Question

$f^{\prime \prime }\left(4\right)=\frac{-4^{\frac{1}{3}}{2}^{\frac{2}{3}}-0}{4^{\frac{2}{3}}{2}^{\frac{4}{3}}}$.

Answer 1

Consider given the deferential equation,

$f^{{}^{\prime \prime }}\left(4\right)=\frac{-4^{\frac{1}{3}}{.2}^{\frac{2}{3}}-0}{4^{\frac{2}{3}}{.2}^{\frac{4}{3}}}$

$f^{{}^{\prime \prime }}\left(4\right)=\frac{-4^{\frac{1}{3}}{.2}^{\frac{2}{3}}}{4^{\frac{2}{3}}{.2}^{\frac{4}{3}}}$

$f^{{}^{\prime \prime }}\left(4\right)=-4^{\frac{1}{3}-\frac{2}{3}}{.2}^{\frac{2}{3}-\frac{4}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-4^{-\frac{1}{3}}{.2}^{\frac{-2}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-{\left(2^2\right)}^{-\frac{1}{3}}{.2}^{\frac{-2}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-2^{-\frac{2}{3}}{.2}^{\frac{-2}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-2^{-\frac{2}{3}-\frac{2}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-2^{-\frac{4}{3}}$

$f^{{}^{\prime \prime }}\left(4\right)=-\frac{1}{2^{\frac{2}{3}}}$

$f^{{}^{\prime \prime }}\left(4\right)=-\frac{1}{4^{\frac{1}{3}}}$

$f^{{}^{\prime \prime }}\left(4\right)=-\frac{1}{\sqrt[3]{34}}$

Hence, this is the answer.