Let $y(u)= \ln\left(\frac {(u^2+1)^5}{\sqrt{1-u} }\right)$. I would like to find $y'(u)$.
This what I did:
Chain rule $ \frac {d}{du}(u^{2}+1)^{5} = 10u(u^2+1)^4$
$ \frac {d}{du}\sqrt{1-u} = \frac {1}{2}(1-u)^{-1/2} $
Quotient rule:
$\ (\sqrt{1-u} \cdot 10u(u^2+1)^4 - {1}{2}(1-u)^{-1/2} \cdot(u^{2}+1)^{5} )/(1-u).$
I don't know what to do next. what about $\ln$?
Try to do as many simplifications as possible before actually differentiating, so keeping in mind the domains of definition of all the involved functions (fill in details here):
$$\log\frac{(u^2+1)^5}{\sqrt{1-u}}=\log(u^2+1)^5-\log\sqrt{1-u}=5\log(u^2+1)-\frac12\log(1-u)$$
and now you only have to take care of two easy, separate functions, and applying the chain rule:
$$\begin{cases}\cfrac d{du}(\log(u^2+1))=2u\cdot\cfrac1{u^2+1}=\cfrac{2u}{u^2+1}\\{}\\\cfrac d{du}(\log(1-u))=(-1)\cdot\cfrac1{1-u}=\cfrac1{u-1}\end{cases}$$
Now just put things together.
