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Math help! Quotient rule, second derivative

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dave44
Posts: 11052 - Karma: 113
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
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
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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E.C.
Posts: 3211 - Karma: 319
^yup do exactly wht you did to get the first derivative.
Nov 7 2010 1:04PM
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Gingeski
Posts: 100 - Karma: 16
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
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Melvs
Posts: 15174 - Karma: 480
http://www.wolframalpha.com/input/?i=second+derivative+of+(1%2B2x)/(1-4x)
Nov 7 2010 1:19PM
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