BAB 6. TURUNAN - BAB 6 TURUNAN
BAB 6. TURUNAN Program Studi Teknik Mesin Fakultas Teknik
1 Turunan
Konsep Turunan Definisi turunan Aturan turunan Aplikasi turunan
1 Turunan
Konsep Turunan
Definisi turunan Aturan turunan Aplikasi turunan Untuk mendefinisikan pengertian garis singgung secara formal, perhatikanlah gambar
Gradien garis singgung tersebut dapat dinyatakan :
1 Turunan
Konsep Turunan
Definisi turunan
Aturan turunan Aplikasi turunan Definisi
1 f x ∈ D
Misalkan sebuah fungsi real dan f
2 f x
Turunan dari di titik , ditulis
f
x h f x( ) − ( ) + ′ f x
( ) = lim h →0 h Definisi
1
f x ∈ D Misalkan sebuah fungsi real dan
f
2 f x
Turunan dari di titik , ditulis
f
x h f x( ) − ( ) + ′ f x
( ) = lim h →0 h Definisi
1
f x ∈ D Misalkan sebuah fungsi real dan
f
2
f x Turunan dari di titik , ditulis f x h f x
( ) − ( ) +
′
f x ( ) = lim
h →0 h Definisi
1
f x ∈ D Misalkan sebuah fungsi real dan
f
2
f x Turunan dari di titik , ditulis f x h f x
( ) − ( ) +
′
f x ( ) = lim
h →0 h Definisi
1
f x ∈ D Misalkan sebuah fungsi real dan
f
2
f x Turunan dari di titik , ditulis f x h f x
( ) − ( ) +
′
f x ( ) = lim
h →0 h
1 Turunan
Konsep Turunan Definisi turunan
Aturan turunan
Aplikasi turunan
x
6 D x [( f . g
[ g ( x )] ( g ( x )
[ f ( x )] . g ( x )− f ( x ) . D x
)] = D x
)( x
7 D x [( f g
)]
( x
D
x [
g( x ).
) + f
( x
( x )]. g
)] = D x [ f
)( x
)]
2 D x [ x
( x
)] ± D x [ g
( x
)] = D x [ f
)( x
5 D x [( f ± g
)]
( x
)] = kD x [ f
( x
4 D x [ kf
] = nx n −1
3 D x [ x n
] = 1
2 )
x
6 D x [( f . g
[ g ( x )] ( g ( x )
[ f ( x )] . g ( x )− f ( x ) . D x
)] = D x
)( x
7 D x [( f g
)]
( x
D
x [
g( x ).
) + f
( x
( x )]. g
)] = D x [ f
)( x
)]
2 D x [ x
( x
)] ± D x [ g
( x
)] = D x [ f
)( x
5 D x [( f ± g
)]
( x
)] = kD x [ f
( x
4 D x [ kf
] = nx n −1
3 D x [ x n
] = 1
2 )
x
6 D x [( f . g
[ g ( x )] ( g ( x )
[ f ( x )] . g ( x )− f ( x ) . D x
)] = D x
)( x
7 D x [( f g
)]
( x
D
x [
g( x ).
) + f
( x
( x )]. g
)] = D x [ f
)( x
)]
2 D x [
( x
)] ± D x [ g
( x
)] = D x [ f
)( x
5 D x [( f ± g
)]
( x
)] = kD x [ f
( x
4 D x [ kf
] = nx n −1
3 D x [ x n
x ] = 1
2 )
x
D
x [
g)( x
)] = D x [ f
( x )]. g
( x
) + f
( x ).
( x
)]
)]
7 D x [( f g
)( x
)] = D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [( f . g
( x
2 D x [
4 D x [ kf
x ] = 1
3 D x [
x
n
] = nx
n −1
( x
)] ± D x [ g
)] = kD x [ f
( x
)]
5 D x [( f ± g
)( x
)] = D x [ f
( x
2 )
x
( x ).
6 D x [( f . g
)( x
)] = D x [ f
( x )]. g
( x
) + f
D
x [
g( x
( x
)]
7 D x [( f g
)( x
)] = D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
)]
)] ± D x [ g
2 D x [
4 D x [
x ] = 1
3 D x [
x
n
] = nx
n −1
kf ( x
( x
)] = kD
x [
f ( x
)]
5 D x [( f ± g
)( x
)] = D x [ f
2 )
x
( x ).
)]
6 D x [( f . g
)( x
)] = D x [ f
( x )]. g
( x
) + f
D
x [
gx [
( x
)]
7 D x [( f g
)( x
)] = D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
g ( x
)] ± D
2 D x [
kf ( x
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
)] = kD
f ( x
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
2 )
x
( x ).
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
D
)]
x [
g ( x
)]
7 D x [( f g
)( x
)] = D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
x
D
f . g
)( x
)] = D
x [
f ( x )]. g
( x
) + f
( x ).
x [
)]
g ( x
)]
7 D x [( f g
)( x
)] =
D x
[ f ( x )] . g ( x )− f ( x ) . D x
[ g ( x )] ( g ( x )
6 D x [(
g ( x
2 D x [
)] = kD
x ] = 1
3 D x [
x
n
] = nx
n −1
4 D x [
kf ( x
x [
x [
f ( x
)]
5 D x [(
f ± g )( x
)] = D
x [
f ( x
)] ± D
2 )
- sinx
2 . sinx
, maka f
. sinx
) = 5 x +1 3 x −2
( x
3 Jika f
2 ) =?
′ ( Q
, maka f
) = x
Contoh
( x
2 Jika f
) =?
′ ( x
, maka f
2
) = 5 x
( x
1 Jika f
′ (1) =?
- sinx
) = x
, maka f
. sinx
) = 5 x +1 3 x −2
( x
3 Jika f
2 ) =?
′ ( Q
, maka f
2 . sinx
( x
Contoh
2 Jika f
) =?
( x
′
, maka f
2
) = 5 x
f ( x
1 Jika
′ (1) =?
- sinx
2 .
, maka f
. sinx
) = 5 x +1 3 x −2
( x
3 Jika f
) =?
2
( Q
′
sinx , maka f
) = x
Contoh
f ( x
2 Jika
) =?
( x
′
, maka f
2
) = 5 x
f ( x
1 Jika
′ (1) =?
- sinx
sinx , maka f
′
, maka f
. sinx
5 x +1 3 x −2
) =
f ( x
3 Jika
) =?
2
( Q
′
2 .
Contoh
) = x
f ( x
2 Jika
) =?
( x
′
, maka f
2
) = 5 x
f ( x
1 Jika
(1) =? Aturan Rantai
Misalkan y = f ( u ) dan u = g ( x ). Jika g terdefinisikan di x dan f terdefinisikan di ◦ ◦ u g x f g f g x f g x
= ( ), maka fungsi komposit , yang didefinisikan oleh ( )( ) = ( ( )),
′ ′ ′
◦ adalah terdiferensiasikan di x dan ( f g ) ( x ) = f ( g ( x )) g ( x ) yakni
′ ′ D ( f ( g ( x ))) = f ( g ( x )) g ( x ) x
- 5)
2 Jika f
′ ( x
), maka f
3 x
2 −
2 ( x
) = sin
( x
) =?
Contoh
′ (
x
3 , maka f
3 x
2 −
) = ( x
( x
1 Jika f
) =?
- 5)
) =?
′ ( x
), maka f
3 x
2 −
2 ( x
) = sin
( x
2 Jika f
( x
Contoh
′
, maka f
3
3 x
−
2
) = ( x
f ( x
1 Jika
) =?
- 5)
2 Jika
( x
′
), maka f
3 x
−
2
( x
2
) = sin
f ( x
) =?
Contoh
( x
′
, maka f
3
3 x
−
2
) = ( x
f ( x
1 Jika
) =? Turunan tingkat tinggi ′
f x f x f Misalkan ( ) sebuah fungsi dan ( ) turunan pertamanya. Turuna kedua dari
2
adalah f ”( x ) = D ( f ). Dengan cara yang sama turunan ketiga , keempat dst. Salah
x
S t satu penggunaan turunan tingkat tinggi adalah pada masalah gerak partikel. Bila ( )
′
v ( t ) = S ( t ) dan menyatakan posisi sebuah partikel, maka kecepatannya adalah
′
a t v t S t percepatannya ( ) = ( ) = ”( )
1 Turunan
Konsep Turunan Definisi turunan Aturan turunan
Aplikasi turunan y=f’(x)
′
1 Gradien g singgung : m = y
2 fungsi naik : y
′ >
3 fungsi turun : y
′ <
4 fungsi stasioner : y
′ = 0
5 kecepatan : v
′ = ds dt
= S
′ ′ dv ′ y=f’(x)
′
1 Gradien g singgung : m = y
2 fungsi naik : y
′ >
3 fungsi turun : y
′ <
4 fungsi stasioner : y
′ = 0
5 kecepatan : v
′ = ds dt
= S
′ ′ dv ′ y=f’(x)
′
1 Gradien g singgung : m = y
2
fungsi naik : y
′
>
3 fungsi turun : y
′ <
4 fungsi stasioner : y
′ = 0
5 kecepatan : v
′ = ds dt
= S
′ ′ dv ′ y=f’(x)
′
1 Gradien g singgung : m = y
2
fungsi naik : y
′
>
3
fungsi turun : y
′ y=f’(x)
<
4 fungsi stasioner : y
′ = 0
5 kecepatan : v
′ = ds dt
= S
′ ′ dv ′
′
1 Gradien g singgung : m = y
fungsi naik : y
′
>
3
fungsi turun : y
′
< y=f’(x)
2
fungsi stasioner : y
′
= 0
5 kecepatan : v
′ = ds dt
= S
′ ′ dv ′
4
′
1 Gradien g singgung : m = y
fungsi naik : y
= 0
′ ′ dv ′
= S
ds dt
=
′
kecepatan : v
5
′
′
2
4
<
′
fungsi turun : y
3
>
fungsi stasioner : y y=f’(x)
′
1 Gradien g singgung : m = y
fungsi naik : y
= 0
′ ′ dv ′
= S
ds dt
=
′
kecepatan : v
5
′
′
2
4
<
′
fungsi turun : y
3
>
fungsi stasioner : y y=f”(x)
Uji jenis
1 maximum : y ” > 0
2 minimum : y
” < 0
3 titik belok : y
” = 0 y=f”(x)
Uji jenis
1
maximum : y ” > 0
2 minimum : y
” < 0
3 titik belok : y
” = 0 y=f”(x)
Uji jenis
1
maximum : y ” > 0
2
minimum : y
” < 0
3 titik belok : y
” = 0 y=f”(x)
Uji jenis
1
maximum : y ” > 0
2
minimum : y
” < 0
3
titik belok : y
” = 0