Answer
$f'\left( { - 1} \right) = - 20$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 5{x^4},{\text{ }}a = - 1 \cr
& {\text{Using the definition of the derivative}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5{{\left( {a + h} \right)}^4} - 5{a^4}}}{h} \cr
& {\text{Expanding }}{\left( {a + h} \right)^4} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5\left( {{a^4} + 4{a^3}h + 6{a^2}{h^2} + 4a{h^3} + {h^4}} \right) - 5{a^4}}}{h} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5{a^4} + 20{a^3}h + 30{a^2}{h^2} + 20a{h^3} + 5{h^4} - 5{a^4}}}{h} \cr
& {\text{Simplifying}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{20{a^4}h + 30{a^3}{h^2} + 20a{h^3} + 5{h^4}}}{h} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {20{a^3} + 30{a^2}h + 20a{h^2} + 5{h^3}} \right)}}{h} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \left( {20{a^3} + 30{a^2}h + 20a{h^2} + 5{h^3}} \right) \cr
& {\text{Evaluate the limit when }}h \to 0 \cr
& f'\left( a \right) = 20{a^3} + 30{a^2}\left( 0 \right) + 20a{\left( 0 \right)^2} + 5{\left( 0 \right)^3} \cr
& f'\left( a \right) = 20{a^3} \cr
& {\text{Find }}f'\left( a \right){\text{ at }}a = - 1 \cr
& f'\left( { - 1} \right) = 20{\left( { - 1} \right)^3} \cr
& f'\left( { - 1} \right) = - 20 \cr} $$
