Answer
$$ y'=2(z+1)^3(2z-1)^2(7z+1)$$
Work Step by Step
Since $ y=(z+1)^4(2z-1)^3$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and the product rule: $(uv)'=uv'+u'v $, the derivative $ y'$ is given by
$$ y'=((z+1)^4)'(2z-1)^3+(z+1)^4((2z-1)^3)'\\= 4(z+1)^3(2z-1)^3+3(z+1)^4(2z-1)^2(2)\\=4(z+1)^3(2z-1)^3+6(z+1)^4(2z-1)^2\\
=2(z+1)^3(2z-1)^2[2(2z-1)^1+3(z+1)^1]\\
=2(z+1)^3(2z-1)^2[4z-2+3z+3]\\
=2(z+1)^3(2z-1)^2(7z+1)
$$
