matematika 1 bab 7 teknik diferensial

8/18/2019 MATEMATIKA 1 BAB 7 Teknik Diferensial

1/53

Teknik Diferensiasi

Definisi Derivative

Aturan Dasar Diferensiasi Asal Aturan Diferensiasi Aturan Fungsi Khusus

©Ruminta, 2006


2/53

Definisi Derivative

0

( ) ( )( ) lim

x

f x x f x f x

x∆ →

+ ∆ −′ =

∆

1

2

Deivative dari fungsi f(x) :

h ) x( f ) h x( f lim ) x( ' f

0 h

=

→

atau

Proses menemukan/mendapatkan derivative dari sebuah

fungsi disebut diferensiasi.

©Ruminta, 2006


3/53

Notasi Derivative

1

2

( ) ) ( ) ( ) ( )00 0 0f

f , , D f , Df d x

x x xdx′

,dx

dy),( x f dx

d ,' y ).(' x f

( )

( ) ( ) ( ) x

dy df f x y

dx dx

d f x Df x D f xdx

′ ′= = =

= =

3

Jika fungsi y = f (x) derivative-nya dinyatakan :

©Ruminta, 2006


4/53

Contoh 1

h

x f h x f Lim x f

h

)()()('

0

−+=

→

h

x xh xh x Lim

h

)()()( 22

0

−−+−+=

→

h

x xh xh xh x Lim

h

+−−−++=

→

222

0

2

h

hh xh Lim

h

−+=

→

2

0

2

)12(0

−+=→

h x Limh

12 −= x

Tentukan derivative dari fungsi f(x) = x2 - x

©Ruminta, 2006


5/53

Contoh 2Tentukan derivative dari fungsi 2 x 3 ) x( f

h x f h x f Lim x f

h)()()('

0−+=

→ h xh x Lim

h232)(3

0+−++=

→

232)(3

232)(3

*

232)(3

0 ++++

+++++−++

= → xh x

xh x

h

xh x

Limh

)232)(3(

23233

0 ++++

−−++= → xh xh

xh x Limh

)232)(3(

3

0++++

=→ xh xh

h Lim

h

)232)(3(

3

0 ++++= → xh x Limh

)23(2

3

+=

x

©Ruminta, 2006


6/53

Grafik Derivative dari Fungsi

Jika f(x) naik

(slope positif)

Jika f(x) turun

(Slope negatif)

Tangen horizontal(slope =0)

©Ruminta, 2006


7/53

Derivative Fungsi Tidak Ada

Sudut Tangen vertikal Diskontinue

©Ruminta, 2006


8/53

Aturan Dasar Diferensiasi

( ) ( ) ( )D fg D f g fD g= +

3

4

Produk

( )D 1x =

1

Derivative f(x )=x adalah 1.

2

( ) ( ) ( )D f g D f D g+ = +

( ) 0d

cdx

= Konstanta

Perkalian konstanta( ) ( )d d cu c udx dx

=

5

Penjumlahan dan perbedaan

( ) ( ) ( )d d d

u v u vdx dx dx

+ = +atau

atau ( ) ( ) ( )d d d

uv u v v u

dx dx dx

= +

©Ruminta, 2006


9/53

( )( )( ) ( ) ( )( ) ( ) ( )

( )( )( )( ) ( )

D f g D f g D g

f g f g g

x x x

d xx x

dx

=

′ ′⇔ =

8

Rantai

6

7

Pangkat( ) 1n nd

x nx

dx

−=

( )( )

-1 1D f

D f

=9

( ) ( )2

gD f fD gf D

g g

−⎛ ⎞=⎜ ⎟

⎝ ⎠

Hasi bagi

Funfsi Invers

2

( ) ( )d d

v u u vd u dx dx

dx v v

−⎛ ⎞ =⎜ ⎟⎝ ⎠

atau

©Ruminta, 2006


10/53

Contoh Konstanta

1 ( ) 5

( ) 0

f x

f x

=

′ =

8( ) 3 f x x=

( )7 7

( ) 3 8 24 f x x x′ = =

2

24( )

5

t f x =

24( )

5

d x f x

dx

⎡ ⎤′⇒ =

⎢ ⎥⎣ ⎦

( )4

25 x=

8

5

x

=

24

5

d x

dx⎡ ⎤=

⎣ ⎦

3

©Ruminta, 2006


11/53

Contoh Penjumlahan dan pengurangan

2 1( ) 17

5 3

x f x

x= − +

1

2 21 1( ) 175 3

f x x x−= − +

32

2 1( )

5 6

f x x

x

′ = +

3

21 1 1

( ) 2* ( )( ) 05 3 2 f x x x

−

′ = + − − +

1

( ) ( ) [ ] [ ]( ) ( )d d d f x g x f x g xdx dx dx

⎡ ⎤± = ±⎣ ⎦

©Ruminta, 2006


12/53

3( ) 4 5 f x x x= − +

3( ) 4 5d d d

f x x xdx dx dx

′⇒ = − +

2( ) 3 4 0 f x x′⇒ = − +

2( ) 3 4 f x x′⇒ = −

2

12( ) 7 f x x= +

11 11( ) 0 12 12 f x x x′ = + =3

©Ruminta, 2006


13/53

Contoh

7

6

( )

( ) 7

f x x

f x x

=

′ =

Pangkat

1

[ ] 1

( ) ( ) ( )n

h x n f x f x−

′ ′= ⋅

2 ( )1 2

2 2( ) 3 4 3 4 f x x x x x= + = +

( ) ( )

1 221( ) 3 4 6 4

2 f x x x x

−′ = + +

2

3 2

3 4

x

x x

+=

+©Ruminta, 2006


14/53

Contoh Produk

1 2 1 3(3 2 )(4 5) y x x x−= − +

2 1 3 3 2 1(3 2 ) (4 5) (4 5) (3 2 )dy d d x x x x x xdx dx dx

− −= − + + + −

2 1 2 3 2(3 2 )(12 ) (4 5)(6 2 )dy x x x x x xdx

− −= − + + +

4 4 2

4 2

36 24 24 8 30 10

60 14 10

dy x x x x x xdx

dy x x x

dx

−

−

= − + + + +

= + +

( ) ( ) [ ] [ ]( ) ( ) ( ) ( )d d d

f x g x f x g x g x f xdx dx dx

⎡ ⎤⋅ = +⎣ ⎦

©Ruminta, 2006


15/53

2 2( ) (3 2 )(5 4 ) f x x x x= − +

2 2

( ) (3 2 ) (5 4 ) (5 4 ) (3 2 )

d d

f x x x x x x xdx dx′ = − + + + −2(3 2 )4 (5 4 )(3 4 ) x x x x= − + + −

224 4 15 x x− + +=

2( ) 3 2 f x x x= − ( ) 3 4 f x x′⇒ = −

( ) 5 4g x x= + ( ) 4g x′⇒ =( ) fg f g fg′ ′ ′= + ( )h x f g fg′ ′ ′∴ = +

2( ) (3 4 )(5 4 ) (3 2 )(4)h x x x x x′⇒ = − + + −2( ) 24 4 15h x x x′⇒ = − + +

3

2

( ) (3 2 )(5 4 )... ( ) Let h x x x x find h x′= − +

©Ruminta, 2006


16/53

Contoh Hasil bagi

12

5 2

1

x y

x

−=

+2 2

2 2

( 1) (5 2) (5 2) ( 1)

( 1)

d d x x x xdy dx dx

dx x

+ − − − +=

+

2

2 2

( 1)(5) (5 2)(2 )

( 1)

dy x x x

dx x

+ − −=+

2 2

2 2(5 5) (10 4 )( 1)dy x x xdx x

+ − −= +

2

2 2

5 4 5

( 1)

dy x x

dx x

− + +

= +

( ) [ ] [ ]

[ ]

2

( ) ( ) ( ) ( )

( ) ( )

d d g x f x f x g x f xd dx dx

dx g x g x

−⎡ ⎤=⎢ ⎥

⎣ ⎦

©Ruminta, 2006


17/53

22

5 2( )

1

x f x

x

−=

+

2 2

2 2

( 1) (5 2) (5 2) ( 1)( )

( 1)

d d x x x x

dx dx f x x

+ − − − +′ =

+

2 2 2

2 2 2 2

2

2 2

( 1)5 (5 2)2 (5 5) (10 4 )

( 1) ( 1)

5 4 5

( 1)

x x x x x x

x

x x

x x

+ − − + − −= = =

+ +

− + +

+

( ) 5 2 f x x= −

2( ) 1g x x= +

( ) 5 f x′⇒ =

( ) 2g x x′⇒ = [ ]2 f f g fg

g g

′ ′ ′⎡ ⎤ −=⎢ ⎥

⎣ ⎦

2

2 2

(5)( 1) (5 2)(2 )

( 1)

f x x x

g x

′⎡ ⎤ + − −=⎢ ⎥ +⎣ ⎦

2

2 2

5 4 5

( ) ( 1)

x x

h x x

− + +

′ = +

3 2

5 2 ( ) ... ( )

1

x Let h x find h x

x

− ′=+

©Ruminta, 2006


18/53

42

5 2( )

1

x f x

x

−=

+

2 2

2 2

( 1) (5 2) (5 2) ( 1)( )

( 1)

d d x x x x

dx dx f x

x

+ − − − +′ =

+

2 2 2

2 2 2 2

2

2 2

( 1)5 (5 2)2 (5 5) (10 4 )

( 1) ( 1)

5 4 5

( 1)

x x x x x x

x

x x

x x

+ − − + − −

= = =+ +

− + +

+

©Ruminta, 2006


19/53

Contoh Rantai

1

[ ]( ( )) ( ( )) ( )d

f g x f g x g xdx

′ ′=

72 1( )

3 5

xG x

x

−⎛ ⎞= ⎜ ⎟

+⎝ ⎠( ) ( )

( )

6

2

3 5 2 2 1 32 1( ) 7 3 5 3 5

x x xG x x x

⎛ ⎞+ − −−⎛ ⎞⎜ ⎟′ = ⎜ ⎟ ⎜ ⎟+⎝ ⎠ +⎝ ⎠

( )

( )

( )

66

2 8

91 2 12 1 13( ) 7

3 5 3 5 3 5

x xG x

x x x

−−⎛ ⎞′ = =⎜ ⎟+⎝ ⎠ + +

5 2 8 2, 7 3 y u u x x= = +

( )3 2 75 56 62

u x x= ⋅ +

( ) ( )

3 28 2 75

7 3 56 62 x x x x= + ⋅ +

( )( )

3 27 8 2

140 15 7 3 x x x x= + +

2

©Ruminta, 2006


20/53

3

4

2

7 6

3 5 3 10

7

u x x x

y

du

dy

du

x

u u

d →

→

= − = −

= =

6

3 10 )7 (

dy

ud x x = −

2 67(3 2 ) (3 10 )dy

x x xdx

= − −

2 71) ( ) (3 5 ) f x x x= −

2

2 1

3 3

1 2

2

3

u x x

d

d y u u

y

du

u

dx

−

= − =

= =

→

→

1

323

(2 )d u yd

x x

−=

12

3

2( 1)3 (2 )

dy

dx x x

−

−=

1

2 3

4

3( 1)

dy x

dx x

=

−

2 232) ( ) ( 1) f x x= −

©Ruminta, 2006


21/53

Aturan Fungsi Khusus

( )( )

sincos

d xx

dx=2

3

1 1,r

r dx rx r dx

−= ∈

( ) ( )cos sind x xdx = −

4 ( )

( )2tan 1

cos

d x

dx x=

5 ( )

2

arcsin 1

1

d x

dx x=

−

6 ( )

2

arctan 1

1

d x

dx x=

+

7e

ex

xd

dx=

8 ( )lnx

xda a adx

=

9 ( )ln 1d xdx x

=

©Ruminta, 2006


22/53

Contoh

1

2 2

sin( ) cos( )

du

d

u x

y y

d

u udu

x

→

→

= =

= =2co 2 (2 )s( )dy u

d cos x

x = =

2

2

1 2

tan( ) sec ( )

u x x

y u dy

du

dudx

u

= − =

= =

→

→

2 2 22 secsec ( )(2 )) ( 1dy

u x x x

xd

= = −

2

4) ( ) sin(2 ) f x x=

25) ( ) tan( 1) f x x= −

©Ruminta, 2006


23/53

Diferensiasi Fungsi Sinus

( ) ( ) ( ) ( ) ( ) ( ) ( )sin sin sin cos cos sin sinx h x x h x h xh h

+ − + −=

( ) ( ) ( ) ( )sin cos 1cos sinh hx xh h

−= +

( )( )

sincos

d xx

dx=

( ) ( )( )

( ) ( )0

0 0

sin sinlim cos since

sin cos 1lim 1 (shown earlier) and lim 0 (exercise).

h

h h

x h xx

hh h

h h

→

→ →

+ −⇒ =

−= =

Ingat bahwa :

dan

©Ruminta, 2006


24/53

Diferensiasi Fungsi Cosinus

( )( )

cossin

d xx

dx

= −

( ) ( )

( )

One gets

Dcos Dsin cos 1 cos2 2 2

sin .

x x x x

x

π π π ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= −

©Ruminta, 2006


25/53

Diferensiasi Fungsi Tangen

( )

( )

2

tan 1

cos

d x

dx x

=

( )

( )( )

sin

costan

xd

xd x

dx dx

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

( ) ( ) ( ) ( )( )

( )( ) ( )

( ) ( )

2

2 2

2 2

cos cos sin sin

cos

cos sin 1

cos cos

x x x x

x

x x

x x

− −=

+= =

©Ruminta, 2006


26/53

Diferensiasi Inverse Fungsi Trigonometri

( )2

arcsin 1

1

d x

dx x=

−

( )( )

( )( ) ( ) ( )2 2

Let sin . B y the Inverse Function R ule,arcsin 1 1 1 1

.cossin 1 sin 1

x yd x

dx yd y y x

dy

=

= = = =− −

Jika

( )2

arctan 1

1

d x

dx x=

+

( )2 21 1

.1 tan 1y x

= =+ +

( )

( )( )

( )2

Let tan . By the Inverse Function Rule,

arctan 1cos

tan

x y

d xy

d ydx

dy

=

= =

Jika x = tan (y)

2

1

©Ruminta, 2006

D f F E l


27/53

Diferensiasi Fungsi Exponensiale

e

x

xd

dx =

0x x

0

0

0

e e e 1 e ee e e

e esince, by the definition of the number e, lim 1

x h x h hx

h

h

h

h h h

h

+

→

→

− − −= = ⎯⎯⎯→

−=Ingat definisi dari bilangan naturan e

( )lnx

xda a adx

=

( ) ( )ln x lnWrite e e and use the Chain Rule.

xa ax

a = =Jika dan menggunakan aturan rantai

( )( ) ( ) ( ) ( )

ln

ln

One gets:

e

e ln ln .

x ax

x a xdda

a a adx dx= = =

Maka

2

1

©Ruminta, 2006


28/53

3

( )h

eelime

dx

d xhx

0h

x −=+

→

h

eeelim

xhx

0h

−=

→

( )h

1eelim

hx

0h

−=

→

( )[ ] ( ) ( )

h

xf hxf lim

dx

d

0h

−+=

→

x f

x x h

0 h

x e )1( e h

1elime =

=

→

©Ruminta, 2006

At Dif i i F i E i l


29/53

Aturan 1:

( ) xx eedxd

=

Aturan 2:

( ) )()()( xfeedxd xfxf ′⋅=

Aturan Diferensiasi Fungsi Exponensial

©Ruminta, 2006


30/53

Contoh

1 Temukan derivative dari f(x) = x2ex .

2xeex(x)f

exf(x)

xx2

x2

+=′

=

( )2xxe(x)f x +=′

Temukan derivative dari f(t) = ( )23

t 2e +

( )

( ) t21

t

23t

e2e23

tf

2etf

+=′

+=

)(

)(

2

©Ruminta, 2006

( )xe


31/53

3 Temukan derivative dari :

Temukan derivative dari f(t) =4

( )2x

exf =

( ) 2xe

x'f

x

=

( ) ( )

4

xx2

x

2xeexx'f

−=

( ) ( )

4

x2x

x

ex2xex'f

−=

( ) ( )

( ) ( )

3

x

4

x

4

xx2

x

2xex'f

x

2xxe

x

2xeexx'f

−=

−=

−=

x3e

3exf

exfx3

x3

⋅=′

=

)(

)(

©Ruminta, 2006

T k d i ti d i 1x2 2

f +)(


32/53

5 Temukan derivative dari :

Temukan derivative dari6

1x2exf +=)(

( )4xe(x)f ef(x)

12x

12x

2

2

+

+

=′=

12x 2

4xe(x)f

+=′

( ) 5xexf =

( ) ( )( ) ( ) ( )

( )

( ) 5x2

5e

x'f

5x2

5ex'f

55x2

1ex'f

5x

d

ex'f

5x

5x

2

1-5x

5x

=

⋅=

⋅=

⋅= dx

©Ruminta, 2006

Diferensiasi Logaritma


33/53

Diferensiasi Logaritma

( )ln 1d xdx x

=

yLet e . Use the Inverse Function Rule to get:x =

( )

( )y

ln 1 1 1.

ee

y

d x

dx xddy

= = =

Jika Maka :

©Ruminta, 2006

Aturan Diferensiasi Fungsi Logaritma


34/53

Aturan 1:

Aturan 2:

Aturan Diferensiasi Fungsi Logaritma

x

1x

dx

d=ln ( )0x ≠

[ ])(

)()(ln

xfxfxf

dxd ′= 0xf >)(

©Ruminta, 2006

C t h


35/53

Contoh

1 Temukan derivative dari f(x)= xlnx.

Temukan derivative dari g(x)= lnx/x2

1lnxx1x(x)f

xlnxf(x)

⋅+=′

=

lnx1(x)f +=′

2x

1lnx

x

1x

(x)g

xlnxg(x)

⋅−

=′

=

2x

lnx1

(x)g

−

=′©Ruminta, 2006

T k d i ti d i ²l


36/53

3 Temukan derivative dari : y = x²lnx .

Temukan derivative dari4

y’ = x² + (lnx)(2x)

y’ = x + 2xlnx

Atau y’ = x(1+2lnx)

⎟ ⎠ ⎞⎜

⎝ ⎛

x1

( ) ( )3xln4xlny −−+=

3x

1

4x

1y'

−−

+=

( )( ) ( )( )

( )( )3x4x

7y'

3x4x

4x

3x4x

3-xy'

−+

−=

−+

+−

−+=

©Ruminta, 2006

5 Temukan derivative dari : ( )( )632 2x1x y ++= ln


37/53

( )( ) ( )[ ] ( ) ( ) ( )2x6ln1xln2xln1xln2x1xlny 32632632 +++=+++=⎥⎦

⎤⎢⎣

⎡ ++=

( )( ) ( )( )

( )( ) ( )( )

( )( )2x1x4x18x20xy

1x2x

18x18x

2x1x

4x2xy

1x2x

1x3x6

2x1x

2x2xy

32

24

23

24

32

4

23

22

32

3

++++=′

++

++

++

+=′

++

++

++

+=′

( )( )y

( )

( )

( )

( )

( )( )

( )( )

( )( )1x2x1x3x6

2x1x

2x2xy

r.denominatocommonaget Now2x

3x6

1x

2xy

2x6ln1xlny

23

22

32

3

3

2

2

32

++

++

++

+=′

+

+

+

=′

+++=

©Ruminta, 2006

6 Temukan derivative dari : ))(( 1x1xx y 2 ++=


38/53

Langkah 1 Buat ln pada dua sisi persamaan

))((lnln 1x1xx y 2 ++=

Langkah 2 Kembangkan persamaan tersebut

)ln()ln(lnln

))((lnln

1x1xx y

1x1xx y2

2

++++=

++=

Langkah 3 Diferensiasi kedua sisi (eksplisitkan ln y )

1xx2

1x1

x1

y

y

1x1xx y

2

2

++++=

′

++++= )ln()ln(lnln

Langkah 4: Pecahkan y ‘. ⎟

⎠

⎞⎜

⎝

⎛

+

+

+

+=′

1x

x2

1x

1

x

1 y y 2

©Ruminta, 2006

Langkah 5: Substitusikan y pada persamaan tersebut.


39/53

))(( 1x1xx y

2 ++=

Langkah 5 Substitusikan y pada persamaan tersebut.

( )

( ) ( ) ( )[ ]⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+++

+

+++

++=′

⎟ ⎠

⎞⎜⎝

⎛

++

++++=′

1x

1)(x1xx2x

1x

1)(x1xx

x

1)(x1xxy

1x

2x

1x

1

x

11)(x1xxy

2

222

2

2

( ) ( ) ( )[ ]

( ) ( )[ ]( )

( )

12x3x4xy

2x2xxx1xxxy

1xx2x1)x(x1)(x1xy

1x

1)(x1xx2x

1x

1)(x1xx

x

1)(x1xx

23

23323

22

2

222

+++=′

+++++++=′

++++++=′

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+++

+

+++

++=′

y

©Ruminta, 2006

Diferensial Implisit


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Diferensial Implisit

33 4 17 y x x= − +

y diekspresikan secara explisit sebagai fungsi x .3 3 1 y xy x+ = +

y diekspresikan secara implisit sebagai fungsi x .

Diferensial dari fungsi y yang dinyatakan secara implisit

disebut diferensial implisit

[ ] [ ]3

( ) ( ) 3 1 f x x f x x+ = +

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M f t Dif i l I li it


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Manfaat Diferensial ImplisitMenemukan slope dari garis tangen dan garis normal

Contoh menemukan slope dari garis tangen dan normal di titik

(2,4)

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Contoh


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Contoh

1 Temukan diferensial implisit dari

[ ] [ ]3

( ) ( ) 3 1 f x x f x x+ = +

[ ]2

3 ( ) ( ) ( ) ( ) 3 f x f x f x xf x′ ′+ + =

2

3 3 y y y xy′ ′+ + =

( )23 3 y y x y′ + = −

2

3

3

y y

y x

−′ =

+

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2 Temukan diferensial implisit dari 2 2 sin( )y x xy= +


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2 Temukan diferensial implisit dari sin( ) y x xy= +

2 2 cos( ) (1)dy dy y x xy x ydx dx

⎛ ⎞= + +⎜ ⎟⎝ ⎠

2 cos( )

2 cos( )

dy x y xy

dx y x xy

+=

−

2 2 cos( )( ) cos( )

dy dy

y x xy x xy ydx dx= + +

2 cos( )( ) 2 cos( )dy dy

y xy x x xy ydx dx

− = +

(2 cos( )) 2 cos( )dy

y x xy x y xydx

− = +

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3 Temukan diferensial implisit dari 3 3 9 0x y xy+ =


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3 Temukan diferensial implisit dari 9 0 x y xy+ − =

2 23 3 (9 9) 0dy dy x y x ydx dx

+ − + =

2 23 9 9 3dy dy

y x y xdx dx

− = −

2 23 3 9 9 ) 0dy dy

x y x y

dx dx

+ − − =

2

2

9 3

3 9 )

dy y x

dx y x

−=

−

2 2(3 9 ) 9 3dy

y x y xdx

− = −

©Ruminta, 2006

4 Temukan diferensial implisit dari 3 2 25 4y y y x+ − − = −


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4 Temukan diferensial implisit dari 5 4 y y y x+ =

[ ]3 2 25 4d d y y y xdx dx

⎡ ⎤+ − − = −⎣ ⎦

[ ] [ ]3 2 2

5 4

d d d d d

y y y xdx dx dx dx dx⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ − − = −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

23 2 5 2 0dy dy dy

y y x

dx dx dx

+ − − =

23 2 5 2dy dy dy

y y xdx dx dx

+ − =

2(3 2 5) 2dy

y y xdx

+ − =2

2

(3 2 5)

dy x

dx y y=

+ −

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Dif nsi l P si l


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Diferensial Parsial

Definisi Derivative Parsial dari Fungsi Dua VariabelJika z = f(x,y), derivative parsian pertama dari f

dinyatakan fx dan f y yaitu :

( )

( )

( ) ( )

0

0

, ( , )

, lim

, ( , ), lim

x x

y y

f x x y f x y

f x y x

f x y y f x y f x y

y

∆

∆

+ ∆ −=

∆+ ∆ −

=

∆

uuur

uuur

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Definisi Derivative Parsial dari Fungsi Tiga Variabel

Jika w=f(x,y,z), maka derivative parsialdinyatakansebagai berikut :

( ) ( )

( ) ( )

( ) ( )

0

0

0

, , ( , , ), , lim

, , ( , , ), , lim

, , ( , , ), , lim

x x

y y

z z

f x x y z f x y zw f x y z

x x

f x y y z f x y zw f x y z y y

f x y z z f x y zw f x y z z z

∆

∆

∆

+ ∆ −∂ = =∂ ∆

+ ∆ −∂ = =∂ ∆

+ ∆ −∂ = =∂ ∆

uuur

uuur

uuur

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Contoh


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Contoh

1 Temukan diferensial parsial fx dan f y dari4 2 2 3( , ) 5 2 f x y x x y x y= − +

4 2 2 3

3 2 2

2 3

( , ) 5 2

( , ) 20 2 6

( , ) 2 2

x

y

f x y x x y x y

f x y x y x yx

f x y x y x

= − +

= − += − +

Solusi

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2 Temukan diferensial parsial fx dan f y dari


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y

Solusi

( , ) (2, 2) xy

f x y at

x y

= −

−

( ) ( )

( ) ( )

( )

( ) ( )

( )

2 2

2 2 2

2

2

2 2

2 2 2

2

2

( , ) (2, 2)

,( ) ( ) ( )

2 4 12, 2

16 4(2 2 )

,( ) ( ) ( )

4 12, 2

16 4( )

x

x

y

y

xy f x y at

x y

x y y xy xy y xy y f x y

x y x y x y

f

x y x xy x xy xy x f x y x y x y x y

x f

x y

= −

−− − − − −

= = =− − −

− − − −− = = =

− −

− + − += = =

− − −

− = = =

−

pada titik (2, -2)

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3 Temukan diferensial parsial fx dan f y dari


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y

Solusi

y x y xy y x f 22 523),( +−=

x y y x f

x xy y x f

y x y xy y x f

x y y x f

xy y y x f

y x y xy y x f

x y x f

x xy y x f

y x y xy y x f

y y x f

xy y y x f

y x y xy y x f

yx

y

xy

x

yy

y

xx

x

106),(

526),(

523),(

106),(

103),(

523),(

6),(

526),(

523),(

10),(

103),(

523),(

2

22

2

22

2

22

2

22

+=

+−=

+−=

+=

+=

+−=

=

+−=

+−=

=

+=

+−=

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Derivative Tingkat Tinggi


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Derivative Tingkat TinggiDerivative fungsi f(x) adalah f ´(x). Jika f ‘(x)mempunyai derivative, disebut derivative tingakt duaatau f ´´(x)

2

2( )

d y f x

dx

′′ =

Derivative tingkat 2 mempunyai derivative tingkat tigadan derivative tingkat tiga mempunyai derivative tingkat

empat dst3

3( )

d y f x

dx

′′′ =4

(4)

4( )

d y f x

dx

= ( ) ( )n

n

n

d y f x

dx

=

Notasi

Notasi

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Contoh


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1

Temukan derivative tingkat dua dari

Temukan f ‘’’(x) dari :2

( )1

x f x

x=

−2 2

( 1)(1) (1) 1( )

( 1) ( 1)

x x f x

x x

− − −′ = =

− −

( )221( ) ( 1)( 1)d d f x xdx x dx

−⎛ ⎞−′′ = = −⎜ ⎟−⎝ ⎠

3

3

2( ) 2( 1) (1)

( 1) f x x

x

− −′′ = − − =

−5 3( ) 3 2 14 f x x x= − +

4 2( ) 15 6 f x x x′ = −

3( ) 60 12 f x x x′′ = − 2( ) 180 12 f x x′′′ = −

©Ruminta, 2006

3 Temukan derivative tingkat dua dari2 1

( )3 2

x f x

x

+=


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Temukan f ‘’’(x) dari :4

3 2 x −

( ) ( )( ) ( )

( ) 22 2

2 3 2 3 2 1 7( ) 7 3 23 2 3 2

x x f x x x x

−− − + −′ = = = − −− −

( ) ( )( )

3

342( ) 14 3 2 3

3 2 f x x

x

−′′ = − =−

4 x 4 x ) x( f 2

4 x 2 ) x( ' f

2 ) x( '' f =

0 ) x( ''' f =

©Ruminta, 2006

matematika 1 bab 7 teknik diferensial

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