<s>[INST] 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?
 [/INST] You calculation of $f'(x)$ is correct, though it can be simplified. To do this, we factor out $4x^3(x+1)^3\require{cancel}$ in the numerator.
$$\begin{align}f'(x) &= \frac{4x^3(x+1)^4-4x^4(x+1)^3}{(x+1)^8}\\\\
&= \frac{\color{blue}{4x^3(x+1)^3}\Big((x+1) - x\Big)}{(x+1)^8}\\\\
&= \frac {(4x^3\cancel{(x+1)^3})(1)}{\cancel{(x+1)^8}^5}\\ \\
&= \frac{4x^3}{(x+1)^5}
\end{align}$$
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