Answer
$f'\left( x \right) = 4{x^3}$
$D_f=(-\infty,\infty)$
$D_{f'}=(-\infty,\infty)$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^4} \cr
& {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr
& {\text{Using the definition of derivative}}{\text{. }} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^4} - {x^4}}}{h} \cr
& {\text{Expand and simplify}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4}} \right) - {x^4}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4} - {x^4}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{h{\left( {4{x^3} + 6{x^2}h + 4x{h^2} + {h^3}} \right)}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {4{x^3} + 6{x^2}h + 4x{h^2} + {h^3}} \right) \cr
& {\text{Evaluate the limit when }}h \to 0 \cr
& f'\left( x \right) = 4{x^3} + 6{x^2}\left( 0 \right) + 4x{\left( 0 \right)^2} + {\left( 0 \right)^3} \cr
& f'\left( x \right) = 4{x^3} \cr
& {\text{The domain of the derivative of the function is }}\left( { - \infty ,\infty } \right) \cr} $$
