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Tugas matematika 2 (semester 2)
- 1. Nama :Nikmah Utami NPM: 0031421 Kelas : 1 Elektronika A Tugas : MATEMATIKA Tentukanlah nilai 𝑑𝑦 𝑑𝑥 dari fungsi berikut ini ! 1. 𝑦 = √ 𝑥5 + 6𝑥2 + 3 2. 𝑦 = √ 𝑥4 + 6𝑥 + 1 3 3. 𝑦 = √ 𝑥2 − 5𝑥 5 4. 𝑦 = 1 √𝑥4+2𝑥 5. 𝑦 = 1 √𝑥2−6𝑥 3 6. 𝑦 = 1 √𝑥2−5𝑥+2 5 7. 𝑦 = sin √ 𝑥2 + 6𝑥 8. 𝑦 = cos √ 𝑥3 + 2 3 9. 𝑦 = sin 1 √𝑥2+2 10. 𝑦 = cos 1 √𝑥2+6 3 Jawaban : 1. 𝑦 = √ 𝑥5 + 6𝑥2 + 3 Misal u = 𝑥5 + 6𝑥2 + 3 , maka 𝑑𝑢 𝑑𝑥 = 5𝑥4 + 12𝑥 𝑦 = √ 𝑢 = 𝑢 1 2 , maka 𝑑𝑦 𝑑𝑢 = 1 2 𝑢− 1 2 = 1 2 (𝑥5 + 6𝑥2 + 3)− 1 2 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = 1 2 (𝑥5 + 6𝑥2 + 3)− 1 2 . (5𝑥4 + 12𝑥) 𝑑𝑦 𝑑𝑥 = 1 2 (5𝑥4 + 12𝑥) (𝑥5 + 6𝑥2 + 3) 1 2 = 1 2 (5𝑥4 + 12𝑥) √ 𝑥5 + 6𝑥2 + 3 2. 𝑦 = √ 𝑥4 + 6𝑥 + 1 3 Misal u = 𝑥4 + 6𝑥 + 1 , maka 𝑑𝑢 𝑑𝑥 = 4𝑥3 + 6 𝑦 = √ 𝑢3 = 𝑢 1 3 , maka 𝑑𝑦 𝑑𝑢 = 1 3 𝑢− 2 3 = 1 3 (𝑥4 + 6𝑥 + 1)− 2 3 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = 1 3 (𝑥4 + 6𝑥 + 1)− 2 3 . (4𝑥3 + 6)
- 2. 𝑑𝑦 𝑑𝑥 = 1 3 (4𝑥3 + 6) (𝑥4 +6𝑥 + 1) 2 3 = 1 3 (4𝑥3 + 6) √(𝑥4 + 6𝑥 + 1)23 3. 𝑦 = √ 𝑥2 − 5𝑥 5 Misal u = 𝑥2 − 5𝑥 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 − 5 𝑦 = √ 𝑢5 = 𝑢 1 5 , maka 𝑑𝑦 𝑑𝑢 = 1 5 𝑢− 4 5 = 1 5 (𝑥2 − 5𝑥)− 4 5 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = 1 5 (𝑥2 − 5𝑥)− 4 5 .(2𝑥 − 5) 𝑑𝑦 𝑑𝑥 = 1 5 (2𝑥 − 5) (𝑥2 − 5𝑥) 4 5 = 1 5 (2𝑥 − 5) √(𝑥2 − 5𝑥)45 4. 𝑦 = 1 √𝑥4+2𝑥 = 1 (𝑥4+2𝑥) 1 2 = (𝑥4 + 2𝑥)− 1 2 Misal u = 𝑥4 + 2𝑥 , maka 𝑑𝑢 𝑑𝑥 = 4𝑥3 + 2 𝑦 = 𝑢− 1 2 , maka 𝑑𝑦 𝑑𝑢 = − 1 2 𝑢− 3 2 = − 1 2 (𝑥4 + 2𝑥)− 3 2 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = − 1 2 (𝑥4 + 2𝑥)− 3 2 .(4𝑥3 + 2) 𝑑𝑦 𝑑𝑥 = − 1 2 (4𝑥3 + 2) (𝑥4 + 2𝑥) 3 2 = −2𝑥3 − 1 √(𝑥4 + 2𝑥)3
- 3. 5. 𝑦 = 1 √𝑥2−6𝑥 3= 1 (𝑥2−6𝑥) 1 3 = (𝑥2 − 6𝑥)− 1 3 Misal u = 𝑥2 − 6𝑥 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 − 6 𝑦 = 𝑢− 1 3 , maka 𝑑𝑦 𝑑𝑢 = − 1 3 𝑢− 4 3 = − 1 3 (𝑥2 − 6𝑥)− 4 3 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = − 1 3 (𝑥2 − 6𝑥)− 4 3 .(2𝑥 − 6) 𝑑𝑦 𝑑𝑥 = − 1 3 . (2𝑥 − 6) (𝑥2 − 6𝑥)− 4 3 = − 1 3 (2𝑥 − 6) √(𝑥2 − 6𝑥)43 6. 𝑦 = 1 √𝑥2−5𝑥+2 5 = 1 (𝑥2−5𝑥+2) 1 5 = (𝑥2 − 5𝑥 + 2)− 1 5 Misal u = 𝑥2 − 5𝑥 + 2 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 − 5 𝑦 = 𝑢− 1 5 , maka 𝑑𝑦 𝑑𝑢 = − 1 5 𝑢− 6 5 = − 1 5 (𝑥2 − 5𝑥 + 2)− 6 5 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = − 1 5 (𝑥2 − 5𝑥 + 2)− 6 5 .(2𝑥 − 5) 𝑑𝑦 𝑑𝑥 = − 1 5 . (2𝑥 − 5) (𝑥2 − 5𝑥 + 2) 6 5 = − 1 5 (2𝑥 − 5) √(𝑥2 − 5𝑥 + 2)65 7. 𝑦 = sin √ 𝑥2 + 6𝑥 Misal u = 𝑥2 + 6𝑥 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 + 6 𝑣 = √ 𝑢 = 𝑢 1 2 , maka 𝑑𝑣 𝑑𝑢 = 1 2 𝑢− 1 2 = 1 2 (𝑥2 + 6𝑥 )− 1 2 𝑦 = sin 𝑣 , maka 𝑑𝑦 𝑑𝑣 = cos 𝑣 = cos √ 𝑢 = cos √ 𝑥2 + 6𝑥
- 4. Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑣 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = cos √𝑥2 + 6𝑥 . 1 2 (𝑥2 + 6𝑥 )− 1 2 .(2𝑥 + 6) 𝑑𝑦 𝑑𝑥 = 1 2 . (2𝑥 + 6) .cos √ 𝑥2 + 6𝑥 (𝑥2 + 6𝑥 ) 1 2 = ( 𝑥 + 3) .cos √ 𝑥2 + 6𝑥 √ 𝑥2 + 6𝑥 8. 𝑦 = cos √ 𝑥3 + 2 3 Misal u = 𝑥3 + 2 , maka 𝑑𝑢 𝑑𝑥 = 3𝑥2 𝑣 = √ 𝑢3 = 𝑢 1 3 , maka 𝑑𝑣 𝑑𝑢 = 1 3 𝑢− 2 3 = 1 3 (𝑥3 + 2)− 2 3 𝑦 = cos 𝑣 , maka 𝑑𝑦 𝑑𝑣 = −sin 𝑣 = −sin √ 𝑢3 = −sin √ 𝑥3 + 2 3 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑣 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = −sin √ 𝑥3 + 2 3 . 1 3 (𝑥3 + 2)− 2 3 .3𝑥2 𝑑𝑦 𝑑𝑥 = 1 3 . 3𝑥2 . −sin √ 𝑥3 + 2 3 (𝑥3 + 2) 2 3 = 𝑥2 . −sin √ 𝑥3 + 2 3 √(𝑥3 + 2)23 9. 𝑦 = sin 1 √𝑥2+2 = sin 1 (𝑥2+2) 1 2 = sin(𝑥2 + 2)− 1 2 Misal u = 𝑥2 + 2 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 𝑣 = 𝑢− 1 2 , maka 𝑑𝑣 𝑑𝑢 = − 1 2 𝑢− 3 2 = − 1 2 (𝑥2 + 2)− 3 2 𝑦 = sin 𝑣 , maka 𝑑𝑦 𝑑𝑣 = cos 𝑣 = cos 𝑢− 1 2 = cos(𝑥2 + 2)− 1 2 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑣 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = cos(𝑥2 + 2)− 1 2 . − 1 2 (𝑥2 + 2)− 3 2 .2𝑥
- 5. 𝑑𝑦 𝑑𝑥 = − 1 2 . 2𝑥 .cos(𝑥2 + 2)− 1 2 (𝑥2 + 2) 3 2 = −𝑥 . cos(𝑥2 + 2)− 1 2 √(𝑥2 + 2)3 10. 𝑦 = cos 1 √𝑥2+6 3 = cos 1 (𝑥2+6) 1 3 = cos(𝑥2 + 6)− 1 3 Misal u = 𝑥2 + 6 , maka 𝑑𝑢 𝑑𝑥 = 2𝑥 𝑣 = 𝑢− 1 3 , maka 𝑑𝑣 𝑑𝑢 = − 1 3 𝑢− 4 3 = − 1 3 (𝑥2 + 6)− 4 3 𝑦 = cos 𝑣 , maka 𝑑𝑦 𝑑𝑣 = −sin 𝑣 = −sin 𝑢− 1 3 = −sin(𝑥2 + 6)− 1 3 Maka 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑣 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑢 𝑑𝑥 = −sin(𝑥2 + 6)− 1 3 . − 1 3 (𝑥2 + 6)− 4 3 .2𝑥 𝑑𝑦 𝑑𝑥 = − 1 3 . 2𝑥 . −sin(𝑥2 + 6)− 1 3 (𝑥2 + 6) 4 3 = 1 3 . 2𝑥 . sin(𝑥2 + 6)− 1 3 √(𝑥2 + 6)43