can someone pleez help me find the derivative of (3x^2+x)/(3x^2-x)?

Question

i just need an answer, it is multiple choice, i tried the quotient rule but i didnt get a right answer, please when you answer have it in simplest form. That might be why i cannot find the answer, thank you

Answer 1

quotient rule for f(x)/g(x): [g(x)f'(x)-f(x)g'(x)] / [g(x)]^2

so [ (3x^2 - x) (6x + 1) - (3x^2 + x) (6x - 1) ] / [ (3x^2 - x)^2 ]

(18x^3 + 3x^2 - 6x^2 - x - 18x^3 + 3x^2 - 6x^2 + x) / [ (3x^2 - x)^2 ]

-6x^2 / [ (3x^2 - x)^2 ]

that's the simplified version

Answer 2

Ok follow the following steps OK:
1) Apply quotient rule. We get

(3x^2 - x)d(3x^2 + x) - (3x^2 + x)d(3x^2 - x)/ (3x^2 - x)^2

2) Then u find the derivatives. We get

(3x^2 - x)(6x + 1) - (3x^2 + x)(6x - 1)/ (3x^2 - x)^2
Next simplify and u get the answer.
TIP: Just concentrate on the problem and follow steps above. U will not go wrong.
PS: Are u in Class 11 CBSE? I am too. I reccomend taking the log book to exam hall it has all this and more-check mathematical constants and formulae.

Answer 3

Derivative of (3x^2+x) is (6x+1)
Derivative of (3x^2-x) is (6x-1)

Product Rule:
((6x-1)(3x^2+x) - (6x+1)(3x^2-x)) / (3x^2-x)^2

Simplifying:
((18x^3 +3x^2 -x) - (18x^3 + -3x^2 -x)) / (3x^2-x)^2
(6x^2) / (3x^2-x)^2
(6x^2) / (9x^4 - 6x^3 +x^2)
6 / (9x^2 - 6x + 1)
6 / (3x - 1)^2

So "6 over the squared quantity three x minus one"

Check my simplification as I did that all in my head.

Hope it works out!