Question

How do you find the derivative of `arcsin(sqrt(1-x^2))`?

Answer 1

`d/dx arcsin(sqrt(1-x^2))= -x/(|x|sqrt(1-x^2))`

Explanation:

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

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

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

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

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

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