# Thread: Find the derivative of teh algebraic function.

1. ## Find the derivative of teh algebraic function.

f(x) = (3-2x-x^2) / (x^2-1)

I know the product rule and can get too

(x^2-1)(-2-2x) - (3-2x-x^2)(2x) / (x^2-1)^2

The answer in the back of the book says its 2/(x+1)^2

I feel like a dumbass, math used to be my strongest subject, but now I can't even come close the answer... If someone could give me a STEP by STEP approach I would greatly appreciate it. I think my algebra / simplifying skills are in need of a serious tune up. Thanks again

2. I also need help with x(1-(4/x+3)) without using the chain rule. b/c the chain rule hasn't been introduced yet...thanks again

3. bump... anyone?

should i rewrite the functions to make it easier to read?

4. Originally Posted by action812
bump... anyone?

should i rewrite the functions to make it easier to read?
do not bump.

$\displaystyle f(x) = \frac{3-2x-x^2}{x^2-1} = \frac{(3+x)(1-x)}{(x+1)(x-1)} = -\frac{3+x}{x+1}$

$\displaystyle f'(x) = - \frac{(x+1) - (3+x)}{(x+1)^2}$

clean up the algebra.

5. Ok sorry about bumping, I have another question... how would I know whether to factor like you did, or just try and use the quotient rule without factoring?

6. Originally Posted by action812
Ok sorry about bumping, I have another question... how would I know whether to factor like you did, or just try and use the quotient rule without factoring?
you do not have to factor ... however, it makes taking the derivative a bit easier.

7. Well when i don't factor I don't get the correct answer... is my algebra to blame...?

8. Originally Posted by action812
Well when i don't factor I don't get the correct answer... is my algebra to blame...?
I can't make that determination w/o seeing your algebra.

9. If we were to do it without factoring first:

$\displaystyle f(x)=\frac{3-2x-x^2}{x^2-1}$

$\displaystyle f'(x)=\frac{(x^2-1)(-2-2x)-(3-2x-x^2)(2x)}{(x^2-1)^2}$

$\displaystyle =\frac{-2x^2+2-2x^3+2x-(6x-4x^2-2x^3)}{(x-1)^2(x+1)^2}$

$\displaystyle =\frac{2x^2-4x+2}{(x-1)^2(x+1)^2}$

$\displaystyle =\frac{2(x-1)^2}{(x-1)^2(x+1)^2}$

$\displaystyle =\frac{2}{(x+1)^2}$

10. Ok thanks a lot fellas, it makes sense now. My factoring is just a little rusty.

If you all could help me on this problem as well i'd greatly appreciate it.

x(1-(4/x+3))

11. Originally Posted by action812
Ok thanks a lot fellas, it makes sense now. My factoring is just a little rusty.

If you all could help me on this problem as well i'd greatly appreciate it.

x(1-(4/x+3))
start a new problem w/ a new thread.

show what you have tried, also.