Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Carilah Turunan pertama dari fungsi berikut:
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y = (cos x)^x
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y = x^cosx
Jawab:
Use Logarithmic differentiation
1. ln y = ln (cos x)^x
ln y = x ln(cos x)
d/dx(ln y)= d/dx(x ln cosx)
d/dy(ln y)dy/dx = d/dx(x ) (ln cosx) + x d/dx(ln cos x)
1/y dy/dx = ln cos x + x 1/cosx d/dx(cos x)
1/y dy/dx = ln cos x + x 1/cosx (- sinx)
1/y dy/dx = ln cos x – x sin x/cos x
1/y dy/dx = ln cos x – x tanx
dy/dx = y ( ln cos x – x tan x )
dy/dx = (cos x)^x ( ln cos x – x tan x )
ln y = ln x^cosx
ln y = cos x lnx
d/dx(ln y) = d/dx(cos x ln x)
d/dy(ln y) dy/dx = d/dx(cos x ) (ln x) + cos x d/dx(ln x)
1/y dy/dx = – sin x ln x + cos x 1/x
1/y dy/dx = cos x/x – sin x ln x
dy/dx = y ( cos x/x – sin x ln x)
dy/dx = x^cos x ( cos x/x – sin x ln x)
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah turunan dari :
y = x^sinx dan y = x^x
Jawab:
Use logarithmic derivative
1. ln y = ln x^sinx
ln y = sinx lnx
d/dx (ln y ) = d/dx(sinx lnx)
d/dy(ln y) dy/dx = [d/dx (sinx)] (lnx) + sin x [d/dx ( lnx)]
1/y dy/dx = cos x ln x + sinx/x
dy/dx = y (cos x ln x + sinx/x )
dy/dx = (x^sinx ) (cos x ln x + sinx/x )
2. ln y = ln x^x
ln y = x lnx
d/dx(ln y) = d/dx(x lnx)
d/dy(ln y) dy/dx = d/dx(x) (lnx) + x d/dx ( ln x)
1/y dy/dx = ln x+ x 1/x
1/y dy/dx = ln x+1
dy/dx = y (ln x + 1)
dy/dx = x^x (ln x + 1)
Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Find the first derivative from
x^y = y^x
Answare:
Use logarithmic derivative to find the first derivative this function
ln x^y = ln y^x
y ln x = x ln y
y = (x lny) / lnx
so:
d/dx (x ln y) = d/dx (x) ln y + x d/dx (ln y)
= ln y + x [d/dy (ln y) dy/dx]
= ln y + x ( 1/y) dy/dx
= ln y + x/y dy/dx
So:
y = (x lny) / ln x
dy/dx = [ d/dx(x lny) ln x – x ln y d/dx (ln x) ] / (ln x)^2
dy/dx = [ ( ln y + x/y dy/dx) ln x – x ln y 1/x] / (ln x)^2
dy/dx = [ ln y ln x + lnx ( x/y) dy/dx – ln y ] / (ln x)^2
ln^2 x dy/dx = ln y ln x + lnx ( x/y) dy/dx – ln y
ln^2 x dy/dx – x/y ln x dy/dx = ln y ln x – ln y
dy/dx (ln^2 x – x/y lnx ) = ln y (lnx – 1)
dy/dx = [ln y ln (x-1)] / [ln^2 x – x/y ln x]
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah integral dari:
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∫ sin 2x cos 3x dx
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∫ sin^2 3x cos 3x dx
Jawab:
1. ∫ sin ax cos bx dx = -1 /2 [ cos(a-b)x / (a-b) + (cos ( a+b)x) /(a+b)] + C
Maka:
∫ sin 2x cos 3x dx = -1 /2 [ cos(2-3)x /(2-3) + (cos ( 2+3)x) /(2+3)] + C
∫ sin 2x cos 3x dx = – 1 /2 [ cos(-x) /(-1) + (cos 5x) /5 ] + C
∫ sin 2x cos 3x dx = -cos 5x /10 + cosx /2 + C
2. Misalkan: u = sin 3x
Maka: du/dx = 3 cos3x
Sehingga: dx = du /(3 cos3x )
∫ sin^2 3x cos 3x dx = ∫ u^2 cos 3x (du/3 cos 3x )
∫ sin^2 3x cos 3x dx = 1 /3 ∫u^2 du
∫ sin^2 3x cos 3x dx = 1/3 1/3 u^3 + C
∫ sin^2 3x cos 3x dx = 1/9 (sin 3x)^3 + C
∫ sin^2 3x cos 3x dx = 1/9 sin^3 3x + C
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Posted by andi telaumbanua on Feb 17, 2018 in
Matematika
Tentukanlah Integral dari :
∫ sin^4 x cos^3 x dx
Jawab:
Misalkan: u = sin x
Maka: du/dx = cosx
Sehingga: dx = du/cosx
∫ sin^4 x cos^3 x dx = ∫u^4 cos^3 x du/cosx
∫ sin^4 x cos^3 x dx = ∫ u^4 cos^3 x (du/cos x)
∫ sin^4 x cos^3 x dx = ∫ u^4 cos^2 x du
Karena: cos^2 x = 1 – sin^2 x
Maka:
∫ sin^4 x cos^3 x dx = ∫ u^4 (1- sin^2 x) du
Karena: u = sin x
Maka:
∫ sin^4 x cos^3 x dx = ∫ u^4 (1- u^2) du
∫ sin^4 x cos^3 x dx = ∫ (u^4- u^6) du
∫ sin^4 x cos^3 x dx = 1 /(4+1) u^(4+1) – 1 /(6+1) u^(6+1) + C
∫ sin^4 x cos^3 x dx = 1 /5 u^5 – 1 /7 u^7 + C
∫ sin^4 x cos^3 x dx = 1 /5 sin^5 x – 1 /7 sin^7 x + C
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