Soal dan penyelesaian limit fungsiFull description
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CONTOH DAN JAWABAN SOAL CERITA FUNGSI KUADRAT.pdfDeskripsi lengkap
CONTOH DAN JAWABAN SOAL CERITA FUNGSI KUADRAT.pdf
Turunan Fungsi Eksponen x = e sin ( 2 x + 3)
1. y
dy dx
=
e sin ( 2 x + 3) x
d
+
dx
sin ( 2 x + 3) .e
x
= e { sin ( 2 x + 3) + 2 cos( 2 x + 3)} x
2. y
dy
= e 5 x 2 + 2 x +1
dy
( 5 x 2 + 2 x + 1) = e 5 x + 2 x +1 dx dx 2 = (10 x + + 2) e 5 x + 2 x +1 2
Turunan Fungsi Trigonometri = sin (2x 2+3) + cos 3x 3. y dy/dx = cos (2x2+3) d/dx (2x 2+3) + (–sin 3x) d/dx (3x) = 4x cos (2x2+3) –3sin 3x
2
4. Y = sec 5x misalkan u = sec 5x → Y = u2 dy/du = 2 u ; v = 5x du/dx = sec v t v → dv/dx = 5 ∴ dy/dx = dy/du ! du/dv ! dv/dx = 10 sec25x t 5x Turunan Fungsi Logaritma = lo 5 sin 2x
5. y
dy dx = lo 5e ! 2 sin x! cos x sin 2x = lo 5e ! 2 cos x sin x = 2 ct x lo 5e
dy
2 x
= "n 3x3 + "n
6. y
1
dx = 3 x 3
1
9 x 2 +
2 x
.
1
x
= 4/x x
7. y
x = 2lo 5
1
dy dx
+2
x = 5
2
+2
In 2
d
(5 ) dx x 2 + 2
#
=
1
(5 ) ! "n 5! 2x x 2 + 2
2
x + 2
In 2 5 = 2x "n 5 "n 2
Invers Fungsi 8. $iketa%ui f ( x ) = x − 1 x + 2 a! &e'iksa aaka% memunyai inve's *! ika ada, tentukan inve'snya a-a* . a!
3 1.( x + 2) − 1.( x − 1) = > 0, ∀ x ∈ Df f ' ( x ) = 2 ( x + 2) 2 ( x 2 ) + a'ena selalu naik(monoton mu'ni) maka memunyai inve's
*! isal, y
x − 1
=
x + 2
x 2 y 1 =− x y x− = −2 y − 1 ⇒ x =
+
xy
−1
f ( y ) =
− 2 y − 1 y − 1
⇒ f −1 ( x) =
− 2 x − 1 x − 1
Turunan Fungsi Invers
= x 5 + 2 x + 1 , tentukan . ( f −1 )' (4)
f ( x ) 9. $iketa%ui, a-a* .
,y=4 ika %anya ika x=1 f ' ( x) = 5 x + 2 4
( f −1 )' (4) =
Fungsi Logaritma Asi
1 f ' (1)
=
1 7
1!. $i*e'ikan f ( x ) = ln(sin(4 x + 2)) aka ika
f ' ( x ) =
1 sin( 4 x + 2)
D x (sin(4 x + 2))= 4 cos( 4 x + 2)
y = ln | x | , x ≠ 0
y = ln x → y ' =
ln x , x > 0 = ln( − x ) , x < 0 adi
d
(ln | x |) =
1
dx x $a'i ini $ie'ole% .
11.
4
∫ x
x 3
0
1 x
y = ln(− x ) → y ' =
, x ≠ 0. 1
∫ x dx = ln | x | + C
2
+2
dx
u = x 3 + 2 → du = 3 x 2 dx
x 2
x 2 du
1
1
1
ln | u | + c ∫ x + 2 ∫ u 3x 3 ∫ u 3 4 11 x ] ln | 2 | dx x = + ∫ x + 2 = 33 ln | x + 2 | +0c = 13 (ln 66 − ln 2) = 13 ln 33. 3
4
dx =
2
2
3
3 3
=
du =
−1 1 = − x x
−
2 y − 1 y − 1
Fungsi Eksponen "mum 2 x +1 + 2sin 2 x 12. f ( x ) = 3
f ' ( x ) = 2.32 x +1 ln 3 + 2.2 sin 2 x cos 2 x ln 2
13.
∫
4
u
x
= x
∫ 4
2
. xdx
2
→
du
=
2 xdx
→
dx
=
1 2 x
du
1 4u
du
4x
2
= + C = + C ∫ 2 2 ln 4 2 ln 4 Fungsi Logaritma "mum x 2
. xdx =4
u
ln( x + 1) 2
14. f ( x)= log( x + 1) = 3
f ' ( x ) =
2
2 x
ln 3
1
x 2 + 1 ln 3
15.
ln( xx +−11 ) x + 1 f ( x ) = log( )= x − 1 ln 4 1 1 x + 1 f ' ( x) = Dx ( ) ln 4 ( x x+−11 ) x − 1 4