1. ## Second derivative

I'm not sure how to find the second derivative of this:

$\displaystyle f(x) = (x) / (x-1)$

For the 1st derivative I got: $\displaystyle (-1) / [ (x-1)^2 ]$

The answer for the 2nd derivative is: $\displaystyle (2) / [ (x-1)^3 ]$

But I can't see how they got it...

2. you can find the second derivative the same way u found the first, by using the quotient rule where $\displaystyle u=-1$ and $\displaystyle v=(x-1)^2$

hope that helps

3. Originally Posted by janedoe
I'm not sure how to find the second derivative of this:

$\displaystyle f(x) = (x) / (x-1)$

For the 1st derivative I got: $\displaystyle (-1) / [ (x-1)^2 ]$

The answer for the 2nd derivative is: $\displaystyle (2) / [ (x-1)^3 ]$

But I can't see how they got it...

(-1)/(x-1)^2

So, you do it in the same way as the first. Quotient rule, and then you've got a chain rule in the denominator.

alternatively, you can bring the whole bottom up to the top,

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

and just do the chain rule, so

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

2/(x-1)^3

4. Originally Posted by iLikeMaths
you can find the second derivative the same way u found the first, by using the quotient rule where $\displaystyle u=-1$ and $\displaystyle v=(x-1)^2$

hope that helps
Well I tried that, and I got 2x-2 for the second derivative. However, the answer is not that. Where am I going wrong..I used the quotient rule like you said which makes sense.

5. Originally Posted by coolguy99
(-1)/(x-1)^2

So, you do it in the same way as the first. Quotient rule, and then you've got a chain rule in the denominator.

alternatively, you can bring the whole bottom up to the top,

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

and just do the chain rule, so

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

2/(x-1)^3

We haven't learned this yet, so I would like to know how to do it without this rule (for ex. using the quotient rule, because I used the q.r. and got the wrong answer)

6. $\displaystyle 2x-2$ is the same as $\displaystyle 2(x-1)$ does that make any more sense.

if u are not using the quotient rule, then what rule are you using

7. Originally Posted by iLikeMaths
$\displaystyle 2x-2$ is the same as $\displaystyle 2(x-1)$ does that make any more sense.

if u are not using the quotient rule, then what rule are you using
Yes I am using the quotient rule, but when I did I got 2x-2...yes that is the same as 2(x-1) but the answer is 2/(x-1)^3.....

8. first $\displaystyle \frac{-1}{(x-1)^2}$derivative

using the quotient rule $\displaystyle u=-1$ and $\displaystyle v=(x-2)^2$ and $\displaystyle \frac{du}{dx}=0$ and $\displaystyle \frac{dv}{dx}=2(x-1)$ and $\displaystyle v^2 =(x-1)^4$

$\displaystyle \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

$\displaystyle \frac{0 -(-2(x-1))}{(x-1)^4}$

$\displaystyle \frac{2(x-1)}{(x-1)^4}$

$\displaystyle \frac{2}{(x-1)^3}$

9. Ahh thank you so much, I was forgetting the bottom half of the quotient rule...so stupid of me!!