Find the derivative using the chain rule and the quotient rule 
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.

Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?

Apparently you complicated things too much.
Starting from $$f(x) = \left(\frac{x}{x+1}\right)^4$$ with the chain rule we get first 
\begin{align}
f'(x) &= 4\left(\frac{x}{x+1}\right)^3\cdot \left(\frac{x}{x+1}\right)^\prime\notag\\
&= 4\left(\frac{x}{x+1}\right)^3\cdot \frac{x'(x+1)-x(x+1)'}{(x+1)^2}\notag\\
&= 4\frac{x^3}{(x+1)^3}\cdot \frac{(x+1)-x}{(x+1)^2}\notag\\
&= \frac{4x^3}{(x+1)^3}\cdot \frac 1{(x+1)^2}\notag\\
&= \frac{4x^3}{(x+1)^5}\notag
\end{align}
