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
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
To find the second derivative, you can use the quotient rule again. Remember that the quotient rule states that for a function f(x) = g(x)/h(x), the second derivative can be found using the formula:
f''(x) = (g''(x) * h(x) - g'(x) * h'(x)) / (h(x))^2
Let's apply this to your function:
f(x) = 3/(1+2x^3)
First, let's find g(x) and h(x):
g(x) = 3
h(x) = 1+2x^3
Now, let's find the first derivative of g(x) and h(x):
g'(x) = 0
h'(x) = 6x^2
Next, let's find the second derivatives of g(x) and h(x):
g''(x) = 0
h''(x) = 12x
Now, let's plug these values into the quotient rule formula:
f''(x) = (0 * (1+2x^3) - 0 * 12x) / (1+2x^3)^2
= 0 / (1+2x^3)^2
= 0
Therefore, the second derivative of f(x) = 3/(1+2x^3) is 0.
I hope this helps! Let me know if you have any further questions.