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.
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 $\displaystyle (x+1)'=1.$)
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