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Tentukan turunan pertama dari sin⁡(xy )+x^2=y^2+ 1

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan turunan pertama dari sin⁡(xy )+x^2=y^2+ 1

Jawab:

d/dx(sin(xy)+d/dx (x^2 )= d/dx (y^2 )+d/dx(1)

Pertama Turunkan: d/dx(sin⁡(xy)

d/dx(sin⁡(xy)=[ (1)(y)+ (x)(dy/dx)][cos⁡xy]

=(y+x dy/dx)(cos⁡(xy)

=y cos⁡(xy)+x dy/dx cos⁡(xy)

Maka:

d/dx(sin(xy)+ d/dx (x^2 )= d/dx (y^2 )+d/dx(1)

y cos⁡(xy)+x dy/dx cos⁡xy + 2x = 2y dy/dx + 0

x dy/dx cos⁡(xy)-2y dy/dx = -y cos⁡(xy)-2x

dy/dx (x cos⁡(xy)-2y)= -y cos⁡(xy)-2x

dy/dx = (-y cos⁡(xy)-2x)/(x cos⁡(xy)-2y)

 
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Tentukan turunan pertama dari tan (xy) – 2y = 0

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan turunan pertama dari tan (xy) – 2y = 0

Jawab:

d/dx(tan(xy)-d/dx(2y)=d/dx(0)

Pertama Turunkan: d/dx(tan⁡(xy)

d/dx(tan⁡(xy)=[(1)(y)+ (x)(dy/dx)][sec^2⁡(xy)]

=(y+x dy/dx)(sec^2⁡ (xy)

=y sec^2⁡(xy) + x dy/dx sec^2⁡(xy)

Maka:
d/dx(tan(xy)-d/dx(2y)= d/dx()

y sec^2⁡(xy) + x dy/dx sec^2(⁡xy) -2 dy/dx = 0

x dy/dx sec^2(⁡xy)-2 dy/dx = -y sec^2⁡xy

dy/dx(x sec^2⁡(xy)-2) = -y sec^2⁡(xy)

dy/dx = -y sec^2⁡(xy))/(x sec^2⁡(xy)-2)

 
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Turunan pertama dari f(x) = (1 + 1/x^2 )(1+1/x^3 )

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Turunan pertama dari f(x) = (1 + 1/x^2 )(1+1/x^3 )

Jawab:

f(x) = U*V

f^’ (x)= U^’ V* UV^’

Maka:

f(x) = (1 + 1/x^2 )(1+1/x^3 )

f^’ (x)=((0-2x)/x^4 )((0-3x)/x^6 )

f^’ (x)=(2x/x^4 )(3x/x^6 )

f^’ (x)=((6x^2)/x^10 )

f^’ (x)=(6/x^8 )

 
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anti turunan F(x) + C dari fungsi a. f(x) = (x^2-x)/(2x^3- 3x^2+ 1) b. f(x) = x^(-3/4)

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

carilah anti turunan F(x) + C bila

a. f(x) = (x^2-x)/(2x^3- 3x^2+ 1)
b. f(x) = x^(-3/4)

Jawab:

∫f (x)dx=F(x)+ C

Maka:

a. f(x) = (x^2-x)/(2x^3- 3x^2+ 1) anti turunannya adalah

∫ ((x^2-x)/(2x^3- 3x^2+ 1))dx

Mis : a = 2x^3- 3x^2+ 1
Maka: dx = da/(6x^2-6x)

Sehingga:

∫ ((x^2-x)/(2x^3- 3x^2+ 1))dx= ∫ ((x^2-x)/a) da/(6x^2-6x)

=1/6 ∫(da/a)

=1/6 ( ln⁡ a)+C

=1/6 ln⁡( 2x^3 – 3x^2+ 1)+C

 

b. f(x) = x^(-3/4) anti turunannya adalah

∫x^(-3/4) dx=1/0,25 x^0,25+ C

 
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carilah anti turunan F(x) + C bila a. f(x) = 3x^2-10x+5 b. f(x) = x^2 (20x^7- 7x^5+6) c. f(x) = 1/x^3 +6/x^7

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

carilah anti turunan F(x) + C bila

a. f(x) = 3x^2-10x+5
b. f(x) = x^2 (20x^7- 7x^5+6)
c. f(x) = 1/x^3 +6/x^7

Jawab:

∫f(x)dx = F(x)+ C

Maka:

a. f(x) = 3x^2-10x+5 anti turunannya adalah

∫ (3x^2-10x+5)dx = 3/3 x^3-10/2 x^2+5x+ C

= x^3-5x^2+5x+C

b. f(x) = x^2 (20x^7- 7x^5+6) anti turunannya adalah

∫ x^2 (20x^7- 7x^5+6) dx = ∫ (20x^9- 7x^7+6x^2 ) dx

= 20/10 x^10-7/8 x^8+6/3 x^3+ C

= 2 x^10-7/8 x^8+ 2x^3+ C

c. f(x) = 1/x^3 +6/x^7 anti turunannya adalah

∫ (1/x^3 +6/x^7 )dx = -1/2 x^(-2)-6/6 x^(-6)

=-1/(2x^2 )-1/x^6

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