Answer
$$A'(v)=\frac{1}{3}\bigg(16v^{5/3}+2v^{-1/3}+4v^{-7/3} \bigg)$$
Work Step by Step
$A(v)=v^{2/3}(2v^2+1-v^{-2})=2v^{8/3}+v^{2/3}-v^{-4/3}$
Differentiating with respect to $v$, using power rule,
$$A'(v)=2\bigg(\frac{8}{3}\bigg)v^{8/3-1}+\frac{2}{3}v^{2/3-1}+\frac{4}{3}v^{-4/3-1}$$
$$A'(v)=\frac{16}{3}v^{5/3}+\frac{2}{3}v^{-1/3}+\frac{4}{3}v^{-7/3}=\frac{1}{3}\bigg(16v^{5/3}+2v^{-1/3}+4v^{-7/3} \bigg)$$
