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MATEMATIKA II
Oleh:
Dr. Parulian Silalahi, M.Pd
http://polmansem3.esy.es/
Bentuk Umum:
f (x,y) = 0
Contoh:
1.x2
+ y3
= 0
2.x3
+ 5xy + y4
+3 = 0
3.2x4
– 3y +5= 2y2
4.dll
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
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
)
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)
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+
Contoh 1:
Tentukanlah dy/dx dari fungsi berikut:
1. y = arc sin (5 + x2
)
2. y = arc tg (5x/9)
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+−
24104
2
24
−−− xx
x
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
TERIMA KASIH
Selamat Belajar

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Pt 3&4 turunan fungsi implisit dan cyclometri

  • 1. MATEMATIKA II Oleh: Dr. Parulian Silalahi, M.Pd http://polmansem3.esy.es/
  • 2. Bentuk Umum: f (x,y) = 0 Contoh: 1.x2 + y3 = 0 2.x3 + 5xy + y4 +3 = 0 3.2x4 – 3y +5= 2y2 4.dll
  • 3. 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
  • 4. 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 )
  • 5. 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)
  • 6. 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+
  • 7. Contoh 1: Tentukanlah dy/dx dari fungsi berikut: 1. y = arc sin (5 + x2 ) 2. y = arc tg (5x/9)
  • 8. 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+− 24104 2 24 −−− xx x
  • 9. 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