Question

please help?
what is f'(x) and g'(x) of (x^3+3x+2)/(x^2-1)

Answer 1

Shall we work together on this?

Answer 2

yes

Answer 3

Awesome! And thanks for being the first person in like half an hour to actually reply. lol

First of all, let's make sure you understand everything we need. Do you know which part is F(x) and which is g(x)?

Answer 4

f(x) i know is (x^3+3x+2)
and g(x) is (x^2-1)

lol yw

Answer 5

Perfect!

Alright, so let's do each part individually.

Let's do f'(x) first. What would you suggest doing first to differentiate

(x^3+3x+2)

Answer 6

g(x)f'(x)-f(x)g'(x)/(g(x))^2 Im trying to plug it in this formula

Answer 7

Yup, the quotient formula.


But I mean what do you think we need to do to differentiate just the f(x)?

As in this: (x^3+3x+2)

Answer 8

would it be 3x^2+3?

Answer 9

Perfect!

Answer 10

:D yay

Answer 11

Now we do g'(x).
(x^2-1)

And that turns into what?

Answer 12

2x?

Answer 13

Exactly! See, you knew what you were doing all along! :D

Answer 14

Now let's just go ahead and do the whole quotient rule to finish up.

Answer 15

=(x^2-1)(3x^2+3)-(x^3+3x+2)(2x)/2x^2 ?

Answer 16

What I always like doing is listing each function first:

f(x) = (x^3+3x+2)
f'(x) = (3x^2 + 3)

g(x) = (x^2-1)
g'(x) = 2x

If you don't already do this, trust me, it's REALLY helpful and avoids silly mistakes on tests.

Answer 17

That sounds right, but lemme double check it too.

Answer 18

yes! I do that too!

Answer 19

g(x)f'(x)-f(x)g'(x)/(g(x))^2

[(x^2 - 1)(3x^2 + 3x + 2)] - [ (x^3 + 3x + 2)(2x)]
-------------------------------------------
(x^2 - 1)^2

Answer 20

OOops! I made some mistakes. Let me do this again. Sorry!

Answer 21

its ok

Answer 22

I made some myself

Answer 23

(Ironically, I messed up because I didn't write the original quotient rule again; I was actually about to recommend that to you. How ironic!)

f(x) = (x^3+3x+2)
f'(x) = (3x^2 + 3)

g(x) = (x^2-1)
g'(x) = 2x

g(x)f'(x)-f(x)g'(x)/(g(x))^2


[(x^2-1)(3x^2 + 3)] - [(x^3+3x+2)(2x)]
----------------------------------
(x^2-1)^2

I *think* that should be right. But please lemme know what you get on your own just so we can both be sure.

Answer 24

I got the exact same thing as you ! :)

Answer 25

It makes sense to me

Answer 26

Awesome!

And just one little thing I wanted to point out:

I noticed earlier that in the denominator you had written it as 2x^2. While I know that you corrected it and didn't put the prime of g the second time, I do need to point out that you should always put your terms in parentheses to be safe.

Technically, writing 2x^2 (even on paper) would mean "two times xsquared". But what you need to do is write (2x)^2, which turns into "4 times xsquared".

Not to be a nitpicker, but it's an important difference. ^_^

Have fun!

Answer 27

yeah I know . I caught on to it after I wrote and looked over it again. Thanks for the tip,and thank you so much for your help :)

Answer 28

And the awesome thing about parentheses is they never hurt; as long as you don't do something like change the order of stuff. I especially overuse parentheses on fractions (as you can see here :P) to always be safe.

And no problem; thanks for replying and working it out with me! It means a lot to me when folks actually come here to learn instead of being like "Give me the answer, I don't care how to do it". :P

Answer 29

Do you think you can help me with this last one …
3sqrtx(sqrtx+3)

Answer 30

lol yeah I know what you mean

Answer 31

Sorry, didn't see your reply! I'd love to help; are we just multiplying it out or differentiating it?

Answer 32

Bummer... I took too long. Feel free to fire off a message whenever you get a chance, I'll try to reply. I'm just not sure if you meant to multiply it or to differentiate it. I'll go ahead and multiply it out:

3sqrtx(sqrtx+3)

you can do it two ways. The "real way" where you directly distribute, or a method I invented called the "substitution method".

real way:

"3x+9sqrtx"

That one is bad because you can't see the steps being done. But my method can show you what I did:

say a = 3sqrtx
b = sqrtx
c = 3

so we have:

a(b+c)
that is easier to work with:

ab + ac

ab = 3sqrtx times sqrtx. The two sqrtx's multiply into an x (this is the trickiest part; I'm guessing you know this, but something like sqrt(5) times sqrt(5) is just 5). So ab turns into 3x.

So we have

3x + bc so far.

Let's do the bc now:
sqrtx*3

Since we can't do anything else to those two terms, all we can do is stick them together:

3sqrt(x)


So our final answer:

3x + sqrtx

Actually, I can go ahead and differentiate that too if you need me to, since it doesn't look like it'll take long.

(3x)' = 3
(sqrtx)' = (x^(1/2))'

Just to clarify: sqrtx is x to the one half power.

So using power rule, we put the 1/2 as a coefficient and lower the power by 1 (or in this case, do (1/2) - (2/2) = (-1/2)

So:

(x^(1/2))' = (1/2)x^(-1/2) Note that the negative 1/2 ONLY affects the x and not the 1/2 in front of x. In fact, if it doesn't make it too confusing, I'd recommend:

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

And that means that

(3x + sqrtx)' = 3 + (1/2)[x^(-1/2)]

Hope that helped; sorry again for not replying earlier!

Answer 33

@uber1337h4xx0r it said to find the derivative of the function