1. ## Derivatives/Relative Rates(unsolved)

Hello, I have a couple questions, i'm not sure what im doing wrong.

1.) find the derivative
(x+1)^2(x^2+1)^-3
so I do the product rule and the chain rule to get
(2x)(x+1)^2-3(x^2+1)^-4+(x+1)(x^2+1)^-3
is this correct so far? If yes, where do I go from here?
I already have the answer I just cant figure out how to get there.

2.) Evaluate dy/dt for the function at the point.
x+y/x-y = x^2 + Y^2 , dx/dt = 12 , x = 1, y = 0

I use the quotient rule on the left side and find the derivative or the right and get
2 dy/dt = 2 dx/dt + 2 dy/dt
the answer is 12, but I dont understand how you get there if you remove the dy/dt's. 2dy/dt - 2 dy/dt to take them to one side?? I can see that if you divide all 3 by 2 the dx/dt is 12 but there's no dy/dt's left in the problem for it to be equal to? do you see what im trying to say?

Thanks

2. Originally Posted by cokeclassic
1.) find the derivative
(x+1)^2(x^2+1)^-3
$\frac{d}{dx}((x+1)^2(x^2+1)^{-3})= (x+1)^2((x^2+1)^{-3})'+((x+1)^2)'(x^2+1)^{-3}$

Now find $((x^2+1)^{-3})'$ and $((x+1)^2)'$

3. Originally Posted by pickslides
$\frac{d}{dx}((x+1)^2(x^2+1)^{-3})= (x+1)^2((x^2+1)^{-3})'+((x+1)^2)'(x^2+1)^{-3}$

Now find $((x^2+1)^{-3})'$ and $((x+1)^2)'$
$((x^2+1)^{-3})'$ and $((x+1)^2)'$

(x+1)^2-3(x^2+1)^-4(2x) + 2x= 2(x+1)(1)(x^2+1)^-3

There we go is that correct? Where do I go after this step?

4. Still need help with this

edit: I think I understand what I did wrong on the second one, however i'm getting -12 for the answer and it's suppose to be 12.

5. anyone?

6. I can recall somewhere in the forum rules a part about no begging.

$((x^2+1)^{-3})' = -6x(x^2+1)^{-4}$

$((x+1)^2)' = (x^2+2x+1)' = 2x+2$

7. I understand what the derivative is.... i just don't understand how to simplify it

8. $\frac{dy}{dx} = \frac{-6x(x+1)^2}{(x^2+1)^{4}}+\frac{2x+2}{(x^2+1)^{3}}$

$\frac{dy}{dx} = \frac{-6x(x+1)^2+(2x+2)(x^2+1)}{(x^2+1)^{4}}$

Expand then simplify the numerator by grouping like terms if you wish.