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Tentukan deret MacLaurin dari f(x) = e^(-x) cos⁡(x) 

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan deret MacLaurin dari f(x) = e^(-x) cos⁡(x)

Jawab:

Deret MacLaurin;
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯

Maka:
f(x) = e^(-x) cos⁡x
f^1 (x)= – e^(-x) cos⁡x-e^(-x) sin⁡x
f^2 (x)= e^(-x) cos⁡x+e^(-x) sin⁡x+e^(-x) sin⁡x-e^(-x) cos⁡x =2e^(-x) sin⁡x
f^3 (x)=-2e^(-x) sin⁡x+2e^(-x) cos x
f^4 (x)= 2e^(-x) sin⁡x-2e^(-x) cos ⁡x-2e^(-x) cos⁡x – 2e^(-x) sin⁡x= -4e^(-x) cos⁡x

Maka:
f(0) = e^(-0) cos⁡0=1
f^1 (0)= -e^(-0) cos⁡0-e^(-0) sin⁡0=-1-0=-1
f^2 (0)=2e^(-0) sin⁡0=0
f^3 (0)= -2e^(-0) sin⁡0+2e^(-0) cos 0=0+2=2
f^4 (0)= -4e^(-0) cos⁡0=-4

Maka: deret macLaurinnya
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯
f(x) = 1 + (-1)(x)+0/2! x^2+2/3! x^3+(-4)/4! x^4+⋯
f(x) = 1 -x+0+1/3 x^3-1/6 x^4+⋯
f(x) = 1 – x+1/3 x^3-1/6 x^4+⋯

 
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Tentukan deret MacLaurin dari f(x) = e^x cos⁡(x)

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan deret MacLaurin dari f(x) = e^x cos⁡(x)

Jawab:

Deret MacLaurin;
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯

Maka:
f(x) = e^x cos⁡(x)
f^1 (x)= e^x cos(⁡x)  -e^x sin⁡(x)
f^2 (x)= e^x cos⁡x-e^x sin⁡x-e^x sin⁡(x) – e^x cos⁡(x) =-2e^x sin⁡(x)
f^3 (x)=-2e^x sin⁡x-2e^x cos (⁡x)
f^4 (x)= -2e^x sin⁡x-2e^x cos (⁡x) – 2e^x cos⁡(x) + 2e^x sin⁡x= -4e^x cos(⁡x)

Maka:
f(0) = e^0 cos⁡0=1
f^1 (0)= e^0 cos⁡0-e^0 sin⁡0 =1- 0 =1
f^2 (0)=-2e^0 sin⁡0=0
f^3 (0)= -2e^0 sin⁡0-2e^0 cos 0 = 0 – 2 =-2
f^4 (0)= -4e^0 cos⁡0=-4

Maka: deret macLaurinnya
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯
f(x) = 1 + (1)(x)+0/2! x^2+(-2)/3! x^3+(-4)/4! x^4+⋯
f(x) = 1 + x+0- 1/3 x^3-1/6 x^4+⋯
f(x) = 1 + x-1/3 x^3-1/6 x^4+⋯

 
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Tentukan deret MacLaurin dari f(x) = e^x sin(x) 

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan deret MacLaurin dari f(x) = e^x sin(x)

Jawab:

Deret MacLaurin;
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯

Maka:

f(x) = e^x sinx
f^1 (x)= e^x sinx+e^x cos⁡x
f^2 (x)= e^x sinx+e^x cos⁡x+e^x cos⁡x-e^x sinx=2e^x cos⁡x
f^3 (x)= 2e^x cos⁡x-2e^x sin⁡x
f^4 (x)= 2e^x cos⁡x-2e^x sin⁡x -2e^x sin⁡x-2e^x cos⁡x= -4e^x sin⁡x

Maka:
f(0) = e^0 sin⁡0 = 0
f^1 (0)= e^0 sin⁡0+e^0 cos⁡0 = 0+1=1
f^2 (0)= 2e^0 cos⁡0= 2
f^3 (0)= 2e^0 cos⁡0-2e^0 sin⁡0=2-0=2
f^4 (0)= -4e^0 sin⁡0=0

Maka: deret macLaurinnya
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯
f(x) = 0 + (1)(x)+2/2! x^2+2/3! x^3+0/4! x^4+⋯
f(x) = 0 + x+x^2+1/3 x^3+0+⋯
f(x) = x+x^2+1/3 x^3+⋯

 
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Tentukan deret MacLaurin dari f(x) = e^(-x) sin(x)

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan deret MacLaurin dari f(x) = e^(-x) sinx

Jawab:

Deret MacLaurin;
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯

Maka:
f(x) = e^(-x) sinx
f^1 (x)= -e^(-x) sinx+e^(-x) cosx
f^2 (x)= e^(-x) sinx – e^(-x) cosx  -e^(-x) cos⁡x – e^(-x) sinx =-2e^(-x) cosx
f^3 (x)= 2e^(-x) cos⁡ (x) + 2e^(-x) sin⁡x
f^4 (x)= -2e^(-x) cos⁡ x -2e^(-x) sin⁡x-2e^(-x) sin⁡ x+2e^(-x) cos⁡x

Maka:
f(0) = e^(-0) sin⁡ 0=0
f^1 (0)= -e^(-0) sin⁡ 0 + e^(-0) cos⁡0=0+1=1
f^2 (0)=-2e^(-0) cos⁡ 0 = -2
f^3 (0)= 2e^(-0) cos⁡ 0+2e^(-0) sin⁡ 0=2+0=2
f^4 (0)= -2e^(-0) cos⁡ 0-2e^(-0) sin⁡ 0 -2e^(-0) sin⁡0+2e^(-0)cos⁡0 = -2-0-0+2=0

Maka: deret macLaurinnya

f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯
f(x) = 0 + (1)(x)+(-2)/2! x^2+2/3! x^3+0/4! x^4+⋯
f(x) = 0 + x-x^2+1/3 x^3+0+⋯
f(x) = x-x^2+1/3 x^3+⋯

 
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Tentukan turunan pertama dari : f(x) = 3^(2x+1)+ 2^sin2x

Posted by andi telaumbanua on Jul 28, 2018 in Matematika

Tentukan turunan pertama dari : f(x) = 3^(2x+1)+ 2^sin2x

Jawab:

Note;

a^x =e^xlna
d/dx (a^x )= d/dx (e^xlna )=(e^xlna )(ln⁡a)= a^x lna

Maka: turunan pertama dari : f(x) = 3^(2x+1)+ 2^sin2x adalah

dy/dx = d/dx (3^(2x+1) )+ d/dx (2^sin2x )

= d/dx ( e^(2x+1)ln3 )+d/dx (e^((sin⁡(2x)ln2) )

= e^(2x+1)ln3 d/dx [(2x+1)ln3] + e^((sin⁡(2x)ln2) d/dx [sin⁡(2x)ln2]

=e^(2x+1)ln3 d/dx [(2xln3+ln3) ]+e^((sin⁡(2x)ln2) d/dx [sin⁡(2x)ln2]

=e^(2x+1)ln3 (2ln3)+e^((sin⁡(2x)ln2) (2cos⁡(2x) ln⁡2

Karena: e^(2x+1)ln3=3^(2x+1) dan e^((sin⁡(2x)ln2)=2^sin2x
Sehingga:

=e^(2x+1)ln3 (2ln3)+e^((sin⁡(2x)ln2) (2cos⁡(2x) ln⁡2)

=3^(2x+1) (2ln3)+2^sin2x (2 cos⁡(2x) ln⁡2)

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