1. ## Second Derrivative

Can someone help me take the 2nd Derivative of this function. I can't seem to get it right.$\displaystyle y=\frac{x}{(x+1)^{2}}$

Thanks.

2. Originally Posted by johntuan
Can someone help me take the 2nd Derivative of this function. I can't seem to get it right.$\displaystyle y=\frac{x}{(x+1)^{2}}$
Before to contemplate a derivative, try to do a make-up first:

$\displaystyle f(x) = \frac{x} {{(x + 1)^2 }} = \frac{{x + 1 - 1}} {{(x + 1)^2 }} = \frac{1} {{x + 1}} - \frac{1} {{(x + 1)^2 }} = (x + 1)^{ - 1} - (x + 1)^{ - 2} .$

So from here first and second derivative are practically done.

$\displaystyle f'(x) = - (x + 1)^{ - 2} + 2(x + 1)^{ - 3} \implies f''(x) = 2(x + 1)^{ - 3} - 6(x + 1)^{ - 4} .$

Now it remains to do the algebra make-up:

$\displaystyle f''(x) = 2(x + 1)^{ - 4} \Big[ {(x + 1) - 3} \Big] = \frac{{2(x - 2)}} {{(x + 1)^4 }}.$

(Observe that we've applied the Chain Rule always, it's just that in this case is pretty trivial, since $\displaystyle (x+1)'=1.$)

3. Thanks for the reply but, I don't quite understand your first step, What's the Make-up?

4. Originally Posted by johntuan
Thanks for the reply but, I don't quite understand your first step, What's the Make-up?
It means try to re-arrange the given expression into something that's easier to work with.

5. Originally Posted by johntuan
Thanks for the reply but, I don't quite understand your first step, What's the Make-up?
He means he's putting it into a nicer form for taking the derivative. You can also do this by the brute force method:
$\displaystyle y=\frac{x}{(x+1)^{2}}$

$\displaystyle y^{\prime} = \frac{1 \cdot (x + 1)^2 - x \cdot 2(x + 1)}{(x + 1)^4}$

$\displaystyle y^{\prime} = \frac{-x^2 + 1}{(x + 1)^4} = -\frac{x - 1}{(x + 1)^3}$

So
$\displaystyle y^{\prime \prime} = -\frac{1 \cdot (x + 1)^3 - (x - 1) \cdot 3(x + 1)^2}{(x + 1)^6}$

$\displaystyle y^{\prime \prime} = 2 \frac{x - 2}{(x + 1)^4}$
after some work.

-Dan

6. In this step $\displaystyle y^{\prime} = \frac{-x^2 + 1}{(x + 1)^4} = -\frac{x - 1}{(x + 1)^3}$ how did u get $\displaystyle -x^2+1$ to cancel out one part of the denominator, in order to cancel out shouldn't it equal $\displaystyle x+1^{x}$

7. Originally Posted by johntuan
In this step $\displaystyle y^{\prime} = \frac{-x^2 + 1}{(x + 1)^4} = -\frac{x - 1}{(x + 1)^3}$ how did u get $\displaystyle -x^2+1$ to cancel out one part of the denominator, in order to cancel out shouldn't it equal $\displaystyle x+1^{x}$
$\displaystyle -x^2 + 1 = 1 - x^2 = (1 - x)(1 + x) = -(x - 1)(x + 1)$ and the (x + 1) is clearly seen to be a common factor of numerator and denominator.