Question

If s=siny, y ∈ [-pi/2,pi/2], what is dy/ds?

Answer 1

I don't even understand what it's asking...

So it wants to know the derivative of y with respect to the derivative of s?
So s`=cosy?
cosy/s`=dy/ds???

Completely guessing, honestly have no idea how to proceed with this...

Answer 2

\[dy/ds=1/\sqrt{1-s ^{2}}\]

Answer 3

s=sin(y) => y=arcsin(s)

but the inverse sine only exists if we restrict the domain of sine from
[-pi/2,pi/2]


so let's look at this s=sin(y)
we want to find the derivative of y with respect to s

so (s)'=1 and (y)'=dy/ds
=> (sin(y))'=dy/ds * cos(y)


so we have
1=dy/ds * cos(y)

solving for dy/ds we have

1/cos(y)=dy/ds

dy/ds=1/cos(y)

assume we have a right triangle
Recall sin(y)=s=s/1 (=opp/hyp)

now we need to find the adj side

we can use the pythagorean thm to do this

adj=sqrt(1-s^2)


so we can rewrite dy/ds in terms of s

dy/ds=1/sqrt(1-s^2)

but remember y =arcsin(s)

so we actually found the derivative of arcsin(s) with the restriction that
-pi/2<=arcsin(s)<=pi/2