Editorial - Matlan - Turunan Fungsi Aljabar

Problem 1

ƒ(x)=3*x2+5*x-6

(ƒ^′)(x)=6*x+5+0=6*x+5

(A)

Problem 2

ƒ(x)=-x4+2*x2+8

(ƒ^′)(x)=-4*x3+4*x

(B)

Problem 3

ƒ(x)=5*x3-3*x2-5*x+3

(ƒ^′)(x)=15*x2-6*x-5

(ƒ^′)(2)=15⋅4-6⋅2-5=43

(D)

Problem 4

ƒ(x)=5/(x2)=5*x(-2)

(ƒ^′)(x)=-10/(x3)

(C)

Problem 5

(ƒ^′)(x)=(2*x)⋅(x3-1)+(x2+1)⋅(3*x2)=2*x4-2*x+3*x4+3*x2

(ƒ^′)(x)=5*x4+3*x2-2*x

(E)

Problem 6

(ƒ^′)(x)=2*x2-x/2=1/2*x*(4*x-1)

(D)

Problem 7

Ya udah pasti 4karena (g^′)(x)=4 (D)

Problem 8

(ƒ^′)(x)=(3√(,x))/2-1/(2*x√(,x))=(3*x2-1)/(2*x√(,x))

(C)

Problem 9

g(x)=3*x(-2)-1/2*x+x(1/2)

(g^′)(x)=-6*x(-3)-1/2+1/2*x(-1/2)

(g^′)(x)=-6/(x3)+1/(2√(,x))-1/2

(E)

Problem 10

ƒ(x)=2*x+1/2*x(-1)

(ƒ^′)(x)=2-1/2*x(-2)

(ƒ^′)(1)=2-1/2=3/2

(B)

Problem 11

ƒ(x)=p*x3-8*x2+3

(ƒ^′)(x)=3*p*x2-16*x

(ƒ^′)(2)=12*p-32=16

12*p=32+16=48

∴p=4

ƒ(1)=4-8+3=-1

4*(ƒ^′)(1)=4⋅(12⋅1-16)=-16

2*p=2⋅4=8

∴ƒ(1)+4*(ƒ^′)(1)+2*p=-1+16+8=23

(D)

Problem 12

(ƒ^′)(x)=(2*x-1)⋅1+2⋅(x+1)=4*x+1

(E)

Problem 13

(y^′)=2/3*((3*x+1)*(1)+(3)*(x-2))=4*x-10/3

(B)

Problem 14

(y^′)=(3*x2-2)*(2)+(6*x)*(2*x+3)=18*x2+18*x-4

(D)

Problem 15

(y^′)=(x-1)*(2*x+1)+(1)*(x2+x+1)

(y^′)=2*x2+x-2*x-1+x2+x+1=3*x2

(B)

Problem 16

(ƒ^′)(x)=2*(2*x3-4)*(6*x2)=12*x2*(2*x3-4)

(D)

Problem 17

(ƒ^′)(x)=4*(3*x2-7)3*(6*x)=24*x*(3*x2-7)3

(C)

Problem 18

(y^′)=(x2+2*x+1)*(2)+(2*x+2)*(2*x-3)

(y^′)=6*x2+2*x-4=(3*x-2)*(2*x+2)

(A)

Problem 19

y=(x+1)*(x2-x+1)

(y^′)=(x+1)*(2*x-1)+(1)*(x2-x+1)=3*x2

ƒ(x)=y2

(ƒ^′)(x)=2*(3*x2)⋅(x+1)*(x2-x+1)=6*x2*=6*x5+6*x2

(E)

Problem 20

ƒ(x)=x⋅(4*x2+4*x+1)

(ƒ^′)(x)=(4*x2+4*x+1)*(1)+(x)*(8*x+4)=12*x2+8*x+1

ƒ(x)=(2*x+1)*(6*x+1)

(C)

Problem 21

y=(4*x-1)(1/3)

(ƒ^′)(x)=1/3⋅(4*x-1)(-2/3)⋅4=4/(3√(3,(4*x-1)2))

(B)

Problem 22

ƒ(x)=√(,(3*x-2)2)=3*x-2

(ƒ^′)(x)=3

(E)

Problem 23

ƒ(x)=(x+1)*(x-1)=x2-1

(ƒ^′)(x)=2*x

(B)

Problem 24

ƒ(x)=(x+2√(,x))(1/2)

y=x+2√(,x)

(y^′)=1/√(,x)+1

(ƒ^′)(x)=1/2*(x+2√(,x))(-1/2)*(1/√(,x)+1)

(ƒ^′)(4)=1/2*(4+4)(-1/2)*(1/2+1)=3/16√(,2)

(C)

Problem 25

(y^′)=(3*(2*x-3)-(3*x+2)*2)/((2*x-3)2)=(-13)/((2*x-3)2)

(A)

Problem 26

(ƒ^′)(x)=(4*(5-2*x)-(4*x-3)*(-2))/((5-2*x)2)=(20-8*x+8*x-6)/((5-2*x)2)=14/((5-2*x)2)

(C)

(A) also because it's mathematically the same

Problem 27

y=3*(x+√(,2))(-1)

(y^′)=-3*(x+√(,2))(-2)=(-3)/((x+√(,2))2)

(D)

Problem 28

Identitas:

(a3-b3)=(a-b)*(a2+a*b+b2)

(a3+b3)=(a-b)*(a2-a*b+b2)

ƒ(x)=((x-2)*(x2+2*x+4))/(x-2)=x2+2*x+4

(ƒ^′)(x)=2*x+2

(E)

Problem 29

Typo?

ƒ(x)=(3-2*x2)/(2*x+3)

(ƒ^′)(x)=((-4*x)*(2*x+3)-(2)*(3-2*x2))/((2*x+3)2)

(ƒ^′)(0)=-6/9=-2/3

(ƒ^′)(-1)=(4-2)/1=2

(ƒ^′)(0)⋅(ƒ^′)(1)=-4/3

(A)

Problem 30

(ƒ^′)(x)=((3*x2-4)*(x2-1)-(2*x)*(x3-4*x-5))/((x2-1)2)

(ƒ^′)(0)=4/1=4

(ƒ^′)(3)=((27-4)*(9-1)-(2⋅3)⋅(27-12-5))/(82)=31/16

ƒ(2)=(8-8-5)/(4-1)=-5/3

(ƒ^′)(0)+(ƒ^′)(3)-ƒ(2)=4+31/16+5/3=365/48

???

Problem 31

(y^′)=((1/(2√(,x)))*(√(,x)-5)-(1/(2√(,x)))*(√(,x)+5))/((√(,x)-5)2)

(y^′)=(1/2-5/(2√(,x))-1/2-5/(2√(,x)))/((√(,x)-5)2)

(y^′)=(-5)/(√(,x)*(√(,x)-5)2)

(B)

Problem 32

(ƒ^′)(x)=((2)*(3*x-4)-(3)*(2*x-5))/((3*x-4)2)

(ƒ^′)(1)=(2⋅(3-4)-3*(2-5))/((3-4)2)=7

(D)

Problem 33

ƒ(x)=u(1/2)

u=x-√(,x+1)

(u^′)=1-1/(2√(,x+1))

(ƒ^′)(x)=1/(2√(,u))*(u^′)=(1-1/(2√(,x+1)))/(2√(,x-√(,x+1)))

(ƒ^′)(3)=(1-1/(2√(,3+1)))/(2⋅√(,(3-√(,3+1))))=3/8

(C)

Problem 34

ƒ(x)=3*(x-1)2+5*(x-1)+7=3*x2-x+5

(ƒ^′)(x)=6*x-1

(ƒ^′)(x-1)=6*x-6-1=6*x-7

6*x-7=-x2

x2+6*x-7=0

(x+7)*(x-1)=0

∴x=-7,1

(E)

Problem 35

(ƒ^′)(x)=2*a*x-a-1

(ƒ^′)(a)=2*a2-a-1=14

2*a2-a-15=0

(2*a+5)*(a-3)=0

∴a=-5/2,3

a>0

∴a=3

(C)