Hitung ∫(x^3+ 1)^10 x^5 dx

1. Hitung ∫(x^3+ 1)^10 x^5  dx

Jawab:

Mis:

a = x^3+ 1

 x^3 = a-1

maka:

da/dx = 3x^2
dx = da/3x^2

sehingga:

∫(x^3+ 1)^10 x^5 dx = ∫(a)^10 x^5 da/3x^(2 )

= 1/3 ∫(a)^10 x^3 da

karena  x^3 = a-1 maka

= 1/3 ∫(a)^10 (a-1) da

= 1/3 ∫a^11- a^10) da

= 1/3 ( 1/12 a^12 – 1/11 a^11 ) + C

= 1/36 a^12 – 1/33 a^11 + C

= 1/36 (x^3+ 1)^12 – 1/33 (x^3+ 1)^11 + C

y = (cos⁡ x)^x dan y = x^cos⁡x

Carilah Turunan pertama dari fungsi berikut:

1. y =  (cos⁡ x)^x

2. y = x^cos⁡x

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/cos⁡x d/dx(cos⁡ x)

1/y dy/dx = ln⁡ cos⁡ x + x 1/cos⁡x (- sin⁡x)

1/y dy/dx = ln⁡ cos⁡ x – x sin⁡ x/cos ⁡x

1/y dy/dx = ln ⁡cos⁡ x – x tan⁡x

dy/dx = y ( ln ⁡cos⁡ x – x tan ⁡x )

dy/dx = (cos⁡ x)^x ( ln⁡ cos⁡ x – x tan⁡ x  )

ln y = ln x^cos⁡x

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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Carilah Turunan pertama dari fungsi berikut: 1. y = (cos⁡ x)^x 2. y = x^cos⁡x

y = x^sin⁡x dan y = x^x

Tentukanlah turunan dari :

y = x^sin⁡x  dan  y = x^x

Jawab:

Use logarithmic derivative

1. ln y = ln x^sin⁡x

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 ( ln⁡x)]

1/y dy/dx = cos⁡ x ln⁡ x + sin⁡x/x

dy/dx = y (cos ⁡x ln⁡ x +  sin⁡x/x )

dy/dx = (x^sin⁡x ) (cos⁡ x ln⁡ x + sin⁡x/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)

for a more clear author please click the link below Tentukanlah turunan dari : 1. y = x^sin⁡x 2. y = x^x

first derivative from x^y = y^x

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 ln⁡y) / ln⁡x

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 ln⁡y) / 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 + ln⁡x ( x/y)  dy/dx –  ln⁡ y ] / (ln⁡ x)^2

ln^2 x dy/dx =  ln⁡ y ln⁡ x + ln⁡x ( 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 ln⁡x ) = ln⁡ y (ln⁡x – 1)

dy/dx = [ln⁡ y ln⁡ (x-1)] / [ln^2 x – x/y ln⁡ x]

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Find the first derivative from x^y = y^x

integral dari (a) ∫ sin⁡ 2x cos⁡ 3x dx (b) ∫ sin^2 3x cos⁡ 3x dx

Tentukanlah integral dari:

1. ∫ sin⁡ 2x cos⁡ 3x dx

2. ∫ 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 + cos⁡x /2  + C

2. Misalkan: u = sin 3x

Maka: du/dx = 3 cos⁡3x

Sehingga: dx = du /(3 cos⁡3x )

∫ 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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Tentukanlah integral dari:

1. ∫ sin⁡ 2x cos⁡ 3x dx

2. ∫ sin^2 3x cos⁡ 3x dx