MATEMATIKA 1 BAB 7 Teknik Diferensial [PDF]

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Teknik Diferensiasi Definisi Derivative Aturan Dasar Diferensiasi Asal Aturan Diferensiasi Aturan Fungsi Khusus ©Ruminta, 2006



Definisi Derivative Proses menemukan/mendapatkan derivative dari sebuah fungsi disebut diferensiasi. Deivative dari fungsi f(x) : 1



f ( x + h )− f ( x ) f ' ( x ) = lim h→ 0 h atau



2



f ′( x) = lim ∆x → 0



f ( x + ∆x) − f ( x) ∆x ©Ruminta, 2006



Notasi Derivative Jika fungsi y = f(x) derivative-nya dinyatakan : d f ( x0 )



1



f ′ ( x0 ) ,



2



dy d , f (x), dx dx



3



dx



, D ( f )( x0 ) , Df ( x0 )



y' ,



f ' ( x).



dy df = f ′ ( x ) = y′ = dx dx d f ( x ) = Df ( x ) = Dx f ( x ) dx ©Ruminta, 2006



Contoh



1



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



f ( x + h) − f ( x ) f ' ( x) = Lim h →0 h



2 xh + h 2 − h = Lim h →0 h



( x + h) 2 − ( x + h) − ( x 2 − x ) = Lim h →0 h



= Lim(2 x + h − 1)



x 2 + 2 xh + h2 − x − h − x 2 + x = Lim h→0 h



= 2x −1



©Ruminta, 2006



h →0



Contoh



2



Tentukan derivative dari fungsi



f ( x ) = 3x + 2 3( x + h) + 2 − 3x + 2 = Lim h →0 h



f ( x + h) − f ( x ) f ' ( x) = Lim h →0 h



3( x + h) + 2 − 3 x + 2 3( x + h) + 2 + 3 x + 2 = Lim * h →0 h 3( x + h) + 2 + 3 x + 2



3 x + 3h + 2 − 3 x − 2 = Lim h →0 h( 3( x + h) + 2 + 3 x + 2 )



3 = Lim h→0 ( 3( x + h) + 2 + 3x + 2)



3h = Lim h →0 h( 3( x + h) + 2 + 3 x + 2 )



3 = 2( 3 x + 2 )



©Ruminta, 2006



Grafik Derivative dari Fungsi



Jika f(x) naik (slope positif) Tangen horizontal (slope =0)



Jika f(x) turun (Slope negatif)



©Ruminta, 2006



Derivative Fungsi Tidak Ada



Sudut



Tangen vertikal



©Ruminta, 2006



Diskontinue



Aturan Dasar Diferensiasi 1



d (c ) = 0 dx



Konstanta



2



d d ( cu ) = c (u ) dx dx



Perkalian konstanta



3



D(x) = 1



Derivative f(x)=x adalah 1.



4



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



d d d atau ( u + v ) = (u ) + (v) dx dx dx



Penjumlahan dan perbedaan 5



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



atau



Produk



d d d ( uv ) = u (v) + v (u ) dx dx dx



©Ruminta, 2006



( )



6



d n x = nx n −1 dx



7



D f ( g ( x ) ) = D ( f ) ( g ( x ) ) D ( g)( x )



(



)



⇔



8



Pangkat



d f (g( x )) dx



Rantai



= f ′ ( g ( x ) ) g′ ( x )



⎛ f ⎞ gD ( f ) − fD ( g) D⎜ ⎟ = g2 ⎝ g⎠



atau



d d v (u ) − u (v) d ⎛u⎞ dx dx = ⎜ ⎟ dx ⎝ v ⎠ v2 Hasi bagi



9



( )



D f



-1



1 = D(f )



Funfsi Invers ©Ruminta, 2006



Contoh 1



2



Konstanta



f ( x) = 5 f ′( x) = 0 f ( x ) = 3 x8



( )



f ′( x) = 3 8 x 7 = 24 x 7 3



4t 2 d ⎡ 4x2 ⎤ 4 d 2 ⎡⎣ x ⎤⎦ f ( x) = = ⇒ f ′( x) = ⎢ ⎥ 5 dx ⎣ 5 ⎦ 5 dx



4 8x = ( 2x) = 5 5 ©Ruminta, 2006



Contoh



Penjumlahan dan pengurangan



d d d ⎡⎣ f ( x ) ± g ( x ) ⎤⎦ = [ f ( x) ] ± [ g ( x) ] dx dx dx 1



x2 1 f ( x) = − + 17 5 3 x 1 2 1 f ( x) = x − x 5 3



−



1 2



+ 17



1 1 1 ′ f ( x) = 2 * x + (− )(− ) x 5 3 2 2 1 ′ f ( x) = x + 3 5 6x 2



−



3 2



©Ruminta, 2006



+0



2



f ( x) = x3 − 4 x + 5 d 3 d d ⇒ f ′( x) = x − 4x + 5 dx dx dx 2 ′ ⇒ f ( x) = 3x − 4 + 0



⇒ f ′( x) = 3 x 2 − 4 3



f ′( x) = 0 + 12 x = 12 x 11



f ( x) = 7 + x



12



©Ruminta, 2006



11



Contoh 1



n −1



Pangkat



f ( x) = x



7



f ′( x) = 7 x 2



h′( x) = n [ f ( x) ]



6



(



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



(



1 2 ′ f ( x) = 3 x + 4 x 2 3x + 2 = 2 3x + 4 x



)



−1 2



2



)



12



(6x + 4)



©Ruminta, 2006



⋅ f ′( x)



Contoh 1



Produk



d d d ⎡⎣ f ( x ) ⋅ g ( x ) ⎤⎦ = [ f ( x)] g ( x) + [ g ( x)] f ( x) dx dx dx



y = (3 x 2 − 2 x −1 )(4 x3 + 5) dy d d = (3 x 2 − 2 x −1 ) (4 x3 + 5) + (4 x3 + 5) (3 x 2 − 2 x −1 ) dx dx dx dy = (3 x 2 − 2 x −1 )(12 x 2 ) + (4 x 3 + 5)(6 x + 2 x −2 ) dx



dy = 36 x 4 − 24 x + 24 x 4 + 8 x + 30 x + 10 x −2 dx dy = 60 x 4 + 14 x + 10 x −2 dx ©Ruminta, 2006



2



f ( x) = (3x − 2 x 2 )(5 + 4 x) d 2 d ′ f ( x) = (3 x − 2 x ) (5 + 4 x) + (5 + 4 x) (3 x − 2 x 2 ) dx dx = (3x − 2 x 2 )4 + (5 + 4 x)(3 − 4 x) = −24 x 2 + 4 x + 15



3



Let h( x) = (3x − 2 x 2 )(5 + 4 x)... find h′( x) f ( x) = 3 x − 2 x 2 ⇒ f ′( x) = 3 − 4 x g ( x) = 5 + 4 x ⇒ g ′( x) = 4 ( fg )′ = f g′ + fg ′ ∴ h′( x) = f ′g + fg ′



⇒ h′( x) = (3 − 4 x)(5 + 4 x) + (3 x − 2 x 2 )(4) ⇒ h′( x) = −24 x 2 + 4 x + 15 ©Ruminta, 2006



Contoh 1



Hasil bagi



d ⎡ f ( x) ⎤ ⎢ ⎥= dx ⎣ g ( x) ⎦



5x − 2 y= 2 x +1



g ( x)



d d f ( x ) f ( x ) g ( x) ] − [ ] [ dx dx [ g ( x)]2



d d 2 ( x + 1) (5 x − 2) − (5 x − 2) ( x + 1) dy dx dx = dx ( x 2 + 1) 2 2



dy ( x 2 + 1)(5) − (5 x − 2)(2 x) = dx ( x 2 + 1) 2 dy (5 x 2 + 5) − (10 x 2 − 4 x) = dx ( x 2 + 1) 2 dy −5 x 2 + 4 x + 5 = dx ( x 2 + 1) 2 ©Ruminta, 2006



2



f ( x) =



5x − 2



x2 + 1 d d 2 2 ( x + 1) (5 x − 2) − (5 x − 2) ( x + 1) dx dx f ′( x) = ( x 2 + 1)2 =



( x 2 + 1)5 − (5 x − 2)2 x ( x + 1) 2



3



2



=



(5 x 2 + 5) − (10 x 2 − 4 x) ( x + 1) 2



5x − 2 ... find h′( x) 2 x +1 f ( x) = 5 x − 2 ⇒ f ′( x) = 5



2



=



−5 x 2 + 4 x + 5 ( x 2 + 1) 2



Let h( x) =



g ( x) = x 2 + 1



⇒ g ′( x) = 2 x



⎡ f ⎤′ (5)( x 2 + 1) − (5 x − 2)(2 x) ⎢g⎥ = 2 2 + ( x 1) ⎣ ⎦ ©Ruminta, 2006



⎡ f ⎤′ f ′g − fg ′ ⎢g⎥ = 2 ⎣ ⎦ [g] 2 − 5 x + 4x + 5 h′ ( x) = ( x 2 + 1) 2



4



f ( x) =



5x − 2 x2 + 1



d d 2 ( x + 1) (5 x − 2) − (5 x − 2) ( x + 1) dx dx f ′( x) = ( x 2 + 1)2 2



=



( x 2 + 1)5 − (5 x − 2)2 x ( x + 1) 2



2



=



(5 x 2 + 5) − (10 x 2 − 4 x) ( x + 1) 2



©Ruminta, 2006



2



=



−5 x 2 + 4 x + 5 ( x 2 + 1) 2



Contoh



d [ f ( g ( x))] = f ′( g ( x)) g ′( x) dx



Rantai 7



1



2



⎛ 2x −1 ⎞ G ( x) = ⎜ ⎟ 3 5 x + ⎝ ⎠ 6⎛ 2 x − 1 ⎞ ( 3 x + 5 ) 2 − ( 2 x − 1) 3 ⎞ ⎛ ⎟ G′( x) = 7 ⎜ ⎟ ⎜⎜ 2 ⎟ ⎝ 3x + 5 ⎠ ⎝ ( 3x + 5) ⎠ 6 6 91( 2 x − 1) 13 ⎛ 2x −1 ⎞ = G′( x) = 7 ⎜ ⎟ 2 8 ⎝ 3x + 5 ⎠ ( 3x + 5) ( 3x + 5) y = u 5 2 , u = 7 x8 + 3 x 2



(



)



5 = u 3 2 ⋅ 56 x 7 + 6 x 2 32 5 8 2 = 7 x + 3x ⋅ 56 x 7 + 6 x = 140 x 7 + 15 x 7 x8 + 3 x 2 2



(



) (



) (



©Ruminta, 2006



)(



)



32



3



1) f ( x) = (3x − 5 x 2 )7 du u = 3x − 5 x 2 → = 3 − 10 x dx dy 7 y=u → = 7u 6 du



4



2) f ( x) = 3 ( x − 1) 2



y



dy = 7(3x − 2 x 2 )6 (3 − 10 x) dx



dy 2 = u dx 3



2



du 2 u = x −1 → = 2x dx 2 = u3



dy = 7u 6 (3 − 10 x) dx



1



dy 2 − 3 → = u du 3



−



1 3



(2 x) −



dy 2 2 = ( x − 1) dx 3 dy 4x = 1 dx 3( x 2 − 1) 3 ©Ruminta, 2006



1 3



(2 x)



Aturan Fungsi Khusus 1



2



dx r = rx r −1, r ∈ dx d sin ( x )



= cos ( x )



dx 3



d cos ( x ) dx



4



d tan ( x ) dx



5



= − sin ( x )



=



d arcsin ( x ) dx



1 cos2 ( x ) =



6



d arctan ( x ) dx



=



1 1+ x 2



7



d ex = ex dx



8



da x = a x ln ( a ) dx



9



1 1− x 2



©Ruminta, 2006



d ln ( x ) dx



1 = x



Contoh 1



4) f ( x) = sin(2 x) du u = 2x → =2 dx dy y = sin(u ) → = cos(u ) du



2



dy = 2cos(u ) = 2cos (2 x) dx



5) f ( x) = tan( x 2 − 1) du = 2x dx dy y = tan(u ) → = sec2 (u ) du



u = x2 − 1 →



dy = sec2 (u )(2 x) = 2 x sec2 ( x 2 − 1) dx



©Ruminta, 2006



Diferensiasi Fungsi Sinus d sin ( x ) dx



= cos ( x )



sin ( x + h ) − sin ( x ) h



=



sin ( h ) h



⇒ lim



h →0



sin ( x ) cos ( h ) + cos ( x ) sin ( h ) − sin ( x )



cos ( x ) + sin ( x )



sin ( x + h ) − sin ( x )



h →0



lim



=



sin ( h ) h



h



h



cos ( h ) − 1 h = cos ( x ) since Ingat bahwa :



= 1 (shown dan earlier) and lim



cos ( h ) − 1



h →0



©Ruminta, 2006



h



= 0 (exercise).



Diferensiasi Fungsi Cosinus d cos ( x ) dx



= − sin ( x )



One gets ⎛π ⎞ ⎛π ⎞ ⎛π ⎞ Dcos ( x ) = D sin ⎜ − x ⎟ = cos ⎜ − x ⎟ ( −1) = − cos ⎜ − x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ = − sin ( x ) .



©Ruminta, 2006



Diferensiasi Fungsi Tangen d tan ( x ) dx



=



1 cos2 ( x )



⎛ sin ( x ) ⎞ d ⎜⎜ ⎟⎟ cos x d tan ( x ) ( )⎠ ⎝ = dx dx = =



cos ( x ) cos ( x ) − sin ( x ) ( − sin ( x ) ) cos2 ( x )



cos2 ( x ) + sin2 ( x ) cos2 ( x )



=



1 cos2 ( x )



©Ruminta, 2006



Diferensiasi Inverse Fungsi Trigonometri 1



d arcsin ( x ) dx



=



1 1− x 2



Jika L e t x = sin ( y ) . B y th e In v e rse F u n ctio n R u le ,



d a rcsin ( x ) dx



=



1 1 = = co s y d ( sin ( y ) ) ( )



1 1 − sin



2



(y )



=



1 1− x



dy 2



d arctan ( x ) dx



=



1 1+ x 2



Let xx == tan tan(y) Jika ( y ) . By the Inverse Function Rule, d arctan ( x ) dx



=



1 = cos2 ( y ) d tan ( y ) dy ©Ruminta, 2006



=



1 1 = . 2 2 1 + tan ( y ) 1 + x



2



.



Diferensiasi Fungsi Exponensial 1



d ex = ex dx h h 0 e x +h − e x e − 1 e − e x = ex = ex ⎯⎯⎯ → e h →0 h h h



e h − e0 Ingat definisi dari bilangan e lim =1 since, by the definition of the naturan number e, h →0 h 2



da x = a x ln ( a ) dx



Jika a = e Write x



One gets: Maka x



da = dx



(



d e



x ln( a )



dx



( ) = e x ln(a ) and dan menggunakan aturan rantai use the Chain Rule.



ln a x



)=e



x ln( a )



ln ( a ) = a x ln ( a ) . ©Ruminta, 2006



3



f (x + h ) − f (x ) d [ f (x )] = hlim →0 h dx



( )



e x+h − e x d x e = lim h →0 h dx



exeh − ex = lim h →0 h



(



)



e x e h −1 = lim h →0 h



h e −1 x = e lim = ex(1) = ex h→ 0 h



©Ruminta, 2006



Aturan Diferensiasi Fungsi Exponensial Aturan 1:



( )



d x x e =e dx Aturan 2:



d f( x ) = e f( x ) ⋅ f′(x) e dx



(



)



©Ruminta, 2006



Contoh 1



Temukan derivative dari f(x) = x2ex .



f(x) = x 2 e x f ′(x) = x 2 e x + e x 2x f ′(x) = xe x (x + 2 )



2



3 2



Temukan derivative dari f(t) = (e + 2) t



(



f(t) = e + 2 t



3 2



)



1 3 t f′(t) = e + 2 2 e t 2



(



)



©Ruminta, 2006



3



Temukan derivative dari : f (x ) = x



e f ' (x ) = 2x f ' (x ) = f ' (x ) =



4



x 2 e x − e x (2x x



4



e x (2x ) − x 2 e x x



4



)



f ' (x ) =



f ' (x ) =



ex x2 x 2 e x − 2xe x4



e x (x − 2 ) x3



Temukan derivative dari f(t) = e 3x



f(x) = e 3x f′(x) = e 3x ⋅ 3 ©Ruminta, 2006



x



=



xe



x



(x − 2 ) x4



5



Temukan derivative dari : f(x) = e



f(x) = e



2x 2 +1 2x 2 +1



f ′(x) = e



f ′(x) = 4xe 6



(4x )



2x 2 + 1



Temukan derivative dari f (x ) = e f ' (x ) = e



5x



f ' (x ) = e



5x



f ' (x ) = e



5x



f ' (x ) =



2 x2 +1



5e



( )



d 5x dx -1 1 ⋅ (5x ) 2 (5) 2 5 ⋅ 2 5x ⋅



5x



2 5x ©Ruminta, 2006



5x



Diferensiasi Logaritma d ln ( x ) dx



1 = x



Jika Let x = e y . Use Makathe : Inverse Function Rule to get:



d ln ( x ) dx



1 1 1 = = y = . y e x d e



( )



dy



©Ruminta, 2006



Aturan Diferensiasi Fungsi Logaritma Aturan 1:



d 1 ln x = dx x



(x ≠ 0 )



Aturan 2: d [ln f(x)] = f′(x) dx f(x)



f(x) > 0



©Ruminta, 2006



Contoh 1



Temukan derivative dari f(x)= xlnx.



f(x) = xlnx 1 f ′(x) = x + lnx ⋅ 1 x



f ′(x) = 1 + lnx 2



Temukan derivative dari g(x)= lnx/x g(x)



g ′ (x)



=



lnx x



1 x − lnx x = x 2



⋅1



g ′ (x)



=



1 − lnx x



©Ruminta, 2006



2



3



Temukan derivative dari : y = x²lnx . ⎛ 1 ⎞ ⎟ + (lnx)(2x) ⎝ x ⎠



y’ = x² ⎜



y’ = x + 2xlnx Atau y’ = x(1+2lnx) 4



Temukan derivative dari y = ln (x + 4 ) − ln (x − 3)



1 1 y' = − x +4 x −3



x -3 x+4 − y' = (x + 4 )(x − 3 ) (x + 4 )(x − 3 ) −7 y' = (x + 4 )(x − 3 ) ©Ruminta, 2006



5



[



Temukan derivative dari : y = ln (x + 1)(x + 2)



[( y = ln (x + 1) + 6ln (x + 2 ) 6 (3x ) 2x y′ = + (x + 1) (x + 2 ) (



)(



)] (



)



2



6 y = ln ⎡ x 2 + 1 x 3 + 2 ⎤ = ln x 2 + 1 + ln x 3 + 2 ⎢⎣ ⎥⎦



2



3



2



2



3



Now get a common denominato r. y′ = y′ =



y′ =



y′ =



(



( )( (x + 1)(x + 2 ) (x + 2 )(x 2x x 3 + 2



2



)



+



3



(



2



(x



)



3



( )( (x + 1)(x + 2 ) (x + 2 )(x 2x x 3 + 2 3



2x 4 + 4x 2



)(



+1 x3 + 2



+



) + 1)



6 3x 2 x 2 + 1 2



) + 1)



6 3x 2 x 2 + 1 3



2



18x 4 + 18x 2



) + (x



3



)(



)



+ 2 x 2 +1



20x 4 + 18x 2 + 4x



(x



2



)(



+1 x3 + 2



)



©Ruminta, 2006



3



)



6



(



6



)



]



(



= ln x 2 + 1 + 6ln x 3 + 2



)



6



Temukan derivative dari : y = x(x + 1)(x2 + 1) Langkah 1 Buat ln pada dua sisi persamaan



ln y = ln x(x + 1)(x2 + 1) Langkah 2 Kembangkan persamaan tersebut



ln y = ln x(x + 1)(x2 + 1) ln y = ln x + ln(x + 1) + ln(x2 + 1) Langkah 3 Diferensiasi kedua sisi (eksplisitkan ln y )



ln y = ln x + ln(x + 1) + ln(x2 + 1) y′ 1 1 2x = + + 2 y x x +1 x +1 1 2x ⎞ ⎛1 Langkah 4: Pecahkan y ‘. y′ = y⎜ + + 2 ⎟ x x + 1 x + 1 ⎠ ⎝ ©Ruminta, 2006



Langkah 5: Substitusikan y pada persamaan tersebut.



y = x(x + 1)(x2 + 1) 1 2x ⎞ ⎛1 + 2 ⎟ y′ = x(x + 1)(x 2 + 1)⎜ + ⎝ x x +1 x +1 ⎠



[



]⎞⎟



⎛ x(x + 1)(x 2 + 1) x(x + 1)(x 2 + 1) 2x x(x + 1)(x 2 + 1) + + y′ = ⎜ ⎜ x x + 1 x 2 +1 ⎝



[



⎟ ⎠



⎛ x (x + 1)(x 2 + 1) x (x + 1)(x 2 + 1) 2x x (x + 1)(x 2 + 1) + y′ = ⎜ + 2 ⎜ x x 1 + x +1 ⎝



(



)



y ′ = (x + 1)(x 2 + 1) + x(x 2 + 1) + 2x [x (x + 1)]



(



y ′ = x 3 + x 2 + x + 1 + x 3 + x + 2x 3 + 2x 2 y ′ = 4x 3 + 3x 2 + 2x + 1 ©Ruminta, 2006



)



]⎞⎟ ⎟ ⎠



Diferensial Implisit y = 3 x3 − 4 x + 17 y diekspresikan secara explisit sebagai fungsi x.



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



[ f ( x) ]



3



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



Diferensial dari fungsi y yang dinyatakan secara implisit disebut diferensial implisit ©Ruminta, 2006



Manfaat Diferensial Implisit Menemukan slope dari garis tangen dan garis normal Contoh menemukan slope dari garis tangen dan normal di titik (2,4)



©Ruminta, 2006



Contoh 1



Temukan diferensial implisit dari



[ f ( x)]



3



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



3 [ f ( x) ] f ′( x) + f ( x) + xf ′( x) = 3 2



3 y 2 y′ + y + xy′ = 3



(



)



y′ 3 y 2 + x = 3 − y y′ =



3− y 3 y2 + x ©Ruminta, 2006



2



Temukan diferensial implisit dari 2y



dy ⎛ dy ⎞ = 2 x + cos( xy ) ⎜ x + y (1) ⎟ dx ⎝ dx ⎠



2y



dy dy = 2 x + cos( xy )( x ) + cos( xy ) y dx dx



2y



dy dy − cos( xy )( x ) = 2 x + cos( xy ) y dx dx



dy (2 y − x cos( xy )) = 2 x + y cos( xy ) dx



dy 2 x + y cos( xy ) = dx 2 y − x cos( xy )



©Ruminta, 2006



y 2 = x 2 + sin( xy )



3



Temukan diferensial implisit dari x3 + y 3 − 9 xy = 0 3x 2 + 3 y 2



dy dy − (9 x + y 9) = 0 dx dx



3x 2 + 3 y 2



dy dy − 9 x − 9 y) = 0 dx dx



3y2



dy dy − 9 x = 9 y − 3x 2 dx dx



dy (3 y 2 − 9 x) = 9 y − 3 x 2 dx dy 9 y − 3 x 2 = 2 dx 3 y − 9 x)



©Ruminta, 2006



4



Temukan diferensial implisit dari



y 3 + y 2 − 5 y − x 2 = −4



d d ⎡⎣ y 3 + y 2 − 5 y − x 2 ⎤⎦ = [ −4] dx dx d d d d d ⎡⎣ y 3 ⎤⎦ + ⎡⎣ y 2 ⎤⎦ − [5 y ] − ⎡⎣ x 2 ⎤⎦ = [ −4] dx dx dx dx dx dy dy dy 3y + 2 y − 5 − 2x = 0 dx dx dx 2



3y2



dy dy dy + 2 y − 5 = 2x dx dx dx



dy (3 y 2 + 2 y − 5) = 2 x dx



dy 2x = dx (3 y 2 + 2 y − 5) ©Ruminta, 2006



Diferensial Parsial Definisi Derivative Parsial dari Fungsi Dua Variabel Jika z = f(x,y), derivative parsian pertama dari f dinyatakan fx dan fy yaitu :



f x ( x, y ) =



uuur 0 ∆x lim



f y ( x, y ) =



uuur 0 ∆y lim



f ( x + ∆x, y ) − f ( x, y ) ∆x f ( x, y + ∆y ) − f ( x, y ) ∆y



©Ruminta, 2006



Definisi Derivative Parsial dari Fungsi Tiga Variabel Jika w=f(x,y,z), maka derivative parsial dinyatakansebagai berikut :



∂w = f x ( x, y , z ) = ∂x



uuur 0 ∆x lim



∂w = f y ( x, y , z ) = ∂y



uuur 0 ∆y lim



∂w = f z ( x, y , z ) = ∂z



uuur 0 ∆z lim



f ( x + ∆x, y, z ) − f ( x, y, z ) ∆x f ( x , y + ∆y , z ) − f ( x , y , z ) ∆y f ( x, y, z + ∆z ) − f ( x, y, z )



©Ruminta, 2006



∆z



Contoh 1



Temukan diferensial parsial fx dan fy dari



f ( x, y ) = 5 x 4 − x 2 y 2 + 2 x 3 y Solusi



f ( x, y ) = 5 x 4 − x 2 y 2 + 2 x 3 y f x ( x, y ) = 20 x3 − 2 y 2 x + 6 yx 2 f y ( x, y ) = −2 x 2 y + 2 x3



©Ruminta, 2006



2



Temukan diferensial parsial fx dan fy dari f ( x, y ) =



xy at (2, −titik 2) (2, -2) pada x− y



Solusi f ( x, y ) =



xy at (2, −2) x− y



f x ( x, y ) =



( x − y ) y − xy ( x − y)



f x ( 2, −2 ) = f y ( x, y ) =



− ( −2 )



2



2



(2 − ( −2 )) 2



=



( x − y ) x + xy



f y ( 2, −2 ) =



( x − y) x2 ( x − y)2



=



2



=



xy − y 2 − xy ( x − y)



2



=



− y2 ( x − y)2



−4 − 1 = 16 4



=



x 2 − xy + xy ( x − y)



4 1 = 16 4 ©Ruminta, 2006



2



=



x2 ( x − y)2



3



Temukan diferensial parsial fx dan fy dari f ( x, y) = 3xy 2 − 2 y + 5 x2 y



Solusi



f ( x, y) = 3 xy 2 − 2 y + 5 x2 y f x ( x, y) = 3 y 2 + 10 xy f xx ( x, y) = 10 y f ( x, y) = 3 xy 2 − 2 y + 5 x2 y f y ( x, y) = 6 xy − 2 + 5 x2 f yy ( x, y) = 6 x f ( x, y) = 3 xy 2 − 2 y + 5 x2 y f x ( x, y) = 3 y 2 + 10 xy f xy ( x, y) = 6 y + 10 x f ( x, y) = 3 xy 2 − 2 y + 5 x2 y f y ( x, y) = 6 xy − 2 + 5 x2 f yx ( x, y) = 6 y + 10 x



©Ruminta, 2006



Derivative Tingkat Tinggi Derivative fungsi f(x) adalah f ´(x). Jika f ‘(x) mempunyai derivative, disebut derivative tingakt dua atau f ´´(x) Notasi



d2y f ′′( x) = 2 dx



Derivative tingkat 2 mempunyai derivative tingkat tiga dan derivative tingkat tiga mempunyai derivative tingkat empat dst Notasi



4 d3y d y (4) f ′′′( x) = 3 f ( x) = 4 dx dx ©Ruminta, 2006



n d y (n) f ( x) = n dx



Contoh 1



Temukan derivative tingkat dua dari f ′( x) =



( x − 1)(1) − x(1) −1 = ( x − 1) 2 ( x − 1) 2



f ′′( x) =



d ⎛ −1 ⎞ d −2 = − ( x 1) ( ) ⎜ ⎟ dx ⎝ ( x − 1) 2 ⎠ dx



x f ( x) = x −1



−2 f ′′( x) = −2( x − 1) (1) = ( x − 1)3 −3



2



f ( x) = 3 x5 − 2 x3 + 14



Temukan f ‘’’(x) dari :



f ′( x) = 15 x − 6 x 4



2



3 ′′ f ( x) = 60 x − 12 x



2 ′′′ f ( x) = 180 x − 12



©Ruminta, 2006



3



2x +1 Temukan derivative tingkat dua dari f ( x) = 3x − 2 2 ( 3x − 2 ) − 3 ( 2 x + 1) −7 −2 f ′( x) = = = −7 ( 3 x − 2 ) 2 2 ( 3x − 2 ) ( 3x − 2 ) f ′′( x) = 14 ( 3 x − 2 )



4



−3



( 3) =



42



( 3x − 2 )



3



Temukan f ‘’’(x) dari : f ( x ) = x 2 + 4 x + 4 f' ( x ) = 2x + 4 f '' ( x ) = 2 f ''' ( x ) = 0 ©Ruminta, 2006