Question

How do I find the derivative of `y = (arcsin(x))^-1`?

Answer 1

`dy/dx=-(arcsin(x))^(-2)* 1/sqrt{1-x^2}=-1/(arcsin^{2}(x)sqrt{1-x^2})`

Explanation:

Use the Chain Rule: `d/dx(f(g(x)))=f'(g(x))*g'(x)` with `f(x)=x^{-1}` and `g(x)=arcsin(x)`. Since `f'(x)=-x^{-2}=-1/x^2` and `g'(x)=1/sqrt{1-x^2}`, it follows that

`d/dx((arcsin(x))^{-1})=-(arcsin(x))^(-2)* 1/sqrt{1-x^2}`

`=-1/(arcsin^{2}(x)sqrt{1-x^2})`