Answer
$f'\left( x \right) = 75{x^{74}} - 1$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{75}} - x + 3 \cr
& {\text{Differentiate the function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{75}} - x + 3} \right] \cr
& \cr
& {\text{Use the sum diffence rules for differentiation }} \cr
& \frac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \frac{d}{{dx}}f\left( x \right) \pm \frac{d}{{dx}}g\left( x \right) \cr
& {\text{then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{75}}} \right] - \frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ 3 \right] \cr
& \cr
& {\text{Compute the derivatives}}{\text{, }}\frac{d}{{dx}}\left[ x \right] = 1{\text{ and }}\frac{d}{{dx}}\left[ c \right] = 0,{\text{ so}} \cr
& f'\left( x \right) = 75{x^{75 - 1}} - 1 + 0 \cr
& {\text{Simplify}} \cr
& f'\left( x \right) = 75{x^{74}} - 1 \cr} $$
