Question

What is the derivative of `sin^-1(x)`?

Answer 1

`1/sqrt(1-x^2)`

Explanation:

Let `y=sin^-1x`,
so `siny=x` and `-pi/2 <= y <= pi/2` (by the definition of inverse sine).

Now differentiate implicitly:

`cosy dy/dx = 1`, so

`dy/dx = 1/cosy`.

Because `-pi/2 <= y <= pi/2`, we know that `cosy` is positive.

So we get:

`dy/dx = 1/sqrt(1-sin^2y) = 1/sqrt(1-x^2)`. (Recall from above `siny=x`.)