This document discusses differentiation rules for simple functions including: - Constant functions have a derivative of 0 - Polynomial functions have derivatives that are the polynomial with the exponent decreased by 1 and multiplied by the exponent - The product rule, quotient rule, and chain rule for differentiation - Examples of applying these rules to differentiate a variety of functions

1. Diferensiasi
fungsi sederhana

2. Kaidah- kaidah Deferensiasi
• Jika Y=K, dimana K adalah konstanta,
maka Y’ =
• Misal Y = 4
0=
dx
dy
0=
dx
dy

3. • Jika Y = xn
• Dimana n adalah konstanta maka
Y’ = . Xn-1
n
dx
dy
=

4. Cara 1
Misal
Y = (x² - 5x )²
= x4
+ 10 x3
+ 25 x2
= 4x3
+ 30 x2
+ 50 x
dx
dy

5. Cara 2
Y = (x² + 5x ) (x² + 5x )
Misal
U = x² + 5x U’ = 2x + 5
V = x² + 5x V’ = 2x + 5
= UV’ + VU’
= (x² + 5x ) (2x + 5 ) + (x² + 5x ) (2x + 5 )
= 2x3
+ 5 x2
+ 10 x2
+ 25x + 2x3
+ 5 x2
+ 10 x2
+ 25x
= 4x3
+ 30 x2
+ 50x
dx
dy

6. Cara 3
Y = (x² + 5x )²
u = x² + 5x u’ = 2x + 5
n = 2
= 2(x² + 5x ) (2x + 5)
= (2x2
+ 10 x) (2x + 5)
= 4x3
+ 10 x2
+ 20 x2
+ 50x
= 4x3
+ 30x2
+ 50x
dx
dy

7. Diferensiasi penjumlahan ( pengurangan ) fungsi
Jika Y = u ± v
Maka = ± = u’ + v’
Misal Y= 4x² + x3
u = 4x² u’ = 8x
v = x3
v’ = 3x²
= 8x + 3x²
dx
du
dx
dy
dx
dv
dx
dy

8. Diferensiasi perkalian fungsi
Jika Y = u . v
Maka = u . v’ + v . u’
Misal Y= 4x² . x3
u = 4x² u’ = 8x
v = x3
v’ = 3x²
= 4x² . 3x² + x3 .
8x
= 12 x4
+ 8x4
= 20 x4
dx
dy
dx
dy

9. Diferensiasi pembagian fungsi
Jika Y =
Maka =
Misal Y=
u = x² + 1 u’ = 2x
v = x + 2 v’ = 1
v
u
dx
dy
2
'.'.
v
vuuv −
2
12
+
+
x
x

10. =
=
=
dx
dy
2
2
)2(
1.1)2)(2(
+
+−+
x
xxx
44
142
2
22
++
−−+
xx
xxx
44
14
2
2
++
−+
xx
xx

11. Hitunglah dari fungsi- fungsi sbb :
1. Y =
2. Y = 3 x4
+ (2x – 1)²
3. Y =
4. Y = (x² - 4) ( 2x – 6 )
5. Y = 2x3
– 4x² + 7x - 5
2
1
x
x
1
dx
dy

12. 6. Y =
7. Y = (x²+2) 3
8. Y = 5x²
9. Y = 2x² + 4x + 1
10. Y = -5 + 3x - - 7x3
5
222
−
+−
x
xx
x²
2
3