Question

What is the derivative of `y=arcsin(x)`?

Answer 1

The answer is:

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

This identity can be proven easily by applying `sin` to both sides of the original equation:

1.) `y = arcsinx`

2.) `sin y = sin(arcsinx)`

3.) `sin y = x`

We continue by using implicit differentiation, keeping in mind to use the chain rule on `siny`:

4.) `cosy dy/dx = 1`

Solve for `dy/dx`:

5.) `dy/dx = 1/cosy`

Now, substitution with our original equation yields `dy/dx` in terms of `x`:

6.) `dy/dx = 1/cos(arcsinx)`

At first this might not look all that great, but it can be simplified if one recalls the identity
`sin(arccosx) = cos(arcsinx) = sqrt(1 - x^2)`.

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

This is a good definition to memorize, along with `d/dx[arccos x]` and `d/dx[arctan x]`, since they appear quite frequently in differentiation problems.