Answer
a. $ f^{\prime}(8)=\displaystyle \frac{10}{3}$
b. $f^{\prime}(0)=2$
Work Step by Step
f is differentiable at x=a if if a is in the domain of $f^{\prime}$,
that is, if $f^{\prime}(a)$ is defined
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$f^{\prime}(x)=[2x+x^{4/3}]^{\prime}=$
... Sum Rule: $[f\pm g]^{\prime}(x)=f^{\prime}(x)\pm g^{\prime}(x)$
... Power Rule$:\ \ \ [x^{n}]^{\prime}=nx^{n-1 }$
... Constant Multiple Rule: $[cf]$'$(x)=cf^{;}(x)$.
$f^{\prime}(x)=2(1)+\displaystyle \frac{4}{3}x^{4/3-1}$
$f^{\prime}(x)=2+\displaystyle \frac{4\sqrt[3]{x}}{3}$
$f^{\prime}(x)$ is defined for all real numbers.
$a.\ \ \ $f is differentiable at x=$8$
$ f^{\prime}(8)=2+\displaystyle \frac{4\sqrt[3]{8}}{3}=2+\frac{2(2)}{3}=\frac{10}{3}$
$b.\ \ \ $f is differentiable at x=$0$
$f^{\prime}(0)=2+\displaystyle \frac{4\sqrt[3]{0}}{3}=2+0=2$
