Differentiation and Second Derivative Watch

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Tut
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#1
Report Thread starter 14 years ago
#1
Given that

y = x^4 - 3/ 2x^2

(a) find dy/dx

(b) show that d^2y/dx^2 = x^4 - 9/ x^4

Tricky :mad:
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El Stevo
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#2
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#2
are there any brackets in it or is it WYSIWYG?
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madhapper
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#3
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#3
i get d2y/dx2 = 12x^2-9/x^4
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jpowell
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#4
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#4
First derivative is 4x^3 -3x^-3, the second derivative is not however what you say it is. It is 12x^2 +9x^-4
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El Stevo
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#5
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#5
(Original post by Tut)
Given that

y = x^4 - 3/ 2x^2

(a) find dy/dx

(b) show that d^2y/dx^2 = x^4 - 9/ x^4

Tricky :mad:
y = (x^4) - (3/2)(x^-2)

dy/dx = 4(x^3) + 3(x^-3)

d''y/dx'' = 12(x^2) - 9(x^-4)

same as the others... nearly...
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Tut
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#6
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#6
Thats what i got but thats the way the question is written in the paper :confused:

hmmmm.. confusing
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jpowell
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#7
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#7
Then the question is wrong.
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Tut
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#8
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#8
I copied out the question rong.. sorry bout that peeps.. it is actually:

y= (x^4 - 3) / 2x^2

Still need help tho so any iz very much appreciated
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nas7232
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#9
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#9
a) 4x^3 / -4x^-3
b) 12x^2 / 12x^-4

probally messed it up in a line of working...

just intergrate it 1nce for a)
then integrate it again for b)
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Gauss
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#10
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#10
y = (x^4 - 3)/ (2x^2)

(From using the Qoutient rule, or otherwise)

y' = 2x - (x^4 - 3)/x^3

y'' = -2 + [3(x^4 - 3)]/x^4
______
Euclid
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Bill.L
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#11
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#11
y'=x+3x^-3
y''=1-9x^-4
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