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?

it is easier to use logarithmic differentiation on this problem. here is how you do this. $$y = \left(\frac x{x+1}\right)^4 \to \ln y = 4 \ln x - 4 \ln x + 1\to \frac{dy}{y} =4\left(\frac{dx}{x} - \frac{dx}{x+1} \right) = \frac{4dx}{x(x+1)} $$ multiplying through by $y$ gives you $$\frac{dy}{dx} = \frac{4x^3}{(x+1)^5} $$
