Question

How do you differentiate # g(x) = sqrtarcsin(x^2-1) #?

Answer 1

`(dy)/(dx) =x/(sqrt(arcsin(x^2-1))sqrt[1 - (1 - x^2)^2])`

Explanation:

Define `y = sqrt(arcsin(x^2-1))` and make
`y^2=arcsin(x^2-1)->x^2-1=sin(y^2)`. Now following with
`f(x,y)=x^2-sin(y^2) = 1`
We know that `(dy)/(dx) = -f_x/(f_y) = (2x)/(2y cos(y^2))`. Now, substituting for `y = sqrt(arcsin(x^2-1))` we have
`(dy)/(dx) = x/(sqrt(arcsin(x^2-1))cos(arcsin(x^2-1))`
Simplifying
`(dy)/(dx) =x/(sqrt(arcsin(x^2-1))sqrt[1 - (1 - x^2)^2])`