Contoh Soal Fungsi Transenden

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 aaka%  memunyai 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  memunyai 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 Asi

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 $ie'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

= =

1  x − 1  x − 1 − ( x + 1) ln 4  x + 1

( x − 1) 2

1 −2 ln 4 ( x + 1)( x − 1)