# Differentiation and Second DerivativeWatch

This discussion is closed.
#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
0
14 years ago
#2
are there any brackets in it or is it WYSIWYG?
0
14 years ago
#3
i get d2y/dx2 = 12x^2-9/x^4
0
14 years ago
#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
0
14 years ago
#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
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...
0
#6
Thats what i got but thats the way the question is written in the paper

hmmmm.. confusing
0
14 years ago
#7
Then the question is wrong.
0
#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
0
14 years ago
#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)
0
14 years ago
#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
0
14 years ago
#11
y'=x+3x^-3
y''=1-9x^-4
0
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