Answer
$g'\left( x \right) = - \frac{1}{2}{x^{ - 3/2}} + \frac{1}{4}{x^{ - 3/4}}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \frac{1}{{\sqrt x }} + \root 4 \of x \cr
& {\text{Rewrite the function}} \cr
& g\left( x \right) = {x^{ - 1/2}} + {x^{1/4}} \cr
& {\text{Differentiate the function}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 1/2}} + {x^{1/4}}} \right] \cr
& {\text{Use the sum diffence rules for differentiation }} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 1/2}}} \right] + \frac{d}{{dx}}\left[ {{x^{1/4}}} \right] \cr
& {\text{Apply the power rule: }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& g'\left( x \right) = - \frac{1}{2}{x^{ - 1/2 - 1}} + \frac{1}{4}{x^{1/4 - 1}} \cr
& g'\left( x \right) = - \frac{1}{2}{x^{ - 3/2}} + \frac{1}{4}{x^{ - 3/4}} \cr} $$
