Derivade of: (x((x^2)+1)^2)/square root of (2+(x^2))

To find the devative of (x(x^2+1))^2/
sqrt(2+x^2)

The given expression is of the form f(x^2) =
{x^2(x^2+1)^2}/sqrt{x^2+2}.

Put sqrt(x^2+2) = t. Then x^2 =
t^2-2.

Therefore  differentiating , we
get:

 2xdx = 2tdt. Or

dt/dx =
x/t

dt/dx =x/sqrt(x^2+2).......(1)

f(t^2-2)
= {(t^2-2)(t^2-1)^2}/ t= (t^4--3t^2+2)/t =
t^3-3t+2/t.

Therefore d/dx {f(x^2) =
{d/dt(t^3-3t+2/t)}dt/dx

d/dx f(x^2) = (3t^2-3t-2/t^2)*
(x/sqrt(x^2+2).

Replace t =
sqrt(x^2+2)

d/dx^2 f(x^2) = {3(x^2+2)-
3sqr(x^2+2)-2/sqrt(x^2+2)}(x/(sqrt(x^2+2))

d/dxf(x^2) =
{3(x^2+2)^(3/2) -3(x^2+2)-2}x/(x^2+2).

Therefore d/dx
{x(x^2+1)/(sqrt(2+x^2)} = {3(x^2+2)^(3/2) -3(x^2+2)-2}x/(x^2+2).