The hardest part of this is obviously the derivative of asinh, so we can calculate that easily with the following procedure.
Call #f(x) = sinh(x)# and #g(x) = sinh^-1(x)# i.e. #f(g(x)) = x#.
So, we can take the derivative of this mess:
#f(g(x)) = x -> d/dx f(g(x)) = 1#
#f'(g(x)) * g'(x) = 1 -> g'(x) = 1/(f'(g(x))) #
So now we need to know the derivative of sinh. I'm not that into memorization, so I will derive that really quickly:
#sinh(x) = (e^x - e^-x)/2 -> d/(dx) sinh(x) = (e^x + e^-x)/2 = cosh(x) #
Therefore, from the above,
#d/(dx) sinh^-1(x) = 1/(cosh(sinh^-1(x)) #
If we know that #sinh(theta) =x#, we know that #cosh(theta) = sqrt(1 + x^2)#, i.e.
#d/(dx) sinh^-1(x) = 1/sqrt(1 + x^2) #
Now we can just use chain rule, product rule, and power rule in the original:
#(dy)/(dx) = d/(dx) [x sinh^-1(x/3)] - d/dx(sqrt(9+x^2)) #
#d/(dx) [x sinh^-1(x/3)] = x d/(dx) [sinh^-1(x/3)] + sinh^-1(x/3)(dx)/(dx)#
#= (x/3)/sqrt(1 + x^2 / 9) + sinh^-1(x/3) #
#d/dx(sqrt(9+x^2)) = 1/2 * (9 + x^2)^(-1/2) * 2x = x/sqrt(9 + x^2) #
Therefore,
#(dy)/(dx) = (x/3)/sqrt(1 + x^2 / 9) + sinh^-1(x/3) - x/sqrt(9 + x^2) #
#(dy)/(dx) = x/sqrt(9 + x^2) + sinh^-1(x/3) - x/sqrt(9 + x^2) #
#(dy)/(dx) = sinh^-1(x/3) #
