# how to find the derivatives

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• Oct 30th 2008, 10:29 AM
henri
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)

• Oct 30th 2008, 10:48 AM
plm2e
Quote:

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)
• Oct 30th 2008, 10:56 AM
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
• Oct 30th 2008, 10:57 AM
plm2e
what is x2 on your 2nd problem
• Oct 30th 2008, 11:00 AM
henri
x^2 is the denominator because it has to be squared
• Oct 30th 2008, 11:06 AM
plm2e
Quote:

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
• Oct 30th 2008, 11:08 AM
plm2e
Quote:

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)
• Oct 30th 2008, 11:11 AM
henri
why is the numerator's derivative x-1 instead of 2x-1 and the same for the denominator
• Oct 30th 2008, 11:12 AM
henri
Sorry for my error but the second one is (X^2 -1)(. . .)
• Oct 30th 2008, 11:17 AM
plm2e
Quote:

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
• Oct 30th 2008, 11:25 AM
henri
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.
• Oct 30th 2008, 11:31 AM
plm2e
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
• Oct 30th 2008, 11:33 AM
plm2e
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