# Thread: how to find the derivatives

1. ## how to find the derivatives

1. f(x)=((x+1)(x-2))/(x^2 -5x +1)
2. f(x)= (x2 − 1)(x^3 + 3x^2/x^2 + 2)

2. Originally Posted by henri
1. f(x)=((x+1)(x-2))/(x^2 -5x +1)
2. f(x)= (x2 − 1)(x^3 + 3x^2/x^2 + 2)

Quotient Rule:

y = f(x)/g(x)
y' = [g(x) f '(x) - f(x) g'(x)]/ [g(x)]^2

in prob 1: do: y = ((x+1)(x-2))/(x^2 -5x +1)

your f(x) is: (x+1)(x-2) or: (x^2 -x -2)

your g(x) is: (x^2 -5x +1)

3. For 1 my answer is: (-14x +16x +9)/(x^2 -5x +1)^2
Is this Correct?
For 2 my answer is: (2x^6 +4x^4 +6x^5 +4x +6x^3 -6x^2-6x)/(x^2 +2)^2

4. what is x2 on your 2nd problem

5. x^2 is the denominator because it has to be squared

6. Originally Posted by henri
For 1 my answer is: (-14x +16x +9)/(x^2 -5x +1)^2
Is this Correct?
For 2 my answer is: (2x^6 +4x^4 +6x^5 +4x +6x^3 -6x^2-6x)/(x^2 +2)^2

I got:

y' = [(x^2 -5x +1)(x-1) - (x^2 -x -2)(x -5)] / [x^2 -5x +1]^2

y' = [(x^3 -6x^2 +6x -1) - (x^3 -6x^2 +3x +10)] / (x^2 -5x +1)^2

. . . i got [3x -11] / [x^2 -5x +1]^2

7. Originally Posted by henri
1. f(x)=((x+1)(x-2))/(x^2 -5x +1)
2. f(x)= (x2 − 1)(x^3 + 3x^2/x^2 + 2)

what is x2 in
1. f(x)= (x2 − 1)(x^3 + 3x^2/x^2 + 2)

8. why is the numerator's derivative x-1 instead of 2x-1 and the same for the denominator

9. Sorry for my error but the second one is (X^2 -1)(. . .)

10. Originally Posted by henri
why is the numerator's derivative x-1 instead of 2x-1 and the same for the denominator

wow, you are correct, thanks, sorry to try and confuse you

11. no problem thanks for all. by the way is there an easier way to put in mathematical symbols in this program? I see other posts with better formats.

12. prob 2:

i got:

y= [x^5 +3x^4 -x^3 -3x^2] / (x^2 +2)

y'= [(x^2 +2)(5x^4 +12x^3 -3x^2 -6x) - (x^5 +3x^4 -x^3 -3x^2)(2x)] / (x^2 +2)^2

.....

= [3x^6 + 6x^5 + 9x^4 + 24x^3 - 6x^2 - 12x] / (x^2 +2)^2

13. not sure about the symbols. Im new to this site using it for stats help. Thought i would try one i knew how to do