Diferensiasifungsi sederhana

Kaidah- kaidah Deferensiasi

• Jika Y=K, dimana K adalah konstanta,

maka Y’ =

• Misal Y = 4

0=dx

dy

0=dx

dy

• Jika Y = xn

• Dimana n adalah konstanta maka

Y’ = . Xn-1 n

dx

dy =

Cara 1

Misal

Y = (x² - 5x )²

= x4 + 10 x3 + 25 x2

= 4x3 + 30 x2 + 50 x

dx

dy

Cara 2Y = (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

Cara 3Y = (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

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

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

Diferensiasi pembagian fungsi

Jika Y =

Maka =

Misal Y=

u = x² + 1 u’ = 2x

v = x + 2 v’ = 1

v

u

dx

dy2

'.'.

v

vuuv −

2

12

++

x

x

=

=

=

dx

dy2

2

)2(

1.1)2)(2(

++−+

x

xxx

44

1422

22

++−−+

xx

xxx

44

142

2

++−+xx

xx

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

6. Y =

7. Y = (x²+2) 3

8. Y = 5x²

9. Y = 2x² + 4x + 1

10. Y = -5 + 3x - - 7x3

5

222

−+−

x

xx

x²2

3