Posted by andi telaumbanua on Feb 13, 2018 in
Matematika
Tentukan turunan pertama dari bentuk implisit y+sin(xy)= 1 !
Jawab :
Turunan bentuk implisit
Pertama kita turunkan dulu sin(xy)
Misalkan : u = xy
maka: du/dx=(1)(y)+ (x)(dy/dx)=y+x dy/dx
Maka: y = sin u
maka: dy/du=cos u=cos(xy)
Sehingga :
dy/dx=(dy/du)(du/dx)
dy/dx=(cos(xy) )(y+x dy/dx)
dy/dx= y cos(xy) + x cos(xy) dy/dx
Maka:
d[y+sin(xy)]/dx = d(1)/dx
dy/dx + d(sin(xy) )/dx=0
dy/dx + y cos(xy)+ x cos(xy) dy/dx=0
dy/dx [1 + x cos(xy)] = – (y cos (xy) )
dy/dx = [-y cos (xy)] / [1 + x cos(xy)]
Posted by andi telaumbanua on Feb 13, 2018 in
Matematika
Tentukan turunan pertama dari bentuk implisit x^3-3x^2 y+ y^2=0
Jawab :
Turunan bentuk implisit
(d(x^3 – 3x^2 y + y^2))/dx = (d(0))/dx
d(x^3 )/dx – d(3x^2 y)/dx + d(y^2 )/dx = 0
3x^2 – [(d(3x^2 )/dx)(y) + (3x^2 )(dy/dx) ]+(d(y^2 )/dy)(dy/dx) = 0
3x^2 – [6xy+(3x^2 )(dy/dx) ]+ 2y (dy/dx)=0
3x^2- 6xy – 3x^2 dy/dx + 2y dy/dx = 0
2y dy/dx – 3x^2 dy/dx = 6xy – 3x^2
dy/dx (2y – 3x^2 )= 6xy – 3x^2
dy/dx = (6xy – 3x^2)/(2y – 3x^2 )
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Posted by andi telaumbanua on Feb 13, 2018 in
Matematika
Tentukan nilai c sehingga f^” (c)=0 bila f(x) = x^3+3x^2-45x-6 !
Jawab:
f^’ (x)= 3x^2+ 6x-45
f^” (x)= 6x+ 6
Maka :
f^” (x)= 0
6x+ 6=0
x= (-6)/6
x = -1
maka:
nilai c adalah – 1 agar f^” (c)= 0
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Posted by andi telaumbanua on Feb 13, 2018 in
Matematika
Tentukan turunan pertama dari y = sin x tan (x^2+1) !
Jawab:
Gunakan aturan rantai
Pertama kita menurunkan tan (x^2+1)
Misalkan : u = x^2+1
maka: du/dx=2x
Sehingga : y = tan u
maka : dy/du= sec^2 u= sec^2 (x^2+1)
Maka:
dy/dx= ( dy/du ) (du/dx)=(sec^2 (x^2+1) )(2x)=2xsec^2 (x^2+1)
Kemudian:
y = ab
maka: y^’=a^’ b+ab^’
Maka:
y^’=(cosx) (tan (x^2+1)) + (sinx)(2xsec^2 (x^2+1))
y^’=cosx tan (x^2+1) + 2x sinx sec^2 (x^2+1)
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Posted by andi telaumbanua on Feb 13, 2018 in
Matematika
Tentukan turunan pertama dari y = cos^4 (4x^2- x) !
Jawab:
Gunakan aturan rantai
Misalkan: a = (4x^2- x)
maka: da/dx=8x-1
Misalkan: b = cos a
maka: db/da=-sina = -sin(4x^2- x)
Sehingga: y = b^4
maka: dy/db=4b^3 = 4(cos a)^3 = 4[cos(4x^2- x)]^3
Maka: dy/db=4 cos^3 (4x^2- x)
Sehingga:
dy/dx = (dy/db)(db/da)(da/dx)
dy/dx = (4 cos^3 (4x^2- x) )(-sin(4x^2- x) )(8x-1)
dy/dx = [8x-1] [-4cos^3 (4x^2- x) sin(4x^2- x)]
dy/dx = [-32x cos^3 (4x^2- x) sin(4x^2- x)] – [-4〖cos〗^3 (4x^2- x) sin(4x^2- x)]
dy/dx = 4cos^3 (4x^2- x) sin(4x^2- x) – 32x cos^3 (4x^2- x) sin(4x^2- x)