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1/x+ 1/y = 1

Posted by andi telaumbanua on Feb 11, 2018 in Matematika

Carilah y^’ dari fungsi 1/x+1/y=1!

Jawab : Gunakan metode turunan implisit

1/x+1/y=1

y = x/(x-1)

maka:

(d (1/x+1/y ))/dx = (d(1))/dx

(d(1/x ))/dx+ d(1/y )/dx = (d( 1))/dx

(((0)(x)- (1)(1))/x^2 )+ (((0)(y)-(1)(dy/dx))/y^2 )=0

-1/x^2 – (dy/dx)/y^2 =0

dy/dx= -y^2/x^2

maka:

dy/dx= -(x/(x-1))^2/x^2

dy/dx= -(x^2/(x^2-2x+1))/x^2

dy/dx= -x^2/(x^4-2x^3+ x^2 )

for more detailed writing click on the following link Carilah y^’ dari fungsi 1/x+1/y=1

 
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xy+2x+3x^2 = 4

Posted by andi telaumbanua on Feb 11, 2018 in Matematika

Carilah y^’ dari fungsi xy+2x+3x^2=4 !

Jawab :

Gunakan metode turunan implisit

xy+2x+3x^2=4

y = (4-2x-3x^2)/x

maka:

(d (xy+2x+3x^2 ))/dx = (d(4))/dx

(d( xy))/dx+ d( 2x )/dx+ (d(3x^2))/dx = (d( 4))/dx

[(d( x)/dx)(y)+ (x)(d(y)/dx) ]+ d( 2x )/dx+ (d(3x^2))/dx = (d( 4))/dx

y+x d(y)/dx+ 2+ 6x = 0

x d(y)/dx= -(y+2+6x)

dy/dx=-(y+2+6x)/x

maka∶

dy/dx=(-((4-2x-3x^2)/x)-2-6x)/x

dy/dx=((-((4-2x-3x^2)/x)-2-6x)/x) (x/x)

dy/dx=(-4+2x+3x^2-2x-6x^2)/x^2

dy/dx=(-4-3x^2)/x^2

for more detailed writing click on the following link Carilah y^’ dari fungsi xy+2x+3x^2=4 !

 
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x^3+y^3=1

Posted by andi telaumbanua on Feb 11, 2018 in Matematika

Carilah y^’ dari fungsi x^3+y^3=1 !

Jawab :

Gunakan metode turunan implisit

(d (x^3+y^3 ))/dx = (d( 1))/dx

(d( x^3))/dx+ (d( y^3))/dx = (d( 1))/dx

(d( x^3))/dx+ (d( y^3 )/dy)(dy/dx) = (d( 1))/dx

3x^2 + (3y^2) ( dy/dx) =0

dy/dx= (-3x^2)/(3y^2 )

dy/dx= – x^2/y^2

for more detailed writing click on the following link Carilah y^’ dari fungsi x^3+y^3=1 !

 
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Turunan kedua dari x^4+ y^4 = 81

Posted by andi telaumbanua on Feb 11, 2018 in Matematika

Carilah y^” dari fungsi x^4+ y^4 = 81 !

Jawab :

Gunakan metode turunan implisit

(d(x^4))/dx + (d(y^4))/dx= ((d(81))/dx

4x^3+(d(y^4 )/dy)(dy/dx )=0

4x^3+ 4y^3 y^’=0

y^’= -x^3/y^3

Maka :

y^”= -((d(x^3 )/dx)(y^3 )-x^3 (d(y^3 )/dx) )/(y^3)^2

y^”= -(3x^2 (y^3 )-x^3 (3y^2 )(y^’))/y^6

y^”= -(3x^2 y^3-x^3 (3y^2 )(-x^3/y^3 ))/y^6

y^”= -(3x^2 y^3+ 3x^6/y)/y^6

y^”=(-(3x^2 y^3+ 3x^6/y)/y^6 )(y/y)

y^”= -(3(x^2 y^4+ x^6) )/y^7

y^”= -(3x^2 (y^4+ x^4) )/y^7

Atau

y^”= -(3x^2 (81) )/y^7 = -243(x^2/y^7 )

for more detailed writing click on the following link Carilah y^” dari fungsi x^4+ y^4 = 81 !

 
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sin (x + y) = y^2 cos⁡ x

Posted by andi telaumbanua on Feb 11, 2018 in Matematika

Carilah dy/dx dari fungsi sin(x + y) = y^2 cos⁡x !

Jawab :

Gunakan metode turunan implisit

(d (sin(x + y) ))/dx = (d(y^2 cos⁡x ))/dx

cos (x + y) ((d(x+ y))/dx) = ((d(y^2))/dx)(cos⁡x)+(y^2 )((d(cos⁡x))/dx)

cos (x + y) [(d(x)/dx)+(dy/dx)]=(d(y^2 )/dy)(dy/dx)(cos⁡x)+ y^2 (- sin x)

cos (x + y) [ 1+dy/dx]=2y dy/dx(cos⁡x)- y^2 (sin x)

cos (x + y)+ dy/dx cos (x+y) =2y dy/dx(cos⁡x)- y^2 (sin x)

cos (x + y) + y^2 sin⁡x = 2y dy/dx cos⁡x – dy/dx cos (x+y)

cos (x + y) + y^2 sin⁡x = dy/dx[2y cos⁡x-cos ⁡(x+y) ]

dy/dx= (cos (x + y) + y^2 sin⁡x)/(2y cos⁡x-cos ⁡(x+y) )

for more detailed writing click on the following link Carilah dy/dx dari fungsi sin(x + y) = y^2 cos⁡x !

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