Let #f(x) = y = arcsin^3(5x) = (arcsin(5x))^3#.

We need to use the chain rule here.

Recall that #(dy)/(dx) = (du)/(dx) * (dy)/(du)#.

First let #u = arcsin(5x)# and #y=u^3#.

To solve for #(dy)/(du)# where #y = u^3#, we can use the power rule:

#(dy)/(du) = 3u^2#

To solve for #(du)/(dx)# where #u = arcsin(5x)#, we can use the inverse sin derivative rule, combined with the chain rule **again**.

Let #u' = 5x#, so #u = arcsin(5x) = arcsin(u')#.

By the inverse sin derivative rule:

#(du)/(du') = 1/(sqrt(1-u'^2))#

By the power rule:

#(du')/(dx) = 5#

Then multiplying together and substituting in for #u'#, we have:

#(du)/(dx) = (du)/(du') * (du')/(dx)#

#(du)/(dx) = 5/(sqrt(1-(5x)^2))#

#(du)/(dx) = 5/(sqrt(1-25x^2))#

Finally, we have:

#(dy)/(dx) = (du)/(dx) * (dy)/(du)#

#(dy)/(dx) = 5/(sqrt(1-25x^2)) * 3u^2#

Plugging back in #u# and simplifying:

#(dy)/(dx) = 5/(sqrt(1-25x^2)) * 3arcsin(5x)^2#

#(dy)/(dx) = (15arcsin(5x)^2)/(sqrt(1-25x^2))#