Differentiate (inverse sin x )^2?

Answer 1

(sin^-1(x))^2=y

Solve for this in terms of x:
(sin^-1(x))=sqrt(y)
x=sin(sqrt(y))

Implicitly Differentiate:
1 = cos(sqrt(y))*(1/2)(y^-(1/2))dy/dx

Now solve in for dy/dx:
dy/dx = 2*sqrt(y) / (cos(sqrt(y))

We know that y=(sin^-1(x))^2

dy/dx = 2(sin^-1(x))/cos(sin^-1(x))

A math rule to remember is if you get cos(arcsin(x)) or sin(arccos(x)) it will be equal to sqrt(1-x^2)

So, finally:

dy/dx = 2(sin^-1(x))/(sqrt(1-x^2))

Answer 2

sin(y^1/2) = x
differentiate implicitely
1/2√y * cos(√y) * dy/dx = 1
dy/dx = 2√y /cos(√y)
= 2arcsin(x) / √(1 - x^2)

Answer 3

set sin^-1(x) = u , then ,

x = sin(u) , dx = cos(u)du

du/dx = 1/cos(u) = 1/root(1-x^2)

its easier now

Answer 4

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