Sumarnijwabtb



RINGKASAN MATEMATIKA

BAB DIFERENSIAL/ TURUNAN

SIMBOL DIFERENSIAL/TURUNAN

y’ atau f’(x) atau [pic]

RUMUS-RUMUS DIFERENSIAL/TURUNAN FUNGSI ALJABAR

1. f(x) = k ( f’(x) = 0, k = konstanta

2. f(x) = x ( f’(x) = 1

3. f(x) = xn ( f’(x) = n.xn-1

4. f(x) = a.xn ( f’(x) = n.a.xn-1

5. f(x) = U.V ( f’(x) = U’.V + V’.U

6. f(x) = [pic] ( f’(x) = [pic]

RUMUS-RUMUS TURUNAN FUNGSI TRIGONOMETRI

1. f(x) = sin x ( f’(x) = cos x

2. f(x) = cos x ( f’(x) = - sin x

3. f(x) = tg x ( f’(x) = sec2x

4. f(x) = sin ax ( f’(x) = a. cos ax

5. f(x) = cos ax ( f’(x) = -a. sin ax

CATATAN :

1. sin2x + cos2x = 1

2. sin 2x = 2 sin x.cos x

3. cos 2x = cos2x – sin2x

4. cos 2x = 1 – 2sin2x

5. cos 2x = 2cos2x - 1

6. tg x = [pic]

7. sec x = [pic]

8. cosec x = [pic]

9. ctg x = [pic]

DALIL RANTAI/ TURUNAN BERANTAI

Jika f(x) = Un maka f’(x) = n.Un-1.U’

CONTOH SOAL- SOAL TURUNAN

Tentukan turunan dari fungsi dibawah ini :

1. f(x) = 3

2. f(x) = 2x

3. f(x) = 3x2

4. f(x) = [pic]

5. f(x) = 3[pic]

6. f(x) = (2x + 5)(x3- 3x + 5)

7. f(x) = [pic]

8. f(x) = 2 cos x

9. f(x) = 5 sin x

10. f(x) = - 3 sin x + 4 cos x

11. f(x) = 2x5 - [pic]sin x

12. f(x) = [pic]

13. f(x) = 5x sin x

14. f(x) = (10x3 – 2x)5

PENYELESAIAN

1. f(x) = 3 ( f’(x) = 0

2. f(x) = 2x ( f’(x) = 2

3. f(x) = 3x2 ( f’(x) = 6x

4. f(x) = [pic] ( f’(x) = 2x3 + 15x2 - 7

5. f(x) = 3[pic] ( f(x) = 3x[pic] ( f’(x) = x[pic] ( [pic]

6. f(x) = (2x + 5)(x3- 3x + 5) ( f(x) = 2x4 – 6x2 + 10x + 5x3 – 15x + 25

f(x) = 2x4 + 5x3 – 6x2 – 5x + 25

f’(x) = 8x3 + 15x2 - 12x - 5

7. f(x) = [pic] ( U = x + 3 , V = 5 – x2

U’ = 1 dan V’ = -2x

f’(x) = [pic] ( f’(x) = [pic]

f’(x) = [pic] ( f’(x) = [pic]

8. f(x) = 2 cos x ( f’(x) = -2 sin x

9. f(x) = 5 sin x ( f’(x) = 5 cos x

10. f(x) = - 3 sin x + 4 cos x ( f’(x) = -3 cos x – 4 sin x

11. f(x) = 2x5 - [pic]sin x ( f’(x) = 10x4 - [pic]cos x

12. f(x) = [pic] ( U = sin x , U’= cos x, V = cos x , V’ = -sin x

f’(x) = [pic] ( f’(x) = [pic]

= [pic]

= [pic] = sec2x

13. f(x) = 5x sin x

U = 5x 14. f(x) = (2x3 – 3)8 ( U = 2x3 - 3

V = sin x U’ = 6x2

U’ = 5 f’(x) = n.Un-1.U’

V’ = cos x = 8(2x3 – 3 )7. 6x2

f’(x) = U’.V + V’.U = 8(6x2)(2x3 – 3)7

= 5(sin x) + (cos x. 5x) = 48x2 (2x3 – 3)7

= 5(sin x + x cos x)

LATIHAN SOAL

Tentukan turunan dari fungsi berikut ini :

1. f(x) = 2x - 5

2. f(x) = [pic]

3. f(x) = 5[pic] - 7x

4. f(x) = (2x + 5)(4x -7)

5. f(x) = (2x2 – 3)(2x2 – 5x + 7)

6. f(x) = [pic]

7. f(x) = 3(2x – 4)2

8. f(x) = 5 cos x - [pic]x2

9. f(x) = 4 sin x – 3 cos x

10. f(x) = x2.sin x

11. f(x) = 7 sin 3x + cos 5x

12. f(x) = -cos 6x – sin 2x

13. f(x) = [pic]

14. f(x) = [pic]

15. f(x) = (2x3 – 3)8

RINKASAN FUNGSI NAIK DAN FUNGSI TURUN

1. FUNGSI NAIK DAN FUNGSI TURUN

Syarat : fungsi f(x) naik jika f’(x) > 0

fungsi f(x) turun jika f’(x) < 0

fungsi stasioner f’(x) = 0

CONTOH SOAL :

1. Diketahui fungsi f(x) = x2 – 8x – 9. Tentukan interval x ketika fungsi f(x) naik dan

fungsi f(x) turun.

Jawab:

f(x) = x2 – 8x – 9

f’(x) = 2x – 8

fungsi naik : f’(x) = 0

2x – 8 = 0

2x = 8 ( x = 4 jadi fugsi naik intervalnya : x > 4

Fungsi turun : f’(x) < 0

2x < 8 maka fugsi turun intervalnya : x < 4

2. Diketahui fungsi f(x) = x3 – 6x2 – 36x + 30 . Tentukan interval x ketika fungsi f(x) naik

dan fungsi f(x) turun.

Jawab :

f(x) = x3 – 6x2 – 36x + 30

f’(x) = 3x2 – 12x – 36

f”(x) = x2 – 4x – 12

fungsi naik : f’(x) > 0

x2 – 4x – 12 = 0

(x – 6)(x + 2) = 0

x = 6 atau x = -2

Fungsi naik : x < -2 atau x > 6

Fungsi turun : -2 < x < 6

Tentukan nilai stasioner , titik stasioner dan jenis titik stasioner dari fungsi berikut :

1. f(x) = 2x2 + 5x - 3

2. f(x) = [pic]x3 – 2x2 – 21x + 7

Penyelesain :

1. f(x) = 2x2 + 5x – 3

syarat stasioner f’(x) = 0

4x + 5 = 0

4x = -5 ( x = - [pic]

Nilai stasioner : f(-[pic]) = 2.( -[pic])2 + 5. -[pic]- 3

= - [pic]

Titik stasioner : (- [pic] , -[pic])

jenis stasioner : titik balik minimum

2. f(x) = [pic]x3 – 2x2 – 21x + 7

syarat stasioner f’(x) = 0

x2 – 4x – 21 = 0

(x - 7 )(x + 3 ) = 0

x = 7 atau x = -3

Nilai stasioner : f(7) = [pic].(7)3 – 2(7)2 – 21.(7) + 7

= 43

f(-3) = [pic](-3)3 – 2(-3)2 – 21.(-3) + 7

= - [pic]

Titik stasioner : ( 7, 43) atau ( -3, - [pic])

Jenis stasioner :

f’’(x) > 0 titik balik minimum

f’’(x) < 0 titik balik maksimun

2x – 4 = 2. -3 – 4

= - 10 ( f’’(x) < 0 maka x = -3 adalah titik balik maksimum ( -3, -[pic])

2x – 4 = 2.7 – 4

= 14 – 4

= 10 ( f’’(x) > 0 maka x = 7 adalah titik balik minimum (7, 43)

LIMIT FUNGI

A. Limit fungsi aljabar

Menghitung limit sebuah fungsi pada dasarnya dengan cara subtitusi langsung. Jika

perhitungan dengan subtistusi langsung didapat bentuk tak tentu, yaitu bentuk [pic]maka

perhitungan nilai limit harus dengan cara lain yaitu diselesaikan dengan cara

pemfaktoran atau turunan

CONTOH SOAL :

1. lim ( x2 – x – 3)

x(0

2. lim [pic]

x(0

3. lim [pic]

x (0

PENYELESAIAN

1. lim ( x2 – x – 3) = (02 – 0 – 3) = -3

x(0

2. lim [pic] = [pic] = [pic]maka bisa dilakukan dengan cara :

x(0

a. cara pemfaktoran : lim [pic] = lim [pic] = [pic] = -[pic]

x(0 x(0

b. Cara turunan

Lim [pic] = lim [pic] = [pic]= -[pic]

x(0 x (0

c. lim [pic] = lim [pic] = [pic] = - [pic] = -1

x (0 x (0

SOAL – SOAL

a. Lim (x3 -2x2 + x – 1)

x(1

b. Lim (2x3 + 3x2- 4x – 2)

x(-1

c. Lim [pic]

x(0

d. Lim [pic]

x(0

e. Lim [pic]

x(5

f. Lim [pic]

x(3

g. Lim [pic]

x(4

h. Lim [pic]

x(-3

i. Lim [pic]

x(-1

j. Lim [pic]

x(-2

LIMIT TAK HINGGA

Bentuk : [pic]

a. Jika pangkat tertinggi f(x) = g(x) maka [pic] = [pic]

b Jika pangkat tertinggi f(x) > g(x) maka [pic] = ~

c. Jika pangkat tertinggi f(x) < g(x) maka [pic] = 0

CONTOH SOAL :

1. Lim [pic] = [pic]

x(~

2. Lim [pic] = ~

x(~

3. lim [pic] = 0

x(~

SOAL : kerjakan soal bab limit tak hingga halaman 6 , pelatihan 2 no. 1 yaitu a - j

LIMIT FUNGSI TRIGONOMETRI

Limit fungsi trigonometri adalah limit pendekatan dari suatu sudut pada fungsi trigonometri. Untuk meghitung nilai limit fungsi trigonometri dapat dilakukan dengan cara subtitusi langsung , jika dalam perhitungan dengan cara subtitusi langsung didapat bentuk tak tentu [pic], [pic]atau ~ - ~ maka perhitungan limit harus dilakukan cara lain.

RUMUS-RUMUS YANG MENDUKUNG

1. sin2x + cos2x = 1

2. sin x = 2 sin [pic]x. cos [pic]x

3. cos x = 1 – 2.sin2[pic]x

4. lim [pic] = 1

x(0

5. lim [pic] = 1 atau lim [pic] = 1

x(0 x(0

6. lim [pic] = 1 atau lim [pic] = 1

x(0 x(0

7. lim [pic] = 1 atau lim [pic] = 1

x(0 x(0

8. lim [pic] = 1 atau lim [pic] = 1

x(0 x(0

9. lim [pic] = 1 atau lim [pic] = 1

x(0 x(0

CONTOH SOAL :

1. lim [pic] = [pic] = 3

x(0

2. lim [pic] = [pic]

x(0

-----------------------

YAYASAN PENDIDIKAN BHINA TUNAS BHAKTI

SMK BHINA TUNAS BHAKTI

SBI ( Sekolah Berbasis Industri )

E-mail : admin@smkbtb-jwa.sch.id

Web Site : smkbtb-jwa.sch.id

Jalan Sunan Ngerang 109, Telp./Fax. (0295) 471132

Juwana-Pati 59185, Provinsi Jawa Tengah, Indonesia

YAYASAN PENDIDIKAN BHINA TUNAS BHAKTI

SMK BHINA TUNAS BHAKTI

SBI ( Sekolah Berbasis Industri )

E-mail : admin@smkbtb-jwa.sch.id

Web Site : smkbtb-jwa.sch.id

Jalan Sunan Ngerang 109, Telp./Fax. (0295) 471132

Juwana-Pati 59185, Provinsi Jawa Tengah, Indonesia

YAYASAN PENDIDIKAN BHINA TUNAS BHAKTI

SMK BHINA TUNAS BHAKTI

SBI ( Sekolah Berbasis Industri )

E-mail : admin@smkbtb-jwa.sch.id

Web Site : smkbtb-jwa.sch.id

Jalan Sunan Ngerang 109, Telp./Fax. (0295) 471132

Juwana-Pati 59185, Provinsi Jawa Tengah, Indonesia

YAYASAN PENDIDIKAN BHINA TUNAS BHAKTI

SMK BHINA TUNAS BHAKTI

SBI ( Sekolah Berbasis Industri )

E-mail : admin@smkbtb-jwa.sch.id

Web Site : smkbtb-jwa.sch.id

Jalan Sunan Ngerang 109, Telp./Fax. (0295) 471132

Juwana-Pati 59185, Provinsi Jawa Tengah, Indonesia

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