Assuming that the range of #sin^(-1)x # is #(-oo, oo)#, is # x sin^(-1) x # differentiable, for #sin^(-1)x in [0, 2pi]#?

For multivalued #y=xsin^-1 x# we can use the equations

#y/x = sin^-1 x#

#sin(y/x) = x#

Implicit differentiation gives:

#cos(y/x)((y'x-y)/x^2) = 1#

#xy'cos(y/x)-ycos(y/x) = x^2#

# y' = (x^2+ycos(y/x))/(xcos(y/x))#

Replacing #y = xsin^-1 x# and #y/x = sin^-1x# gets us

#y' = (x+sin^-1xcos(sin^-1x))/cos(sin^-1x)#

I am uncertain about rewriting #cos(sin^-1x)# in this situation.

It looks to me like the derivative will exist provided that #x != +-1#

For what it's worth, here is Socratic's graph of #sin(y/x) = x# (I have not restricted values of #x# or #y# in any way.)
It's hard to see, but the graph is a sequence of loops getting longer and "more perpendicular". (I just invented that technical description.)

graph{sin(y/x)=x [-1.92, 2.406, -0.684, 1.478]}

What I mean by my assumption that #sin^(-1)x in (-oo, oo)# is as

follows.

If #y=f(x)=sin x#, then the graphs of the wave

#y=sin x#, in infinitude, and the the graph of the inverse #x =

sin^(-1)y# are one and the same.

And then, for the locally bijective f(x), #f^(-1)f(x)=x#, everywhere.

Accordingly, it is indubitable that #sin^(-1)sin (kpi) = kpi =0#,

only when k = 0...

I am aware that some students, at Middle School Level, might find

this difficult to follow. So, I use in my answer the conventional

#sin^(-1)x in [-pi/2, pi/2]#. I have chosen to define y piecewise for the

chosen problem.

Here, # sin^(-1)x in [-pi/2, pi/2], (sin^(-1)x)'=1/sqrt(1-x^2)#, with #|x|<=1#

Using the notion of the general value of inverse sine, define

#0<=y<=pi/2, y= x sin^(-1)x#. x increases from 0 to 1

#-3/2pi<=y<=pi/2, y=x(pi-sin^(-1)x)#. x decreases from 1 to #-1#

#-3/2pi<=y<=0, y=x(2pi+x sin^(-1)x)#. x increases from #-1# to 0

The graph comprises two loops with origin as the common point.

The loops are in between the parallels #x=+-1#

Output from #ad# #hoc# computer program reveals that the first

loop is in #Q_1#. It rises from O, touches x= 1 at #(1, pi/2)#, rebound

to the turning point (0.8970, 1.820) at a higher level, for y'=0. .

The second loop is in #Q_3#.. The arm of of this loop is longer

For my piecewise definition of y, y is continuous and differentiable

everywhere, including the points at which two neighboring pieces

meet.

It will be a graph in grandeur, with loops getting elongated, in

infinitude, in between #x=+-1#. It will be is a kind of oscillation for y,

with amplitude increasing without limit...

I welcome a graph for the data given below, for just two loops.
(x, y): (0, 0) (.2, .0427) (.4, .6146) (.6, .3861) (.8, .7418) (1, 1.571) (8, 1.771)
(.6, 1.499) (.4, 1.092) (.2, 0.580) (0, 0) (-.2, 0.6686) (-.4, -1.421) (-.6, -2.271) ( -.8, -3.255) (-1, -4.712) (-.8, -4.285 (-.6, -3.384) (-.4, -2.349), (-.2, -1.216) (0, 0)

In the first graph, #0 <= x <=1# and #sin^-1x# is multi-valued in #[0,pi]#

In the second graph, #-1 <= x <= 0# and #sin^-1x# is multi-valued in #[pi, 2 pi]#
graph{(y-xarcsin(x))(sqrt(x-x^2)/sqrt(x-x^2))(y-x(pi-arcsinx))=0 [-3.385, 4.414, -0.96, 2.937]}

graph{(y-x(pi-arcsinx))(y-x(2*pi + arcsinx))sqrt(x^2-x)/sqrt(x^2-x)=0 [-4.273, 6.83, -5.08, 0.47]}

Using the Desmos graphing website, I got this graph using

#y=xsin^-1 x " for " 0 <= x <= 1# #color(red)" RED"#
#y=x(pi-sin^-1 x) " for " 0 <= x <= 1# #color(blue)" BLUE"#
#y=x(pi-sin^-1 x) " for " -1 <= x <= 0# #color(green)" GREEN"#
#y=x(2pi+sin^-1 x) " for " -1 <= x <= 0# #color(purple)" PURPLE"#

Unfortunately these will not zoom or scroll here.

I built them at https://www.desmos.com/calculator