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)#