Working out the first and second derivative

[Willjohnc]

Let f(x)=x^2/(1-x^2 )
a) Find f'(x)
b) Find f"(x)

For the answer to a) they give f'(x)=2x/〖(1-x^2)〗^2
and for b) f"(x)=2 (1+〖3x〗^2)/〖(1-x^2)〗^3

Now after many rounds of trying i have not been able to get an answer remotely close to what they have given. i don;t know if it is due to me over working a simple problem or what. the same applies to b), taking the second derivative.

For a) i applied the quotient rule:

f(x)=x^2/(1-x^2 )
f'(x)=(2x(1-x^2)-(1-x^2)x^2)/(1-x^2)^2

then it stops there as i don't know where to proceed, i am not entirely sure if what i have done is correct but other methods result in something similar.

Then for b)

f'(x)=2x/〖(1-x^2)〗^2

Using their answer and trying to work with it to see if i fared any better for the 2nd derivative proved that i was lost. any help with a) would be appreciated as it means i could work out b) and any similar problems in the future.

 

[theCoker]

You seem to have made an error in applying the quotient rule. You should have
f'(x)=[g'(x)h(x)-h'(x)g(x)]/[h(x)]^2 where g(x)=x^2 and h(x)=1-x^2. After you differentiate h(x) where I suggested, multiply the resulting numerator all out and combine like terms.

I confirmed the answer to a) but I think the answer for b) should be
f"(x)=2 (1+3〖x〗^2)/〖(1-x^2)〗^3

Personally, I never use the quotient rule. Note that you can write any function that readily permits the quotient rule (i.e. f(x)=g(x)/h(x)) as f(x)=g(x)[h(x)]^(-1) and then apply the product rule. However, it is important to realize the inner workings of the quotient rule for problems that specifically call for its use (tends to happen a lot on calculus midterms//finals//what have you).

 

[Willjohnc]

Thank you, for some reason something wasn't clicking.
Found using the product rule with these problems really simplifies them so, thanks again.