Question

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

Answer 1

I'm not sure I've answered your question, but this may help.

Explanation:

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]}

Answer 2

See the explanation

Explanation:

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)

Answer 3

I can't get the Socratic graphing utility on one graph, but here are the two loops.

Explanation:

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