Determine using calculus dy/dx if y =arcsin x/(1-x^2)?

Since the given function is a fraction, we'll apply the
quotient rule to determine the first derivative.

(u/v)' =
(u'*v - u*v')/v^2

We'll put u =
arcsinx

du/dx = d(arcsinx)/dx =
sqrt(1-x^2)

We'll put v =
(1-x^2)

dv/dx =
d(1-x^2)/dx

dv/dx = -2x

Now,
we'll differentiate the function:

dy/dx =
[(1-x^2)sqrt(1-x^2) +
2x(arcsinx)]/(1-x^2)^2

The first derivative
of the given function is: dy/dx = [(1-x^2)sqrt(1-x^2) +
2x(arcsinx)]/(1-x^2)^2.