Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Turunan pertama dari fungsi hiperbol trigonometri berikut!
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y = x sinh x
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y = x^2/cosh x
Jawab:
1. dy/dx=1(sinh x) – x d/dx(sinh x)
dy/dx = sinh x + x cosh x
Atau
dy/dx = (e^x- e^(-x) )/2+ x ((e^x+ e^(-x) )/2)
2. dy/dx = (2x(cosh x) – x^2 (sinh x)) / (cosh x)^2
Atau :
dy/dx = (2x((e^x+ e^(-x) )/2)-x^2 ((e^x- e^(-x) )/2) )/((e^x+ e^(-x) )/2)^2
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Rumus Turunan Pertama dari fungsi Invers Trigonometri
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d/dx(arc sin u )= 1/√(1- u^2 ) (du/dx)
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d/dx(arc cos u )= (-1)/√(1- u^2 ) (du/dx)
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d/dx ( arc tg u)= 1/(1+ u^2 ) (du/dx)
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d/dx ( arc sec u)= 1/(|u| √(u^2- 1)) (du/dx)
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d/dx ( arc ctg u)= (-1)/(1+ u^2 ) (du/dx)
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d/dx ( arc cosec u)= (-1)/(|u| √(u^2- 1)) (du/dx)
Fungsi Hiperbol Trogonometri
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Sinh = (e^x- e^(-x) )/2 cosh = (e^x+ e^(-x) )/2
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tanh = (e^x- e^(-x) )/(e^x+ e^(-x) )
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coth = (e^x+ e^(-x) )/(e^x- e^(-x) )
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sech = (2 )/(e^x+ e^(-x) )
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d/dx(sinh u ) = -coshu (du/dx) = (e^x+ e^(-x) )/2 (du/dx)
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d/dx(cosh u ) = sinhu (du/dx) = (e^x- e^(-x) )/2 (du/dx)
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d/dx(tanh u ) = sech^2 u (du/dx) = ((2 )/(e^x+ e^(-x) ))^2 (du/dx)
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d/dx(coth u ) =Rumus Turunan Pertama dari fungsi hiperbol Trigonometri – cosech^2 u (du/dx) = -((2 )/(e^x- e^(-x) ))^2 (du/dx)
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d/dx(sech u ) = – sech u tanh u (du/dx) = – ((2 )/(e^x+ e^(-x) ))^2 ((e^x- e^(-x) )/(e^x+ e^(-x) ))(du/dx)
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d/dx(cosech u ) = – cosech u coth u (du/dx) = – ((2 )/(e^x- e^(-x) ))^2 ((e^x+ e^(-x) )/(e^x- e^(-x) ))(du/dx)
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukan turunan pertama dari :
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y = arc tg (x – 1)/(x +1) !
Jawab:
Karena: d/dx ( arc tg u)= 1/(1+ u^2 ) (du/dx)
d/dx (arc tg (x – 1)/(x +1))= 1/(1+ ((x – 1)/(x +1))^2 ) (d((x – 1)/(x +1))/dx)
d/dx (arc tg (x – 1)/(x +1))= 1/(((x+1)^2+ (x-1)^2)/(x+1)^2 ) [(1(x+1)- (x-1)1)/(x+1)^2 ]
d/dx (arc tg (x – 1)/(x +1))= (x+1)^2/((x+1)^2+ (x-1)^2 ) (2/(x+1)^2 )
d/dx (arc tg (x – 1)/(x +1))= 2/((x+1)^2+ (x-1)^2 )
d/dx (arc tg (x – 1)/(x +1))= 2/(( x^2+ 2x+1)+ (x^2- 2x+1))
d/dx (arc tg (x – 1)/(x +1))= 2/(2x^2+ 2)
d/dx (arc tg (x – 1)/(x +1))= 2/(2(x^2+ 1))
d/dx (arc tg (x – 1)/(x +1))= 1/(x^2+ 1)
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Tentukan turunan pertama dari y = arc tg (x – 1)/(x +1)
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah turunan pertama dari y = arc ctg 2/x+ arc tg x/2 !
Jawab:
Karena:
d/dx ( arc ctg u)= (- 1)/(1+ u^2 ) (du/dx)
Maka :
d/dx ( arc ctg 2/x)= (- 1)/(1+ (2/x)^2 ) (d(2/x)/dx)
d/dx ( arc ctg 2/x)= (- 1)/(1+ 4/x^2 ) ((-2)/x^2 )
d/dx ( arc ctg 2/x)= (-1)/((x^2+4)/x^2 ) ((-2)/x^2 )
d/dx ( arc ctg 2/x)=((-x^2)/(x^2+ 4)) ((-2)/x^2 )
d/dx ( arc ctg 2/x)=(2/(x^2+ 4))
Karena :
d/dx ( arc tg u) = 1/(1+ u^2 ) (du/dx)
Maka:
d/dx ( arc tg x/2)= 1/(1+ (x/2)^2 ) (d(x/2)/dx)
d/dx ( arc tg x/2)= 1/(1+ x^2/4) (2/4)
d/dx ( arc tg x/2)=(4/(4+ x^2 )) (2/4)
d/dx ( arc tg x/2)=(2/(4+ x^2 ))
Maka turunan dari y = arc ctg 2/x+ arc tg x/2 adalah
dy/dx= d/dx ( arc ctg 2/x)+ d/dx ( arc tg x/2)
dy/dx= (2/(x^2+ 4))+ (2/(4+ x^2 ))
dy/dx= ((2+2)/(x^2+ 4))
dy/dx= (4/(x^2+ 4))
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukan turunan pertama dari y = x^4 sin (1/x^2) !
Jawab:
y = uv maka : y^’= u^’ v+uv^’
Pertama cari dy/dx dari sin (1/x^2)
Maka:
dy/dx = cos (1/x^2) d(1/x^2 )/dx
dy/dx = – 2x/x^4 (cos 1/x^2 )
Kemudian carilah dy/dx dari y = x^4 sin (1/x^2 )
dy/dx = d(x^4)/dx sin (1/x^2 ) + x^4 d(sin(1/x^2 )/dx
dy/dx = 4x^3 sin (1/x^2) + x^4 [- 2x/x^4 (cos 1/x^2 )]
dy/dx = 4x^3 sin (1/x^2 ) – 2x cos 1/x^2
dy/dx = 2x(2x^2 sin (1/x^2 ) – cos (1/x^2 )
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