How is the derivative of the function sin[inverse sin(x)]not equal to one.?

Answer 1

It is equal to one.

Proof:
d(sin( arcsin(x)) ) /dx
(apply chain rule)
= d( sin(arcsin(x)) ) / darcsin(x) · d(arcsin(x))/dx
= cos(arcsin(x) / √(1 - x²)
(because sin²(z) + cos²(z) =1 <=> cos(z) = √(1 - sin²(z))
= √(1 - sin²(arcsin(x))) / √(1 - x²)
= √(1 - x²) / √(1 - x²)
= 1
q.e.d

Answer 2

Look at what happens when pi/2 arcsin[sin(x)]=pi-x, so the derivative there is -1, not 1. You will also find that the derivative does not exist at x=pi/2.

The point of this exercise is to realize that arcsin(x) is not a perfect inverse function for sin(x). It couldn't be since sin(x) is not one-to-one.

Answer 3

Ok now i'm curious..

How is the derivative of the function sin[inverse sin(x)]not equal to one?

i read that as sin(arcsin(x))

I standby that it is 1 but there seems to be some disagreement. If it's not 1, i'm not sure where i'm going wrong.

Two ways to show it.

Method One
Simplify sin(arcsin(x)) first
arcsin(x) reads the angle.. whose sin is x.. if you take the sine of that angle.. you're just going to get x..

sin(arcsin(x)) = x
Take the derivative of x which is 1

Second Method
Don't simplify it and apply chain rule to take the derivative
cos(arcsin(x)) * 1/(sqrt(1-x^2))

Now arcsin(x) gives you the angle whose sin is x.. and sine is just a ratio of two sides of a triangle, meaning if you draw a triangle
opp = x
and hyp = 1
adj = sqrt(1-x^2)

so cos(arcsin(x)) = sqrt(1-x^2)/1
So we end up with
sqrt(1-x^2)/(sqrt(1-x^2))

which is just 1

Answer 4

If it is with respect to x, the derivative is 1. If it is with respect to any oyher alphabet, say u, it is treated as a constant. Hence the derivative is 0.

Answer 5

if Ur question is sin inverse[sin(x)]
then its one
but if its sin[sin inverse(x)]
let sin(y)=x
then sin inverse(x)=y
so Ur question changes to sin(y)
which is not one

Answer 6

if it was arsine[sine(x)] then its derivative would equal 1.

Answer 7

y = sin [ sin^(-1) x ]
let u = sin^(-1) x
du /dx = 1 / √ (1 - x ² )
dy/du = cos u
dy/dx = (dy/du) x (du/dx)
dy/dx = (cos u) / √(1 - x ²)
dy/dx = cos [ (sin^(-1) x) ] / √ (1 - x² )