# Math CONTINUATION OF QUESTION FROM bobpursley

posted by on .

Thank you for answering the question before bobpursley,
I understand how you did the deriv of inside, outside, but how do I now do the second derivative since you have the extra -18x^2 in the numerator??
Thank you very much! (I suppose I have to brush up on my derivatives)

Posted by KIKSY on Thursday, March 10, 2011 at 8:52pm.

What is the first and second derivative of 3/(1+2(x^3)) ?
im tempted to do the quotient rule, but this looks weird to me...any advice would be greatly appreciated!!

Math - bobpursley, Thursday, March 10, 2011 at 8:55pm
let u=(1+2x^3)

then f= 3u^-1
f'= -3 *u^-2 *u'

but u'= 6x^2

so f'= -3*6x^2 /(1+2x^3)^2

• Math CONTINUATION OF QUESTION FROM bobpursley - ,

I don't think bobpursley is on line, so permit me to continue ...

from Bob, y' = -18x^2(1+2x^3)-2

so using the product rule and chain rule combination

y'' = -18x^2)(-2)(1+2x^3)^-3(6x^2) + (1+2x^2)^-2(-36x)
= 216x^4(1+2x^3)^-3 - 36x(1+2x^3)^-2
= 36x(1+2x^3)^-3 [ 6x^3 - (1+2x^3)]
= 36x(1+2x^3)^-3 (4x^3-1)
or
= 36x(4x^3-1)/(1+2x^3)^3