f(x) = sin ((x/(x^2-1))^(3/2))
Ok, so let's break your function down into a chain of functions:
f(u) = sin(u)
u(v) = v^(3/2)
v(x) = x/(x^2-1)
As next, you need to differentiate those three functions.
f(u) = sin(u) color(white)(xxx) => f'(u) = cos(u)
u(v) = v^(3/2) color(white)(xxxxxx) => u'(v) = 3/2 v^(1/2)
To differentiate v(x), you can use the quotient rule:
if f(x) = g(x) / (h(x)), the derivative is f'(x) = (g'(x) * h(x) - h'(x) * g(x)) / (h^2(x)).
So, in your case, the derivative is:
v(x) = x/(x^2-1) color(white)(x) => v'(x) = (1 * (x^2-1) - 2x * x)/(x^2-1)^2 = (-x^2 - 1)/(x^2-1)^2
Now, to compute the derivative of f(x), you just need to multiply your three derivatives! :-)
f'(x) = f'(u(v(x))) * u'(v(x)) * v'(x)
color(white)(xxxx) = cos((x/(x^2-1))^(3/2)) * 3/2 (x/(x^2-1))^(1/2) * (-x^2 - 1)/(x^2-1)^2
color(white)(xxxx) = 3/2 cos((x/(x^2-1))^(3/2)) (x/(x^2-1))^(1/2) * (-x^2 - 1)/(x^2-1)^2
