1. ## Differentiate

s= 3t/(t^3-1)

i have worked so far into this:

(3t)(t^3-1)^-1
(-3t)(t^3-1)^-2
(-3t)/(t^3-1)^2

I am not sure if I am doing this right.
Could some one check so far and finish it off for me.
I am not sure where the top (1+2t^3) comes from please check.

p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks

2. Originally Posted by Awsom Guy

s= 3t/(t^3-1)

i have worked so far into this:

(3t)(t^3-1)^-1
(-3t)(t^3-1)^-2
(-3t)/(t^3-1)^2

I am not sure if I am doing this right.
Could some one check so far and finish it off for me.
I am not sure where the top (1+2t^3) comes from please check.

p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
Have you learnt the quotient rule?

Check it yourself here: differentiate 3t/(t^3-1) - Wolfram|Alpha

Click on Show steps.

3. Originally Posted by Awsom Guy

s= 3t/(t^3-1)

i have worked so far into this:

(3t)(t^3-1)^-1
(-3t)(t^3-1)^-2
(-3t)/(t^3-1)^2

I am not sure if I am doing this right.
Could some one check so far and finish it off for me.
I am not sure where the top (1+2t^3) comes from please check.

p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
Use the quotient rule:

$s = \frac{3t}{t^3 - 1}$

$\frac{ds}{dt} = \frac{(t^3 - 1)\frac{d}{dt}(3t) - 3t\frac{d}{dt}(t^3 - 1)}{(t^3 - 1)^2}$

$= \frac{3(t^3 - 1) - 3t(3t^2)}{(t^3 - 1)^2}$

$= \frac{3t^3 - 3 - 9t^3}{(t^3 - 1)^2}$

$= \frac{-6t^3 - 3}{(t^3 - 1)^2}$

$= \frac{-3(2t^3 + 1)}{(t^3 - 1)^2}$.

4. oh my,,, I am so sorry about that I have forgotten my old work thank you for reminding me . I have correctly done the question thanks.

5. ## Thanks you again.

Originally Posted by Prove It
Use the quotient rule:

$s = \frac{3t}{t^3 - 1}$

$\frac{ds}{dt} = \frac{(t^3 - 1)\frac{d}{dt}(3t) - 3t\frac{d}{dt}(t^3 - 1)}{(t^3 - 1)^2}$

$= \frac{3(t^3 - 1) - 3t(3t^2)}{(t^3 - 1)^2}$

$= \frac{3t^3 - 3 - 9t^3}{(t^3 - 1)^2}$

$= \frac{-6t^3 - 3}{(t^3 - 1)^2}$

$= \frac{-3(2t^3 + 1)}{(t^3 - 1)^2}$.

Thanks again but I figured it out sorry for troubling you

6. Originally Posted by Awsom Guy

s= 3t/(t^3-1)

i have worked so far into this:

(3t)(t^3-1)^-1
You can do it this way, using the product rule rather than the quotient rule- but you didn't use the product rule.
(fg)'= f'g+ fg'
$(3t(t^3- 1)^{-1})'= (3t)'(t^3-1)^{-1}+ (3t)((t^3-1)^{-1})'$
$(3)(t^-1)^{-1}+ (3t)(-1(t^3- 1)^{-2}(3t^2)$

(-3t)(t^3-1)^-2
(-3t)/(t^3-1)^2

I am not sure if I am doing this right.
Could some one check so far and finish it off for me.