Vtyjkyuk

While trying to find the derivative of
$$f(x)=arcsinleft(2xsqrt1-x^2right)$$
we arrive at two different answers by substituting $x=sin t$ or $x=cos t$. The aim is to simplify it as $arcsin(sin2t)$ and further simplify it as $2t$ and take its derivative. Proceeding in that manner, I arrived at two different answers for the derivative. I could understand that it was because the function $f$ is defined differently in different intervals, and as I have solved the problem without considering the intervals, I hit this answer. Also, I thought that since $sin t$ and $cos t$ will coincide only when $t=dfracpi4$, $x=pmdfrac1sqrt2$ will be the points where the definition changes. I got some more clarity when I looked at the graph of $f(x)$ but I could not figure out a way to systematically figure out what answer suits what interval.

MAJOR EDIT: I made a mistake in the function itself that I had given. My question earlier read:
$$f(x)=arcsinleft(2xsqrtx^2-1right)$$
But I wanted only for what I've now altered it as. My sincere apologies...

EDIT: My question is how to figure out the interval in which the substitution applies for such problems in general. I don't need the expression for the derivative of $f(x)$.

Note: I am new to this community, so please point out any deviation from the policy, if I have deviated.
Also, I couldn't post my working as I do not have $10$ reputation.

$$f(x)=arcsin(2xsqrtx^2-1)$$
$$sin(f(x))=2xsqrtx^2-1$$
$$f'(x) cos(f(x))=2sqrtx^2-1+2xdfrac2x2sqrtx^2-1$$
$$f'(x)sqrt1-sin^2f(x)=2dfrac2x^2-1sqrtx^2-1$$
$$f'(x)=2dfrac2x^2-1sqrtx^2-1sqrt1-4x^4+4x^2$$

$$ f(x) = arcsin(2xsqrtx^2-1) $$

Replace x=cosh$theta$

then f(x) = arcsin(sinh$2theta$)

by differentiating f(x),

f '(x) = $frac2cosh(2theta)sqrt1-sinh^2(2theta)sqrtx^2-1$

Replacing $ cosh(2theta) = 2x^2-1 $

f '(x) = $frac4x^2-2sqrt1-4x^4+4x^2sqrtx^2-1$