Answer
a. $ f^{\prime}(1)=\displaystyle \frac{1}{5}$
b. f is not differentiable at x=$0$
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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$f^{\prime}(x)=[x^{1/5}+5]^{\prime}=$
... Sum Rule: $[f\pm g]^{\prime}(x)=f^{\prime}(x)\pm g^{\prime}(x)$
... Power Rule$:\ \ \ [x^{n}]^{\prime}=nx^{n-1 }$
... Derivative of a Constant: $\displaystyle \frac{d}{dx}(c)=0$
$f^{\prime}(x)=\displaystyle \frac{1}{5}x^{-4/5}+0=\frac{1}{5\sqrt[5]{x^{4}}}$
$f^{\prime}(x)$ is defined for positive real numbers.
$a.\ \ \ $f is differentiable at x=$1$
$f^{\prime}(1)=\displaystyle \frac{1}{5\sqrt[5]{1^{4}}}=\frac{1}{5(1)}=\frac{1}{5}$
$b.\ \ \ $f is not differentiable at x=$0$, because
$f^{\prime}(x) $ is not defined for x=0 (yields zero in the denominator)
