Find y prime using logarithmic differentiation?

TheGreatDerived1

From: US
Posts: 18
Votes: 54
How do you solve these using logarithmic differentiation? I'm looking to find y prime. 1. y = √(x^2 + sin x) / √(x^2 + ln x) 2. y = x^(3x^2)

 

Answer No.1

MrRoger

From: US
Posts: 6
Votes: 12
Y = x^(3x²) ln(y) = 3x² ln(x) y'/y = (6x ln(x) + 3x²(1/x)) y' = y (6x ln(x) + 3x) y' = x^(3x²) (6x ln(x) + 3x)

Answer No.2

DrDan

From: US
Posts: 29
Votes: 58
1. lny = (1/2) ln[x^2+sinx] - (1/2) ln[x^2 +lnx] So (1/y)y' = (1/2)[2x+cosx] - (1/2)1/(x^2+lnx)(2x+(1/x)) Clearly y' = y{(1/2)[2x+cosx] - (1/2)1/(x^2+lnx)(2x+(1/x))} Now try the second one. Its not a different technique.

Answer No.3

Anonymous

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1. y = √(x^2 + sin x) / √(x^2 + ln x) dy/dx ...........................2x+cos(x) =√(x^2+ln(x))d/dx --------------------- ............................2√(x^2+sin… ..............................(2x+1/x) .-√(x^2+sin(x)) d/dx-------------------- ..............................2√(x^2+l… --------------------------------------… ............(√(x^2+ln(x))^2 ..√(x^2+ln(x))(2x+cos(x).... =[---------------------------------- .......2√(x^2+sin(x) ....√(x^2+sin(x))(2x+1/x).......1 - -----------------------------------] -------------- ........2√(x^2+ln(x)).............x^2+… ...2x+cos(x) =------------------------------------- ...2√(x^2+ln(x))√(x^2+sin(x) ....(2x+1/2)√(x^2+sin(x)) - --------------------------------- answer// .........2(x^2+ln(x))^3/2 2. y = x^(3x^2) ln(y)=3x^2ln(x) dy/dx ----------=3[ln(x) /d/dx 2x+x^2d/dx 1/x] ....y dy/dx=y[6xln(x)+3x] y'=x^(3x^2)[6xln(x)+3x] answer// You revealed this well! aturan pakai cialis 20mg cialis prices expiration date on cialis buy cialis online

Answer No.4

Anonymous

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