just for teaching

1. MATEMATIKA
Oleh:
Dr. Parulian Silalahi, M.Pd
http://polmansem3.esy.es/

2. Rumus Dasar:
y = ex
 y’ = ex
y = e-x
 y’ = - e-x
y = eax
 y’ = a. eax
y = e-ax
 y’ = -a e-ax
Contoh 5:
Tentukan dy/dx dari fungsi hiperbolik berikut:
1. y = ecos5x
2. y = (e4x
– e5x
)4

3. Jawab:
1. y = ecos5x
mis u = cos 5x  du/dx = - 5.sin 5x
y = eu
 dy/du = eu
= ecos5x
dy/dx = du/dx . dy/du = - 5.sin 5x. ecos5x
2. y = (e4x
– e5x
)4
mis u = (e4x
– e5x
)  du/dx = 4e4x
–5 e5x
y = u4
dy/du = 4u3
= 4(e4x
– e5x
)3
dy/dx = du/dx . dy/du = (4e4x
–5 e5x
). 4(e4x
– e5x
)3
= 4(4e4x
–5 e5x
). (e4x
– e5x
)3

4. Rumus Dasar:
1. y = a
log x y’ =1/a. a
log e
2. y = ln x  y’ = 1/x e
log e =1 /x
3. y = ax
 y’ = ax.
ln a
Contoh 2.6:
Tentukan dy/dx dari fungsi logaritma berikut:
1. y = ln (x2
+ 5)
2. y =
)6( 2
3 xx +

5. Jawab:
1. y = ln (x2
+ 5)
mis: u = x2
+ 5  du/dx = 2x
y = ln u  dy/du = 1/u = 1/(x2
+ 5)
dy/dx = du/dx . dy/du = 2x . 1/(x2
+ 5) = 2x/(x2
+ 5)
2. y =
mis: u = x2
+ 6x  du/dx = 2x + 6
y = 3u
dy/du = 3u .
ln 3 = . ln 3
dy/dx = du/dx . dy/du = (2x + 6) . ln 3
)6( 2
3 xx +
)6( 2
3 xx +
)6( 2
3 xx +

6. Rumus Dasar:
1. y = sinh x  y’ = cosh x
2. y = cosh x  y’ = sinh x
Contoh 7:
Tentukan dy/dx dari fungsi hiperbolik berikut:
1. y = sinh 7x
2. y = cosh3
(1-x)

7. Jawab:
1. y = sinh 7x
mis u = 7x  du/dx =7
y = sinh u  dy/du = cosh u = cosh 7x
dy/dx = du/dx . dy/du = 7. cosh 7x
2. y = cosh3
(1-x)
mis u = 1 – x  du/dx = -1
t = cosh u  dt/du = sinh u = sinh (1-x)
y = t3
dy/dt = 3 t2
= 3 cosh2
(1-x)
dy/dx = du/dx . dt/du. dy/dt
= -1. sinh (1-x). 3 cosh2
(1-x)
= -3 sinh (1-x). cosh2
(1-x)

8. Bentuk Umum:
f (x,y) = 0
Contoh:
1.x2
+ y3
= 0
2.x3
+ 5xy + y4
+3 = 0
3.2x4
– 3y +5= 2y2
4.dll

9. Tentukanlah dy/dx dari fungsi implisit berikut ini:
1. x3
+ y4
= 0
2. x5
+ xy + y3
+4 = 0
3. x4
– 3y +5xy= 4y2
Jawab:
1. x3
+ y4
= 0
d/dx (x3
+ y4
)= d/dx (0)
3x2
dx/dx + 4y3
dy/dx = 0
3x2
+ 4y3
dy/dx = 0
dy/dx = -3x2
/ 4y3

10. 2. x5
+ xy + y3
+4 = 0
d/dx(x5
+ xy + y3
+4 )= d/dx (0)
5x4
dx/dx + 1 dx/dx. y +x. dy/dx + 3y2
dy/dx + 0 = 0
5x4
+ y +x. dy/dx + 3y2
dy/dx + 0 = 0
x. dy/dx + 3y2
dy/dx = - 5x4
- y
(x + 3y2
) dy/dx = - (5x4
+ y)
dy/dx = -(5x4
+ y)/(x + 3y2
)

11. 3. x4
– 3y +5xy= 4y2
d/dx(x4
– 3y +5xy) = d/dx (4y2
)
4x3
dx/dx – 3 dy/dx + 5 dx/dx. y + 5x dy/dx = 8y dy/dx
4x3
– 3 dy/dx + 5y + 5x dy/dx = 8y dy/dx
-3 dy/dx + 5x dy/dx – 8y dy/dx = - 4x3
– 5y
( -3 + 5x – 8y) dy/dx = - (4x3
+ 5y)
dy/dx = -( 4x3
+ 5y)/(-3 + 5x – 8y)

12. Rumus Dasar
1.y = arc sin x y’ =
2.y = arc cos x y’ = -
3.y = arc tg x y’ =
4.y = arc cot x y’ = -
2
x1
1
−
2
x1
1
−
2
1
1
x+
2
1
1
x+

13. Contoh 1:
Tentukanlah dy/dx dari fungsi berikut:
1. y = arc sin (5 + x2
)
2. y = arc tg (5x/9)

14. Jawab:
1. y = arc sin (5 + x2
)
misalkan: u = 5 + x2
 du/dx = 2x
y = arc sin u  dy/du = =
dy/dx = du/dx . dy/du = 2x .
=
2
1
1
u−
22
)5(1
1
x+−
22
)5(1
1
x+−
2410
2
24
−−− xx
x

15. 2. y = arc tg (5x/9)
misalkan: u = 5x/9  du/dx = 5/9
y = arc tg u  dy/du =
dy/dx = du/dx . dy/du = 5/9 .
=
2
9
5
1
1






+
x
2
9
5
1
1






+
x
2
9
5
1
9/5






+
x

16. TERIMA KASIH
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