Question

If f(x)=3x^2+x/3x^2-x then f'(x) is

Answer 1

I used the quotient rule and I got 1/(3x^2-x) but it's none of my answer choiced :(

Answer 2

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

Answer 3

can you please show your steps so that I can spot where you may have made a mistake?

Answer 4

Sure,

\[\frac{(3x^2-x)(6x+1)-(3x^2+x)(6x-1)}{(3x^2-x)^2}\]

Answer 5

hmm I see what I did wrong lel

Answer 6

let me fix it

Answer 7

correct so far :)

Answer 8

yeah, it was my subtraction

Answer 9

I mean the addition, when combining the like terms.

Answer 10

so I am assuming you got the correct answer this time?

Answer 11

well, idk, now I get 3x^2-2x/(3x^2-x)^2

Answer 12

show me how you simplified the numerator please

Answer 13

\[18x^3+3x^2-6x^2-x-18x^3-3x^2+6x^2-x\]

Answer 14

you have a sign error there

Answer 15

D:

Answer 16

\[(3x^2-x)(6x+1)-(3x^2+x)(6x-1)=(6x)(3x^2-x) + (1)(3x^2-x)-\]\[(6x)(3x^2+x)-(-1)(3x^2+x)\]

Answer 17

\[=(6x)((3x^2-x)-(3x^2+x))+3x^2-x+3x^2-x\]\[=6x(3x^2-x-3x^2-x)+6x^2\]\[=6x(-2x)+6x^2\]\[=-12x^2+6x^2\]\[=-6x^2\]

Answer 18

did you follow the above?

Answer 19

no :( but I'm seeing what I did wrong now :)

Answer 20

I simplify it and I get -6/(3x-1)^2

Answer 21

you could have actually simplified the problem 1st as follows:\[\frac{3x^2+x}{3x^2-x}=\frac{x(3x+1)}{x(3x-1)}=\frac{3x+1}{3x-1}\]and then differentiated this :)

Answer 22

your final answer above is correct :)

Answer 23

thank you :)

Answer 24

Wait a minute, did you used the product rule after you used the quotient rule?

Answer 25

no - I just expanded the numerator that you wrote above