Help With Derivative (Calculus)!!!!! Please...?

Question

f(x) = (x^3 -1) / (x^3 +1)
I am getting the first derivative as 6(x^2) / (x^3 +1)^2
Need Help with the second derivative. Please Help!!!
Thanks in advance.....
Pleaseeeeeeee explain....

Answer 1

You have f'(x) correctly:
f'(x) = 6x^2 / (x^3 + 1)^2

To get f''(x) there's not much we can do except bite the bullet and jump into the quotient rule:
f''(x) = [(12x) (x^3 + 1)^2 - (6x^2) (2) (x^3 + 1) (3x^2)] / (x^3 + 1)^4
= [(12x (x^3 + 1) - 36x^4] / (x^3 + 1)^3
= (12x - 24x^4) / (x^3 + 1)^3
= (12x) (1 - 2x^3) / (x^3 + 1)^3.

Answer 2

The first derivative is right.

As for the second derivative, just repeat the process.

Use the quotent function rule and the chain rule.

The quoteint rule is this:
When f(x) is the function and g(x) is the numerator and h(x) is the denominator,

f(x)= g(x)/h(x)

f'(x) = (h(x)*g'(x)-g(x)*h'(x))/(h(x))^2

The chain rule is hard to explain so I will just show examples.

If g(x)=(h(x))^2
then g'(x)= 2(h(x))*h'(x)

You have to take the derivative of the outermost part and then multiply by the derivative of the inner part.

If (in this case) h(x)= (x^3 + 1)^2
then h'(x)= 2(x^3+1)*3x^2

Therefore- in your problem

f''(x)= (6x^2*2(x^3+1)*3x^2-(x^3 + 1)^2*12x)/(x^3 + 1)^4

f''(x)= (36x^4*(x^3+1)-12x(x^3+1)^2)/(x^3 + 1)^4

Here pull out common factors 12x(x^3 + 1)

f''(x)=12x(x^3 + 1)(3x^3-x^3+1)/(x^3 + 1)^4

f''(x)=12x(2x^3+1)/(x^3 + 1)^3

and that should be your final answer.