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# Math help! Quotient rule, second derivative

Posts: 11095

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"

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dave

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Nov 7 2010 12:48PM

Posts: 7763

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

Posts: 100

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

Posts: 3216

^yup do exactly wht you did to get the first derivative.

ski the eastTM

Nov 7 2010 1:04PM

Posts: 100

solved out using the quotient rule:

y'=6/(1-4x)^2

y''=[-6(2)(1-4x)^1(-4)]/(1-4x)^4

y''=[48(1-4x)]/(1-4x)^4

y''=48/(1-4x)^3

Nov 7 2010 1:13PM

Posts: 15303

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