Math help! Quotient rule, second derivative

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Posts: 11052 - Karma: 113

Karma: 113

Posts: 11052

Crazy Fool

Okay, I'm fine with doing the first derivative using the quotient rule, but some of the homework we've been set asks for the second derivative using the quotient rule, which I haven't come across yet.

I've tried finding the derivative, then finding the derivative of that, but something is going wrong cause the answer comes out wrong every time.

The question is

"Show that if y = 1+2x/1-4x

then d2y/dx2 = k/(1-4x)^3

for constant k, and find the value of k"

Nov 7 2010 12:48PM

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Stickity

Posts: 7750 - Karma: 769

Karma: 769

Posts: 7750

Crazy Fool

he question is

"Show that if y = 1+2x/1-4x

then d2y/dx2 = k/(1-4x)^3

for constant k, and find the value of k

first derivative is y' = [(1-4x)(2) - (-4)(1+2x)]/(1-8x+16x^2)

simplified y' = 5/(1-8x+16x^2)

second derivative is y'' = [0 - 5(2(1-4x)(-4))]/(1-4x)^4

simplified is y'' = 40/(1-4x)^3

boom

Nov 7 2010 1:03PM

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Gingeski

Posts: 100 - Karma: 16

Karma: 16

Posts: 100

Pro

k=48

Although I couldnt get that answer by using the quotient rule, I just used product and chain.
y'=6/(1-4x)^2 which can be re-written as
y'=6*(1-4x)^-2
using product and chain rule for y''
y''=6(-2)(1-4x)^-3(-4)
y''=48/(1-4x)^-3

Nov 7 2010 1:04PM

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