Question

What is the derivative of this function `y=sin^-1(1-x)^(1/2)`?

Answer 1

`dy/dx = -1/(2sqrt(x)sqrt(1-x)`

Explanation:

Let `y = arcsin(sqrt(1-x))`
Then `siny = sqrt(1-x) = (1-x)^(1/2)`

Differentiating Implicitly and applying the chain rule;
`cosydy/dx = (1/2)(1-x)^(-1/2)(-1)`
`:. cosydy/dx = -1/(2sqrt(1-x)) ... [1]`

Then using the fundamental trig identity `sin^2A+cos^2A-=1` we have:

`cos^2y=1-sin^2y`
`cos^2y=1-((1-x)^(1/2))^2`
`:. cos^2y=1 - (1-x)`
`:. cos^2y=x`
`:. cosy=sqrt(x)`

Substituting into [1] we get:
`:. sqrt(x)dy/dx = -1/(2sqrt(1-x))`

Leading to the solution:
`:. dy/dx = -1/(2sqrt(x)sqrt(1-x)`