Help with quotient/chain rule, please, I'm stuck!!?

Question

What is the derivative of 
((sin^-1)(x^1/2))
------------
(1-x)^1/2

Any help would be amazing, thank you!

Answer 1

= (sin^-1√x) / [√(1 - x)]

Derivative formulas:

Recall: quotient / chain rule f(x)·g(x) = [ f'(x)·g(x) - f(x)·g'(x) ] / g²(x)

The Derivative of the Inverse Sine. (d/dx)(sin^-1x) = 1/√(1-x²)

The Power Rule (Variable raised to a constant). (d/dx)(Aⁿ) = n A^n-1 and (d/dx)(√x) = 1/(2√x)

= [(sin^-1√x)'·√(1 - x) - (sin^-1√x)·√(1 - x)' ] / [√(1 - x)]²

= {[1/√(1-√x²)][1/(2√x)][√(1 - x)] - (sin^-1√x) [1/(2√1-x)](-1)} / (1 - x)         (inner derivative)

= {[1/√(1- x)][1/(2√x)][√(1 - x)] + (sin^-1√x) [1/(2√1-x)]} / (1 - x)                         (Simplify)

= {[1/(2√x)] + (sin^-1√x) [1/(2√1-x)]} / (1 - x)

Take out denominator terms.

= [1/2(1-x)][1/(√x) + (sin^-1√x) / (√1-x)]

There fore the derivative of (sin^-1√x) / [√(1 - x)] = [1/2(1-x)][1/(√x) + (sin^-1√x) / (√1-x)].