑡Δ𝑡 − 2Δ𝑡 2 + 3Δ𝑡
3
1
Δ𝑠 = Δ𝑡(𝑡 2 + 𝑡Δ𝑡 + Δ𝑡 2 −4𝑡 − 2Δ𝑡 + 3)
3
Δ𝑠
1
= lim (𝑡 2 + 𝑡Δ𝑡 + Δ𝑡 2 −4𝑡 − 2Δ𝑡 + 3)
Δ𝑡 Δ𝑡→0
3
Δ𝑠
1
= (𝑡 2 + 𝑡(0) + (0)2 −4𝑡 − 2(0) + 3)
Δ𝑡
3
Δ𝑠 =
𝑉 = 𝑡 2 −4𝑡 + 3
𝑉 = 0 → 0 = 𝑡 2 −4𝑡 + 3
0 = (𝑡 − 3)(𝑡 − 1) → 𝑡 = 3⁄𝑡 = 1
𝑻 𝒂𝒅𝒂𝒍𝒂𝒉 (𝟑𝒔)𝒅𝒂𝒏 (𝟏𝒔)𝒑𝒂𝒅𝒂 𝒔𝒂𝒂𝒕 𝒗 = 𝟎
Halaman : 8
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
............................................................................................................
Asisten : .........................................................
LEMBAR KERJA PRAKTIKUM :
11. Tentukan turunan dari fungsi 𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟑 𝒑𝒂𝒅𝒂 𝒙 = 𝟐
𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟑
𝛥𝑦 = (𝑥 + 𝛥𝑥)2 − 5(𝑥 + 𝛥𝑥) + 3 − 𝑦
𝛥𝑦 = (𝑥 + 𝛥𝑥)2 − 5(𝑥 + 𝛥𝑥) + 3 − 𝑥 2 + 5𝑥 − 3
𝛥𝑦 = 𝑥 2 + 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝑥 − 5𝛥𝑥 + 3 − 𝑥 2 + 5𝑥 − 3
𝛥𝑦 = 2𝑥𝛥𝑥 + 𝛥𝑥 2 − 5𝛥𝑥
𝛥𝑦 = 𝛥𝑥(2𝑥 + 𝛥𝑥 − 5)
𝛥𝑦
= lim (2𝑥 + 𝛥𝑥 − 5)
𝛥𝑥 𝛥𝑥→0
𝛥𝑦
= (2𝑥 + (0) − 5)
𝛥𝑥
𝑓 ′ (𝑥) = (2𝑥 − 5)
𝑓 ′ (2) = 2(2) − 5
𝑓 ′ (2) = (−1)
12. Sebuah partikel bergerak setelah t detik menempuh jarak yang dinyatakan dengan rumus 𝒔 = 𝟐𝟒𝒕 +
𝟏
𝟔𝒕𝟐 − 𝟑 𝒕𝟑 . Tentukan besarnya kecepatan pada saat t = 3 sekon.
𝟏
𝟑
𝒔 = 𝟐𝟒𝒕 + 𝟔𝒕𝟐 − 𝒕𝟑
1
𝛥𝑠 = 24(𝑡 + 𝛥𝑡) + 6(𝑡 + 𝛥𝑡 )2 − (𝑡 + 𝛥𝑡)3 − 𝑠
3
1
1
𝛥𝑠 = 24(𝑡 + 𝛥𝑡) + 6(𝑡 + 𝛥𝑡 )2 − (𝑡 + 𝛥𝑡)3 − 24𝑡 − 6𝑡 2 + 𝑡 3
3
3
1
1
1
𝛥𝑠 = 24𝑡 + 24𝛥𝑡 + 6𝑡 2 + 12𝑡𝛥𝑡 + 6𝛥𝑡 2 − 𝑡 3 − 𝑡 2 𝛥𝑡 − 𝑡𝛥𝑡 2 − 𝛥𝑡 3 − 24𝑡 − 6𝑡 2 + 𝑡 3
3
3
3
1
𝛥𝑠 = 24𝛥𝑡+12𝑡𝛥𝑡 + 6𝛥𝑡 2 − 𝑡 2 𝛥𝑡 − 𝑡𝛥𝑡 2 − 𝛥𝑡 3
3
𝟏
𝒔 = 𝟐𝟒𝒕 + 𝟔𝒕𝟐 − 𝒕𝟑 pada saat
𝛥𝑠
1 2
𝟑
2
= lim 24+12𝑡 + 6𝛥𝑡 − 𝑡 − 𝑡𝛥𝑡 − 𝛥𝑡
t= 3 sekon yaitu
𝛥𝑡 𝛥𝑡→0
3
(-51𝒎⁄𝒔𝟐 )
1
𝛥𝑠
2
2
= 24+12𝑡 + 6(0) − 𝑡 − 𝑡(0) − (0)
3
𝛥𝑡
𝛥𝑠
= 24 + 12𝑡 − 𝑡 2
𝛥𝑡
𝑉(𝑡) = 24 + 12𝑡 − 𝑡 2 → 𝑉(3) = 24 + 12(3) − (3)2 , 𝑉(3) = 51 𝑚⁄𝑠 2
Halaman : 9
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
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 = (𝑥 2 + 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
𝟒
𝟑
14. 𝐽𝑖𝑘𝑎 𝒇(𝒙) = 𝟑√𝒙𝟐 + 𝟖√𝒙𝟓 , 𝑚𝑎𝑘𝑎 𝑡𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝒇′(𝟏)
2
3
𝑚𝑖𝑠𝑎𝑙 𝑢 = 3√𝑥 2 𝑎𝑡𝑎𝑢 𝑢 = 3𝑥 3 , 𝑢′ =
5
4
𝑣 = 8√𝑥 5 𝑎𝑡𝑎𝑢 𝑣 = 8𝑥 4 , 𝑣 ′ =
𝑓 ′ (𝑥) = 𝑢′ + 𝑣 ′
1
5
4
1
3𝑥 −3
1
8𝑥 4
1
𝑓 ′ (𝑥) = 2𝑥 −3 + 10𝑥 4
1
2
3
1
𝑓 ′ (𝑥) = 2(1)−3 + 10(1)4
𝑓 ′ (1) = 12
Halaman : 10
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Paraf :
Pertemuan Ke :
Materi
: ...................................................................
Asisten : .........................................................
............................................................................................................
LEMBAR KERJA PRAKTIKUM :
15. 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑛𝑖𝑙𝑎𝑖 𝑓𝑢𝑛𝑔𝑠𝑖 𝒇(𝒙) =
𝑢 = (3𝑥 + 1)2 , 𝑢′ = 6(3𝑥 + 1)
(𝟑𝒙+𝟏)𝟐
(𝒙−𝟑)𝟑
, 𝑢𝑛𝑡𝑢𝑘 𝑥 = (−1)
𝑣 = (𝑥 − 3)3 , 𝑣 ′ = 3(𝑥 − 3)2
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) =
𝑢′ 𝑣 − 𝑢𝑣′
𝑣2
6(3𝑥 + 1). ((𝑥 − 3)3 ) − ((3𝑥 + 1)2 . (3(𝑥 − 3)2 ))
((𝑥 − 3)3 )2
𝑓 ′ (−1) =
𝑓 ′ (−1) =
𝑓 ′ (−1) =
6(3(−1) + 1). (((−1) − 3)3 ) − ((3(−1) + 1)2 . (3((−1) − 3)2 ))
(((−1) − 3))6
6(−2). ((−4)3 ) − ((−2)2 . (3((−4)2 ))
(−4)6
(−12). (−64) − ((4). (48))
(−4)6
𝑓 ′ (−1) =
𝑓 ′ (−1) =
9
64
768−192
4096
→ 𝑓 ′ (−1) =
576
4096
Halaman : 11
Nama: Nur lutfi H
Kelas : TI 1 semester 4
Pertemuan Ke :
Materi
Paraf :
: ...................................................................
Asisten : .........................................................
............................................................................................................
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 = (3𝑥 4 − 2), 𝑣 ′ = 12𝑥 3
𝑓 ′ (𝑥) = 𝑢′ 𝑣 + 𝑢𝑣 ′
𝑓 ′ (𝑥) = (6𝑥 2 )(3𝑥 4 − 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)
𝑓 𝑢𝑜′ (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) = 3𝑥 2 − 5𝑥 + 2 , 𝑓 ′ (𝑥) = 6𝑥 − 5
𝑔(𝑥) = 𝑥 2 + 3𝑥 − 3 , 𝑔′ (𝑥) = 2𝑥 + 3
ℎ(𝑥) = 𝑓(𝑥) − 2𝑔(𝑥)
ℎ(𝑥) = 3𝑥 2 − 5𝑥 + 2 − 2(𝑥 2 + 3𝑥 − 3 )
ℎ(𝑥) = 3𝑥 2 − 5𝑥 + 2 − 2𝑥 2 − 6𝑥 + 6)
ℎ(𝑥) = 𝑥 2 − 11𝑥 + 8
ℎ′ (𝑥) = 2𝑥 − 11
Halaman : 12