Answer
$f'\left( x \right) = \frac{5}{{\sqrt {1 - 25{x^2}} }}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\sin ^{ - 1}}\left( {5x} \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {5x} \right)} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }}\left( {p223} \right) \cr
& {\text{and the chain rule}}{\text{. }}\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}},\;u = 5x \cr
& f'\left( x \right) = \frac{1}{{\sqrt {1 - {{\left( {5x} \right)}^2}} }}\frac{d}{{dx}}\left[ {5x} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( x \right) = \frac{1}{{\sqrt {1 - 25{x^2}} }}\left( 5 \right) \cr
& f'\left( x \right) = \frac{5}{{\sqrt {1 - 25{x^2}} }} \cr} $$
