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December 18, 2014

December 18, 2014

Posted by **KIKSY** on Thursday, March 10, 2011 at 9:05pm.

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 -
**Reiny**, Thursday, March 10, 2011 at 11:02pmI 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

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