Assuming by #sin^(-5)(x)# you mean #arcsin^5(x)# we have
#y = arcsin^5(x)/x - (xcos^3(x))/3#
By saying
#y = u - v#
We can break it down into smaller derivatives and doing that such that
#dy/dx = (du)/dx - (dv)/dx#, so for
#u = arcsin^5(x)/x#, we have, by the product rule that
#(du)/dx = 1/x*d/dx(arcsin^5(x)) + arcsin^5(x)d/dx(1/x)#
#(du)/dx = 1/x*d/dx(arcsin^5(x)) - arcsin^5(x)/x^2#
By the chain rule, we have that #d/dx(arcsin^5(x)) = 5arcsin^4(x)*arcsin^'(x)# where #arcsin^'(x) = 1/sqrt(1-x^2)#
#(du)/dx = (5arcsin^4(x))/(xsqrt(1-x^2)) - arcsin^5(x)/x^2#
And for
#v = (xcos^3(x))/3#
We have, by the product rule and the chain rule
#(dv)/dx = cos^3(x)d/dx(x/3) + x/3d/dx(cos^3(x))#
#(dv)/dx = cos^3(x)/3 + x/3 * 3cos^2(x) * -sin(x)#
#(dv)/dx = cos^3(x)/3 - xcos^2(x)sin(x)#
Adding it all up we have
#dy/dx = (du)/dx - (dv)/dx#
#dy/dx = (5arcsin^4(x))/(xsqrt(1-x^2)) - arcsin^5(x)/x^2 - cos^3(x)/3 + xcos^2(x)sin(x)#
